Math 441 - Abstract Algebra - Schedule

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January 6

RSA public key encryption

This is found on page 164 of your text

Receiver
1. Pick very large primes $p$ and $q$ and compute $n=pq$.
2. Compute the least common multiple of $p-1$ and $q-1$; call it $m$.
3. Pick $r$ relatively prime to $m$.
4. Find $s$ such that $rs \text{ mod } m =1$.
5. Publicly announce $n$ and $r$.
Sender
1. Convert the message to strings of digits.
2. Break up the message into uniform blocks of digits; call them $M_1$, $M_2$, $\ldots$, $M_k$.
3. Calculate and send $R_i=M_i^k \text{ mod }n$.
Receiver
1. For each received message $R_i$, calculate $R_i^s\text{ mod }n$.
2. Convert the string of digits back to a string of characters.

Much of the development of modern abstract algebra began by studying how to solve algebraic equations, in particular polynomial equations of the form $a_0+a_1x+a_2x^2+a_3x^3+\cdots a_nx^n=0$. If the coefficients are integers, then what can we say about the solutions? What can we say about the solutions if the coefficients are instead rational numbers, real numbers, or complex numbers?

The quadratic formula gives all solutions to $a_0+a_1x+a_2x^2=0$.
In the 1500's, Girolamo Cardano (1501-1576) published a solution to the general cubic equation.
Shortly afterwards, his student Lodovico Ferrari (1522-1565) published a solution to the general quartic equation.

At this point in mathematical history, the next obvious goal was to find the solution to the general quintic equation. This problem remained open into the 1800s, where finally Abel showed that obtaining a formula using radicals was in general impossible. Galois invented group theory to describe the symmetries of solutions to polynomial equations. Most of our modern view of algebra has its roots in the history of solving the general quintic.

A lot of new words have been created since the 1500s to describe different types of number systems. To start this semester, I'd like us to explore the properties of the rationals, and then from them develop the concepts of field, ring, integral domains, and more.

The rational numbers have two group structures. Under addition, the set of rational numbers $(\mathbb{Q},+)$ forms an Abelian group. Under multiplication, we almost have another Abelian group with one problem. The number 0 does not have a multiplicative inverse. We'll let $\mathbb{Q}^*$ denote the set of rational numbers, excluding zero, and then we see that $(\mathbb{Q}^*,\cdot)$ is another Abelian group. When we look at how addition and multiplication interact together, the properties of addition and multiplication satisfy the distributive laws $a(b+c)=ab+ac$ and $(b+c)a=ba+ca$.

Definition (Field)

A field is a set $F$ together with two binary operations $+$ and $\cdot$ that satisfies the following properties:

$(F,+)$ is an Abelian group. Let $F^*$ be the set $F$ with the additive identity removed.
$(F^*,\cdot)$ is an Abelian group.
The distributive laws hold, namely we have $a(b+c)=ab+ac$ and $(b+c)a=ba+ca$.

Problem 1 (Listing The Properties Of A Field)

Suppose that $F$ is a field. Since $(F,+)$ and $(F^*,\cdot)$ are Abelian group, this means there are several properties that the binary operations $+$ and $*$ must satisfy. Make a list of all these properties, which together with the distributive laws, should give you a list of 9 properties that characterize a field. Then state at least two other sets that you know satisfy these properties.

Problem 2 (Algebraic Properties Of Modular Arithmetic)

Consider the sets $\mathbb{Z}_n$ for each positive integer $n$, together with modular addition and multiplication.

Give three different integers $n$ so that $\mathbb{Z}_n$ is a field.
Give an integer $n$ where $\mathbb{Z}_n$ is not a field. List the properties of being a field that are not satisfied.
Find an integer $n$ and elements $a,b\in \mathbb{Z}_n$ with $a\neq b$ such that $ab=0$ but neither $a$ nor $b$ is zero.
For each integer $k\geq 2$, find an integer $n$ and element $a\in \mathbb{Z}_n$ so that $a^k=0$ but $a^{k-1}\neq 0$.
State all $n$ for which $\mathbb{Z}_n$ is a field.

Definition (Polynomial Rings Over A Field)

Let $F$ be a field. We denote by $F[x]$ the set of all polynomial in the variable $x$ with coefficients in $F$ together with polynomial addition and multiplication (if needed, see page 294 in your text for a formal description of something you are very familiar with). The set $F[x]$ is called the polynomial ring over $F$ (in the indeterminate $x$). We can write each element $p(x)\in F[x]$ in the form $$p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_2x^2+a_1x+a_0$$ for some $n$ where $a_i\in F$ for each $i$ and $a_n\neq 0$ if $n\geq 1$.

Problem 3 (Algebraic Properties Of Polynomial Rings Over A Field)

Let $F$ be a field (think the rationals $\mathbb{Q}$). Which of the properties of being a field does $F[x]$ not satisfy? Give an example of each property (of the 9 total) that is not satisfied.

The previous problem shows that polynomial rings over a field are not a field. Algebra's roots are in finding solutions to polynomial equation of the for $p(x)=0$. The number $\sqrt{2}$ is a solution to the equation $x^2-2=0$, and so is the root of some polynomial. We say that $\sqrt{2}$ is an algebraic number. The number $\pi$ is not the root of any polynomial (a very nontrivial thing to prove), and such numbers we call transcendental. They transcend algebra. One of our goals this semester will be to understand the set of algebraic numbers.

Definition (Algebraic Number)

We say that a number $x$ is algebraic if it is a root of a polynomial with rational coefficients. In symbols, we say that $x$ is algebraic if $x$ satisfies $p(x)=0$ for some $p(x)\in \mathbb{Q}[x]$.

Problem 4 (Algebraic Numbers)

Let $\mathbb{Z}[x]$ be the set of all polynomials with coefficients in $\mathbb{Z}$, together with the usual properties of polynomial addition and multiplication. Show that $x$ is an algebraic number if and only if $x$ satisfies $q(x)=0$ for some $q(x)\in \mathbb{Z}[x]$.

Let's now compare the field properties to operations with matrices and vectors. In the problems below, you'll be asked to list the properties of a field that are not satisfied. There are 9 properties to consider in each case, as there are 4 from each Abelian group and one more from the distributive laws. You should be prepared to prove each property that is satisfied (or at least state why you know it is satisfied).

Problem 5 (Algebraic Properties Of Square Matrices)

Let $\text{M}_2(\mathbb{Q})$ bet the set of 2 by 2 matrices with entries in $\mathbb{Q}$, together with the usual properties of matrix addition and matrix multiplication. Recall that $\text{GL}(2,\mathbb{Q})$ is the set of invertible 2 by 2 matrices with entries in $\mathbb{Q}$.

Which of the properties of being a field does $\text{M}_2(\mathbb{Q})$ not satisfy? Give an example of each property that is not satisfied.
Which of the properties of being a field does $\text{GL}(2,\mathbb{Q})$ not satisfy? Give an example of each property that is not satisfied.

Problem 6 (Algebraic Properties Of 3 D Vectors And The Cross Product)

Consider the set $\mathbb{R}^3$, together with the two binary operations of vector addition and the cross product. Which of the properties of being a field does $\mathbb{R}^3$ not satisfy? Give an example of each property that is not satisfied.

For fun, please see the wikipedia page on magmas. People have given lots of names to algebraic structures that satisfy various properties.

January 10

Definition (Ring)

A ring $R$ is an Abelian group $(R,+)$ together with an additional associative binary operation (multiplication) that satisfies the left and right distributive laws, namely $a(b+c)=ab+ac$ and $(b+c)a=ba+ca$.

Definition (Commutative Ring)

A commutative ring is a ring in which $ab=ba$ (multiplication commutes).

Definition (Unity And Unit)

A unity in a ring is a nonzero element that is an identity under multiplication. A nonzero element of a ring does not need to have a multiplicative inverse. When it does, we say that element is a unit of the ring.

Problem 7 (Recognizing Rings)

Which of the following are rings? Which rings have a unity? Which are commutative?

For each $n$, the set $\mathbb{Z}_n$ together with modular addition and multiplication.
The set $2\mathbb{Z}$ together with regular addition and multiplication.
The set $\mathbb{Q}[x]$ of all polynomials in the variable $x$ with coefficients in $\mathbb{Q}$ together with polynomial addition and multiplication.
Let $M_2(\mathbb{Z})$ bet the set of 2 by 2 matrices with entries in $\mathbb{Z}$, together with the usual properties of matrix addition and matrix multiplication.
Consider the set $\mathbb{R}^3$, together with the two binary operations of vector addition and the cross product.
The set of all real valued functions $f:\mathbb{R}\to\mathbb{R}$, together with function addition $(f+g)(x)=f(x)+g(x)$ and function multiplication $(fg)(x)=f(x)g(x)$.

Problem 8 (Rules Of Multiplication)

Let $R$ be a ring. Prove each of the following:

$a0=0a=0$
$a(-b)=(-a)b=-ab$
$(-a)(-b)=ab$
$a(b-c)=ab-ac$.
If $R$ has a unity, then $(-1)a = -a$.
If $R$ has a unity, then $(-1)(-1)=1$.

Definition (Subring)

A subset $S$ of a ring $R$ is a subring of $R$ if $S$ is itself a ring with the operations of $R$.

Problem 9 (Subring Test)

Let $R$ be ring. Prove that a nonempty subset $S$ of a ring $R$ is a subring of $R$ if $S$ is closed under subtraction and multiplication - that is, if $a-b$ and $ab$ are in $S$ whenever $a$ and $b$ are in $S$.

Problem 10 (Examples Of Rings Different Than The Integers)

For each item below, give an example of a ring $R$ and elements in the ring that satisfy the requested property.

$ab=0$ but neither $a$ nor $b$ equals $0$.
$ab=ac$ and $a\neq 0$ but $b\neq c$.
$ab=0$ but $ba\neq 0$.
$a^2=a$ but $a$ is neither 0 or 1.

Definition (Zero Divisor)

A zero-divisor is a nonzero element $a$ of a commutative ring $R$ such that there is a nonzero element $b\in R$ with $ab=0$.

Definition (Integral Domain)

An integral domain is a commutative ring with unity and no zero-divisors.

Problem 11 (Integral Domains Have The Cancellation Law)

Let $R$ be a commutative ring with unity. Prove that the following two statements are equivalent:

The ring $R$ is an integral domain.
For every $a,b,c\in R$ with $a\neq 0$, the equality $ab=ac$ implies $b=c$.

Definition (Gaussian Integers)

The set $\mathbb{Z}[i]=\{a+bi\mid a,b\in\mathbb{Z}\}$ is called the Gaussian integers.

Problem 12 (The Gaussian Integers Is An Integral Domain)

Use the subring test to show that the Gaussian Integers is a subring of the complex numbers $\mathbb{C}$. Then show that the Gaussian integers is an integral domain.

January 13

Make sure you finish problems 5 and 6 from Friday. Then please read chapters 12 and 13 in the text. Before Monday, please make sure you submit a written solution to one of the problems from this week.

In class I gave you a challenge. Find a ring $R$ and a subring $S$ so that that both $R$ and $S$ have a unity, but the unity in $S$ is not the unity in $R$.
We also asked the following in class:If a ring has a unity, must that unity be unique? If a ring element has a multiplicative inverse, must that inverse be unique? Prove these, or disprove them.

If you want to master the material we are learning, I strongly suggest that each week you pick some additional problems from the text to help reinforce what we are doing.

Chapter 12: 1,2,3,4,8,12,13,14, 18, 20. I'll let you read the others. If you see one that looks interesting, then do it. Challenge yourself.
Chapter 13: You can do any of them, though the problems before 30 are related to what we did this week.

See you Monday.

In class Work

I'll be putting a few notes in here for in class Monday.

Work with $\mathbb{Z}_p[i]=\{a+bi \mid a,b\in \mathbb{Z}_p \}$ for several values of $p$. Any guess as to when is it a field? Look at $p=2,3,5$. For each, construct a multiplication table.

Show that $\mathbb{Z}[\sqrt{d}]$ is an integral domain. Show that $\mathbb{Q}[\sqrt{d}]$ is a field.

Show that $\phi:\mathbb{Z}\to\mathbb{Z}_n$ is a ring homomorphism.

Let $R$ be a ring with unity with characteristic $c$. Show that $\phi:\mathbb{Z}_c\to R$ is a ring homomorphism.

Suppose $R$ is a ring and $A$ is a subring of $R$. We know that $R/A$ is already an Abelian group. What would it take for $R/A$ to be a ring? What properties must we satisfy? What would we need to guarantee that $R/A$ is an integral domain? What would we need to guarantee that $R/A$ is a field?

Define nilpotent and idempotent. Find a nilpotent 2 by 2 nonzero matrix. Find a nonidentity, nonzero, matrix that is idempotent. If you are in an integral domain, what are the nilpotent and idempotent elements?

Under what conditions will $\mathbb{Z}_p[\sqrt{d}]$ be a field? Show that if $p=7$ and $d=3$, then this is a field.

When is it true that $(x+y)^n=x^n+y^n$?

If $a$ is nilpotent in a ring with unity, show that $1-a$ is a unit. If

January 15

Today we'll get more practice with rings, and introduce ring homomorphisms and factor rings. First, let's show that any time we start with an integral domain, then the ring of polynomials with coefficients in that integral domain will still be an integral domain. In other words, let's show that any time you multiply two polynomials and obtain the zero polynomial, then one of the polynomials must have been zero to start with.

Problem 16 (A Polynomial Ring Over An Integral Domain Is An Integral Domain)

If $D$ is an integral domain, show that $D[x]$, the set of all polynomials with coefficients in $D$, is an integral domain.

Just as we defined a group homomorphism to be a map from one group another that preserved the group operation, we'll define a ring homomorphism to preserve both ring structures.

Definition (Ring Homomorphism And Isomorphism)

A ring homomorphism $\phi$ from a ring $R$ to a ring $S$ is a mapping from $R$ to $S$ that preserves the two ring operations. In other words $\phi(a+b)=\phi(a)+\phi(b)$ and $\phi(ab)=\phi(a)\phi(b)$. A bijective ring homomorphism is called a ring isomorphism.

The kernel of a group homomorphism is the collection of group elements that map to zero. The same holds true for the kernel of a ring homomorphism.

Definition (Kernel Of A Ring Homomorphism)

Let $\phi:R\to S$ be a ring homomorphism. The kernel of $\phi$ is the set $$\ker \phi = \{r\in R\mid \phi(r)=0\}.$$

When studying groups, we invented the word normal subgroups to parallel the properties of kernel. A subgroup of a group is normal precisely when it is the kernel of a group homomorphism. We then used these properties to create factor groups. We need to do the exact same thing now with rings. We'll first look at some properties of the kernel, and then we'll turn our attention to factor rings.

Problem 18 (Kernels Are Closed Under Multiplication By Arbitrary Elements)

Let $\phi:R\to S$ be ring homomorphism with kernel $K$. Show the following:

The kernel $K$ is a subring of $R$.
If $r\in R$ and $k\in K$, then we have $rk\in K$ and $kr\in K$.

The two properties above are precisely the key for characterizing when we can create a factor ring. We use the word ideal to describe these subrings. The next problem shows that you can create a factor ring, provided you have an ideal.

Definition (Ideal)

A subring $A$ of $R$ is called an ideal if $ra\in A$ and $ar\in A$ whenever $a\in A$ and $r\in R$.

Problem 19 (Ideals Give Us Factor Rings)

Let $R$ be a ring and let $A$ be a subring of $R$. Show the following are equivalent.

The set of cosets $\{ r+A\mid r\in R\}$ is a ring under the operations $(s+A)+(t+A) = (s+t)+A$ and $(s+A)(t+A) = st+A$.
The subring $A$ is an ideal of $R$.

When you assume the first bullet above, you get to assume that the operations are binary operation. However, when you assume the second bullet, you must prove that the operations are binary operations. If we let $x,y\in R/A$, then there exists $s,t\in R$ such that $x=s+A$ and $y=t+A$. However, the choice of $s$ and $t$ is not unique. Suppose also that $x=s'+A$ and $y=t'+A$. The operation given for computing products then says that $xy=(s+A)(t+A) = st+A$, but it also says that $xy=(s'+A)(t'+A) = s't'+A$. However, if $st+A\neq s't'+A$, then we have a problem, as then $xy$ is not well-defined. This is what you must show. You must prove that if $A$ is an ideal, then you are guaranteed that $(s'+A)(t'+A) = s't'+A$.

Definition (Factor Ring)

Let $R$ be a ring and let $A$ be an ideal of $R$. The set of cosets $\{ r+A\mid r\in R\}$ together with the binary operations $(s+A)+(t+A) = (s+t)+A$ and $(s+A)(t+A) = st+A$ is called the factor ring of $R$ by $A$, or just the factor ring $R/A$.

We'll return to factor rings more next time. They happen to be a key tool in our future study, and we'll spend plenty of time becoming comfortable with them. For the rest of today, the next few problems have you practice with some of the definitions we've been building up over the last week, as well as adding in the characteristic of a field and the ideal generated by something.

Problem 20 (Finite Integral Domains Are Fields)

Suppose that $R$ is a finite integral domain. Prove that $R$ is a field.

Definition (Characteristic Of A Ring)

The characteristic of a ring $R$, written $\text{char } R$, is the smallest positive integer $n$ such that $n\cdot a = a+a+\cdots+a=0$ for all $a\in R$. If no such integer exists, then we say $R$ has characteristic zero.

Problem 22 (Characteristics And Rings With Unity)

Suppose that $R$ is a ring with unity.

Show that the characteristic of $R$ is equal to the least positive integer $n$ such that $n\cdot 1=0$, provided the characteristic is not zero.
Show that the characteristic of an integral domain is either zero or prime.

Definition (Principle Ideal Generated by $a$, or $\left<a\right>$)

Let $R$ be a commutative ring with unity and let $a\in R$. The set $\left<a\right> = \{ra| \ r\in R\}$ is an ideal of $R$ called the principle ideal generated by $a$.

Definition (Ideal Generated By $a_1, \ldots, a_n$, or $\left<a_1,a_2,\ldots,a_n\right>$)

Let $R$ be a commutative ring with unity, and let $a_1, \ldots, a_n \in R$. The set $I=\left<a_1,a_2,\ldots,a_n\right> = \{r_1a_1+\cdots +r_na_n|r_i\in R \}$ is called the ideal generated by $a_1, \ldots, a_n$. Any other ideal that contains $a_1, \ldots, a_n$ must contain $I$.

Was moved to Problem.TheIdealGeneratedByASubsetIsAnIdeal - got rid of the extra "By" in the name.

January 17

The first few examples today are designed to help us get a better handle on factor rings. Recall that $\left<a\right>$ is called the principle ideal generated by $a$.

Problem 24 (A Matrix Factor Ring)

Consider the set of two by two matrices $R=M_2(\mathbb{Z})$ with integer coefficients. Let $A$ the set of matrices whose coefficients are multiples of 3.

Show that $A$ is an ideal of $R$.
Find three different elements of $R$ that are in the coset $\begin{bmatrix}5&-2\\7&9\end{bmatrix}+A$.
How many distinct elements are in this factor ring? Justify your answer.
Challenge: This factor ring is isomorphic to another matrix ring we have already seen. What is this ring? You don't need to prove your answer.

Problem 25 (A Factor Ring Of The Gaussian Integers)

Let $R=\mathbb{Z}[i]$, $a=3-i$, and let $A=\left<a\right>$. We would like to examine the factor ring $R/A$, determine how many elements are in this factor ring, and find a simple ring to which $\mathbb{Z}[i]/\left<3-i\right>$ is isomorphic.

Why do we know $3+A=i+A$? Use this to explain why $10+A=0+A$.
Explain why every element of $R/A$ must be of the form $k+A$ where $k\in \{0,1,2,\ldots 9\}$, so there are at most 10 elements.
Suppose the integer $k$ is an element of $A = \left<a\right> = \{ra\mid r\in R\}$. In other words, suppose that there exists an element $r\in R$ such that $k+0i = ra = (c+di)(3-i)$. Use this to show that $k$ must be a multiple of 10.
State a much simpler ring to which $R/A$ is isomorphic, you don't need to prove your answer.

After completing these exercises (or if you get stuck), I strongly recommend that you take a few minutes and read the examples on pages 263-266. They summarize many of the key ideas we need to focus on with factor rings.

Problem 26 (Factor Rings Of $\mathbb{Z}$ And $\mathbb{Z}[x]$)

Answer the following questions as they pertain to the integral domains $\mathbb{Z}$ and $\mathbb{Z}[x]$. You are welcome to rapidly make claims about factor rings, without proof, as we have seen many of these as factor groups in the past.

We know that the ideals of $\mathbb{Z}$ are of the form $n\mathbb{Z} = \left< n\right>$, the set of multiples of $n$.
- For which $n$ is the factor ring $\mathbb{Z}/\left<n\right>$ an integral domain?
- For which $n$ is the factor ring $\mathbb{Z}/\left<n\right>$ a field?
Now consider the ring of polynomials $\mathbb{Z}[x]$. The ideal $\left<n\right>$ is now the set of polynomials whose coefficients are multiples of $n$. It should not be a surprise that $\mathbb{Z}[x]/\left<n\right> \approx \mathbb{Z}_n[x]$.
- For which $n$ is the factor ring $\mathbb{Z}[x]/\left<n\right>$ an integral domain?
- Show that the factor ring $\mathbb{Z}[x]/\left<n\right>$ is never a field, regardless of which $n$ you pick.
We can look at other ideals of $\mathbb{Z}[x]$.
- Show that $\mathbb{Z}[x]/\left<x\right>$ is an integral domain, but not a field.
- Show that $\mathbb{Z}[x]/\left<3,x\right>$ is a field.

As seen in the previous exercise, sometimes when we create a factor ring, we obtain an integral domain, and sometimes we obtain a field. We would like some words to describe ideals for which the corresponding factor ring is an integral domain, or a field. It would be nice to have a characterization that we could check in the ideal itself, without having to actually look at the factor ring.

Problem 27 ($R/A$ Is An Integral Domain Iff $A$ is prime)

Let $R$ be a commutative ring with unity, and let $A$ be a proper ideal of $R$. Prove that the following are equivalent.

$R/A$ is an integral domain.
If $a,b\in R$ and $ab\in A$ then $a\in A$ or $b\in A$. (We say that $A$ is a prime ideal).

Why do we call an ideal a prime ideal if the second condition above is satisfied? Recall in the integers that if $p$ is a prime, then if $ab$ is a multiple of $p$, then either $a$ or $b$ must be a multiple of $p$. So $ab\in \left<p\right>$ forces either $a\in \left<p\right>$ or $b\in \left<p\right>$. This condition forces the number to be prime, so we extend the notation of prime numbers to prime ideals. Any time a product is divisible by a prime, one of the two factors must be divisible by the prime. We'll extend this to say that any time a product is in a prime ideal, one of the factors must be in the prime ideal.

We now have a simple way to check if $R/A$ is an integral domain. We just have to make sure that if $ab\in A$, then either $a\in A$ or $b\in A$. To obtain a field, we need even more. We need to make sure that if $b\notin A$, then there exists $c\notin A$ such that $bc\in 1+A$, or that $1-bc\in A$, or equivalently $1\in bc+A$. So given any element $b\notin A$, if we can show that $B=\{bc+a\mid c\in R, a\in A\}$ must contain 1, then we'd win. Let's first show that this set $B$ is actually an ideal, and then show that $1\in B$ forces $B=R$.

Problem 29 (Obtaining A New Ideal By Adding One Element)

Let $A$ be an ideal of the commutative ring $R$ with unity. We'd like to increase the size of $A$ by adding in a single element $b$. Show the following:

Pick $b\in R$. The set $B=\{bc+a\mid c\in R,a\in A\}$ is an ideal of $R$ that contains both the element $b$ and subset $A$.
We know $1\in A$ if and only if $A=R$.

The set $B$ described above is the smallest ideal of $R$ that contains $b$ and every element in $A$, so we could call it the ideal generated by $b$ and $A$.

Given a proper ideal $A_1$, we can use the idea above to create an ideal $A_2$ that properly contains $A_1$ (add one element not in $A_1$). If this ideal is not the entire ring, we could then continue this process, possibly indefinitely, to obtain a sequence of ideals that properly contain the previous. Such a sequence $A_1\subset A_2\subset A_3\subset \cdots \subset A_n\subset \cdots$ is called an ascending chain of ideals. Once one of these ideals contains the number 1, the sequence terminates. Emmy Noether used this idea to create what we now call Noetherian domains, an integral domain in which any ascending chain of ideals must be finite in length (so the process must stop, no matter what ideal $A$ you start with). We'll come back to Noetherian domains later. See page 275 for more information about Emmy Noether.

Problem 30 (An Ascending Chain Of Ideals In The Integers)

Let $R=\mathbb{Z}$. Consider the ideal $A_1=\left<56\right>.$

Find an ascending chain of ideals consisting of three ideals $A_1\subsetneq A_2\subsetneq A_3$ with $A_3=R$.
What is the largest $n$ such that $A_1\subsetneq A_2\subsetneq A_3\subsetneq \cdots\subsetneq A_n=R$ is an ascending chain of ideals? How many such chains are there of this length?
If $n\in \mathbb{Z}$, describe a process we could use to find a longest possible ascending chain of ideals with $\left<n\right>\subsetneq A_2\subsetneq \cdots\subsetneq A_n=\mathbb{Z}$.

The two facts from the problem Obtaining A New Ideal By Adding One Element will help us characterize the properties of an ideal $A$ that will result in $R/A$ being a field. Remember, the following:

We need to make sure that if $b\notin A$, then there exists $c\in R$ such that $bc\in 1+A$, or equivalently $1\in bc+A$.
So given any element $b\notin A$, if we can show that the ideal $B=\{bc+a\mid c\in R, a\in A\}$ must contain 1, which means $B=R$, then we'd win.
This is basically the same as showing that given any ideal $B$ trapped between $A$ and $R$, so $A\subset B\subset R$, then if $B$ is not equal to $A$, it must equal $R$. We call an ideal such as $A$ a maximal ideal.

The next problem has you carefully show that $R/A$ is a field if and only if $A$ is a maximal ideal.

Problem 31 ($R/A$ Is A Field Iff $A$ Is Maximal)

Let $R$ be a commutative ring with unity, and let $A$ be a proper ideal of $R$. Prove that the following are equivalent.

$R/A$ is a field.
Whenever $B$ is an ideal of $R$ and $A\subseteq B\subseteq R$, then either $B=A$ or $B=R$. (We say that $A$ is a maximal ideal.)

The previous problems give us the following definitions of prime and maximal ideals. Basically, these are now characteristics of an ideal that we can check to determine if $R/A$ is an integral domain or a field, without having to ever consider elements of the factor ring.

Definition (Prime Ideal And Maximal Ideal)

A prime ideal $A$ of a commutative ring $R$ is a proper ideal of $R$ such that $a,b\in R$ and $ab\in A$ imply $a\in A$ or $b\in A$.
A maximal ideal $A$ of a commutative ring $R$ is a proper ideal of $R$ such that, whenever $B$ is an ideal of $R$ and $A\subseteq B\subseteq R$, then $B=A$ or $B=R$.

Let's end today with another example of some interesting things that can happen in a ring.

Definition (Idempotent And Nilpotent)

Let $R$ be a ring.

We say that $a$ is an idempotent of $R$ if $a^2=a$.
We say that $a$ is nilpotent if $a^n=0$ for some integer $n$.

Problem 32 (Finding Idempotent And Nilpotent Elements In A Matrix Ring)

Complete the following:

In the ring $M_2(\mathbb{Z})$, the matrices $\begin{bmatrix}0&0\\0&0\end{bmatrix}$, $\begin{bmatrix}1&0\\0&1\end{bmatrix}$, $\begin{bmatrix}1&0\\0&0\end{bmatrix}$, $\begin{bmatrix}0&0\\0&1\end{bmatrix}$, are idempotents. Find two others.
Show that $\begin{bmatrix}0&t\\0&0\end{bmatrix}$ is nilpotent in $M_2(\mathbb{Z})$ and that $\begin{bmatrix}0&t&0\\0&0&t\\0&0&0\end{bmatrix}$ is nilpotent in $M_3(\mathbb{Z})$, where $t\in\mathbb{Z}$.
Let $X=\begin{bmatrix}0&t&0&0&0\\0&0&t&0&0\\0&0&0&t&0\\0&0&0&0&t\\0&0&0&0&0\end{bmatrix}$. Compute $X^k$ for each positive integer $k$. What patterns do you see? Is $X$ a nilpotent element of $M_5(\mathbb{Z})$?

The matrices above are central to finding the Jordan form of a matrix, and then using the Jordan form to compute the matrix exponential. Come see me out of class if you would like more information on this topic, or consider taking Math 495 next fall, where the topic will be linear algebra.

January 20

January 22

Please complete the problems from Friday's prep. We finished problem 1 in class. Here they are.

The first few examples today are designed to help us get a better handle on factor rings. Recall that $\left<a\right>$ is called the principle ideal generated by $a$.

Problem 24 (A Matrix Factor Ring)

Consider the set of two by two matrices $R=M_2(\mathbb{Z})$ with integer coefficients. Let $A$ the set of matrices whose coefficients are multiples of 3.

Show that $A$ is an ideal of $R$.
Find three different elements of $R$ that are in the coset $\begin{bmatrix}5&-2\\7&9\end{bmatrix}+A$.
How many distinct elements are in this factor ring? Justify your answer.
Challenge: This factor ring is isomorphic to another matrix ring we have already seen. What is this ring? You don't need to prove your answer.

Problem 25 (A Factor Ring Of The Gaussian Integers)

Why do we know $3+A=i+A$? Use this to explain why $10+A=0+A$.
Explain why every element of $R/A$ must be of the form $k+A$ where $k\in \{0,1,2,\ldots 9\}$, so there are at most 10 elements.
Suppose the integer $k$ is an element of $A = \left<a\right> = \{ra\mid r\in R\}$. In other words, suppose that there exists an element $r\in R$ such that $k+0i = ra = (c+di)(3-i)$. Use this to show that $k$ must be a multiple of 10.
State a much simpler ring to which $R/A$ is isomorphic, you don't need to prove your answer.

Problem 26 (Factor Rings Of $\mathbb{Z}$ And $\mathbb{Z}[x]$)

We know that the ideals of $\mathbb{Z}$ are of the form $n\mathbb{Z} = \left< n\right>$, the set of multiples of $n$.
- For which $n$ is the factor ring $\mathbb{Z}/\left<n\right>$ an integral domain?
- For which $n$ is the factor ring $\mathbb{Z}/\left<n\right>$ a field?
Now consider the ring of polynomials $\mathbb{Z}[x]$. The ideal $\left<n\right>$ is now the set of polynomials whose coefficients are multiples of $n$. It should not be a surprise that $\mathbb{Z}[x]/\left<n\right> \approx \mathbb{Z}_n[x]$.
- For which $n$ is the factor ring $\mathbb{Z}[x]/\left<n\right>$ an integral domain?
- Show that the factor ring $\mathbb{Z}[x]/\left<n\right>$ is never a field, regardless of which $n$ you pick.
We can look at other ideals of $\mathbb{Z}[x]$.
- Show that $\mathbb{Z}[x]/\left<x\right>$ is an integral domain, but not a field.
- Show that $\mathbb{Z}[x]/\left<3,x\right>$ is a field.

Problem 27 ($R/A$ Is An Integral Domain Iff $A$ is prime)

Let $R$ be a commutative ring with unity, and let $A$ be a proper ideal of $R$. Prove that the following are equivalent.

$R/A$ is an integral domain.
If $a,b\in R$ and $ab\in A$ then $a\in A$ or $b\in A$. (We say that $A$ is a prime ideal).

Problem 29 (Obtaining A New Ideal By Adding One Element)

Let $A$ be an ideal of the commutative ring $R$ with unity. We'd like to increase the size of $A$ by adding in a single element $b$. Show the following:

Pick $b\in R$. The set $B=\{bc+a\mid c\in R,a\in A\}$ is an ideal of $R$ that contains both the element $b$ and subset $A$.
We know $1\in A$ if and only if $A=R$.

The set $B$ described above is the smallest ideal of $R$ that contains $b$ and every element in $A$, so we could call it the ideal generated by $b$ and $A$.

Problem 30 (An Ascending Chain Of Ideals In The Integers)

Let $R=\mathbb{Z}$. Consider the ideal $A_1=\left<56\right>.$

Find an ascending chain of ideals consisting of three ideals $A_1\subsetneq A_2\subsetneq A_3$ with $A_3=R$.
What is the largest $n$ such that $A_1\subsetneq A_2\subsetneq A_3\subsetneq \cdots\subsetneq A_n=R$ is an ascending chain of ideals? How many such chains are there of this length?
If $n\in \mathbb{Z}$, describe a process we could use to find a longest possible ascending chain of ideals with $\left<n\right>\subsetneq A_2\subsetneq \cdots\subsetneq A_n=\mathbb{Z}$.

The two facts from the problem Obtaining A New Ideal By Adding One Element will help us characterize the properties of an ideal $A$ that will result in $R/A$ being a field. Remember, the following:

We need to make sure that if $b\notin A$, then there exists $c\in R$ such that $bc\in 1+A$, or equivalently $1\in bc+A$.
So given any element $b\notin A$, if we can show that the ideal $B=\{bc+a\mid c\in R, a\in A\}$ must contain 1, which means $B=R$, then we'd win.
This is basically the same as showing that given any ideal $B$ trapped between $A$ and $R$, so $A\subset B\subset R$, then if $B$ is not equal to $A$, it must equal $R$. We call an ideal such as $A$ a maximal ideal.

The next problem has you carefully show that $R/A$ is a field if and only if $A$ is a maximal ideal.

Problem 31 ($R/A$ Is A Field Iff $A$ Is Maximal)

Let $R$ be a commutative ring with unity, and let $A$ be a proper ideal of $R$. Prove that the following are equivalent.

$R/A$ is a field.
Whenever $B$ is an ideal of $R$ and $A\subseteq B\subseteq R$, then either $B=A$ or $B=R$. (We say that $A$ is a maximal ideal.)

Definition (Prime Ideal And Maximal Ideal)

A prime ideal $A$ of a commutative ring $R$ is a proper ideal of $R$ such that $a,b\in R$ and $ab\in A$ imply $a\in A$ or $b\in A$.
A maximal ideal $A$ of a commutative ring $R$ is a proper ideal of $R$ such that, whenever $B$ is an ideal of $R$ and $A\subseteq B\subseteq R$, then $B=A$ or $B=R$.

Let's end today with another example of some interesting things that can happen in a ring.

Definition (Idempotent And Nilpotent)

Let $R$ be a ring.

We say that $a$ is an idempotent of $R$ if $a^2=a$.
We say that $a$ is nilpotent if $a^n=0$ for some integer $n$.

Problem 32 (Finding Idempotent And Nilpotent Elements In A Matrix Ring)

Complete the following:

In the ring $M_2(\mathbb{Z})$, the matrices $\begin{bmatrix}0&0\\0&0\end{bmatrix}$, $\begin{bmatrix}1&0\\0&1\end{bmatrix}$, $\begin{bmatrix}1&0\\0&0\end{bmatrix}$, $\begin{bmatrix}0&0\\0&1\end{bmatrix}$, are idempotents. Find two others.
Show that $\begin{bmatrix}0&t\\0&0\end{bmatrix}$ is nilpotent in $M_2(\mathbb{Z})$ and that $\begin{bmatrix}0&t&0\\0&0&t\\0&0&0\end{bmatrix}$ is nilpotent in $M_3(\mathbb{Z})$, where $t\in\mathbb{Z}$.
Let $X=\begin{bmatrix}0&t&0&0&0\\0&0&t&0&0\\0&0&0&t&0\\0&0&0&0&t\\0&0&0&0&0\end{bmatrix}$. Compute $X^k$ for each positive integer $k$. What patterns do you see? Is $X$ a nilpotent element of $M_5(\mathbb{Z})$?

January 24

Problem 29 (Obtaining A New Ideal By Adding One Element)

Let $A$ be an ideal of the commutative ring $R$ with unity. We'd like to increase the size of $A$ by adding in a single element $b$. Show the following:

Pick $b\in R$. The set $B=\{bc+a\mid c\in R,a\in A\}$ is an ideal of $R$ that contains both the element $b$ and subset $A$.
We know $1\in A$ if and only if $A=R$.

The set $B$ described above is the smallest ideal of $R$ that contains $b$ and every element in $A$, so we could call it the ideal generated by $b$ and $A$.

Problem 30 (An Ascending Chain Of Ideals In The Integers)

Let $R=\mathbb{Z}$. Consider the ideal $A_1=\left<56\right>.$

Find an ascending chain of ideals consisting of three ideals $A_1\subsetneq A_2\subsetneq A_3$ with $A_3=R$.
What is the largest $n$ such that $A_1\subsetneq A_2\subsetneq A_3\subsetneq \cdots\subsetneq A_n=R$ is an ascending chain of ideals? How many such chains are there of this length?
If $n\in \mathbb{Z}$, describe a process we could use to find a longest possible ascending chain of ideals with $\left<n\right>\subsetneq A_2\subsetneq \cdots\subsetneq A_n=\mathbb{Z}$.

The two facts from the problem Obtaining A New Ideal By Adding One Element will help us characterize the properties of an ideal $A$ that will result in $R/A$ being a field. Remember, the following:

We need to make sure that if $b\notin A$, then there exists $c\in R$ such that $bc\in 1+A$, or equivalently $1\in bc+A$.
So given any element $b\notin A$, if we can show that the ideal $B=\{bc+a\mid c\in R, a\in A\}$ must contain 1, which means $B=R$, then we'd win.
This is basically the same as showing that given any ideal $B$ trapped between $A$ and $R$, so $A\subset B\subset R$, then if $B$ is not equal to $A$, it must equal $R$. We call an ideal such as $A$ a maximal ideal.

The next problem has you carefully show that $R/A$ is a field if and only if $A$ is a maximal ideal.

Problem 31 ($R/A$ Is A Field Iff $A$ Is Maximal)

Let $R$ be a commutative ring with unity, and let $A$ be a proper ideal of $R$. Prove that the following are equivalent.

$R/A$ is a field.
Whenever $B$ is an ideal of $R$ and $A\subseteq B\subseteq R$, then either $B=A$ or $B=R$. (We say that $A$ is a maximal ideal.)

Definition (Prime Ideal And Maximal Ideal)

A prime ideal $A$ of a commutative ring $R$ is a proper ideal of $R$ such that $a,b\in R$ and $ab\in A$ imply $a\in A$ or $b\in A$.
A maximal ideal $A$ of a commutative ring $R$ is a proper ideal of $R$ such that, whenever $B$ is an ideal of $R$ and $A\subseteq B\subseteq R$, then $B=A$ or $B=R$.

In class we looked at the factor ring $\mathbb{Z}[x]/\left<x^2\right>$, showed as a set that it was equal to $\{a_1x+a_0\mid a_1,a_0\in \mathbb{Z}\}$, and showed that it is not an integral domain as $$(x+\left<x^2\right>)(x+\left<x^2\right>) = (x^2+\left<x^2\right>)=(0+\left<x^2\right>).$$ We also showed that $\mathbb{Z}[x]/\left<x^2-1\right>$ was equal to the same set, however multiplication is different in this field, yet it is not an integral domain because $(x^2-1)=(x-1)(x+1)$. The next problem has you examine this question if we instead use the polynomial $x^2+1$.

Problem 39 (Some Polynomial Factor Rings)

Complete the following:

Use the first isomorphism theorem to prove that $\mathbb{Z}[x]/\left<x^2+1\right>$ is isomorphic to $\mathbb{Z}[i]$. From this we know that $\mathbb{Z}[x]/\left<x^2+1\right>$ is an integral domain and not a field.
Is $\mathbb{Q}[x]/\left<x^2+1\right>$ an integral domain? a field?
Is $\mathbb{R}[x]/\left<x^2+1\right>$ an integral domain? a field?
Is $\mathbb{C}[x]/\left<x^2+1\right>$ an integral domain? a field?

Problem 35 (When Is A Polynomial Factor Ring An Integral Domain)

Let $R=\mathbb{Z}[x]$. Suppose that $A=\left<p(x)\right>$ is a principle ideal generated by a single polynomial.

Find a polynomial $p(x)$ of degree 2 such that $R/A$ is not an integral domain. Then find one of degree 3 and degree 4.
Find a polynomial $p(x)$ of degree 3 such that $R/A$ is an integral domain.
If you know that $R/A$ is not an integral domain, what do you know about $p(x)$?
State a simple condition on $p(x)$ that is equivalent to $R/A$ being an integral domain.

The last two problems for today are designed as review problems, to help you practice with the definitions. We may not prove all of these in class. They should be very similar to problems we completed last semester.

Problem 36 (The Ideal Test)

Suppose that $A$ is a nonempty subset $A$ of a ring $R$. Prove that $A$ is an ideal of $R$ if

$a-b\in A$ whenever $a,b\in A$, and
$ra$ and $ar$ are in $A$ whenever $a\in A$ and $r\in R$.

In other words, prove that an ideal is a nonempty subset that is closed under subtraction and multiplication by an arbitrary ring element.

Problem 37 (Properties Of Ring Homomorphisms)

Let $\phi:R\to S$ be a ring homomorphism. Prove the following properties.

For any $r\in R$ and any positive integer $n$, $\phi(n\cdot r) = n\phi(r)$ and $\phi(r^n) = (\phi(r))^n$.
If $A$ is a subring of $R$, then $\phi (A) = \{\phi(a)|a\in A\}$ is a subring of $S$.
If $A$ is an ideal, then $\phi(A)$ is an ideal of $\phi(R)$.
If $B$ is an ideal of $S$, then $\phi^{-1}(B) = \{r\in R|\phi(r)\in B\}$ is an ideal of $R$.
If $R$ is commutative, then $\phi(R)$ is commutative.
If $R$ has a unity 1, if $S\neq \{0\}$, and if $\phi$ is onto, then $\phi(1)$ is the unity of $S$.
We know $\phi$ is an isomorphism if and only if $\phi$ is onto and $\ker\phi = \{r\in R|\phi(r)=0\} = \{0\}$.
If $\phi$ is an isomorphism from $R$ onto $S$, then $\phi^{-1}$ is an isomorphism from $S$ onto $R$.

Let me know which parts you are ready to prove in class. We'll split this one up and have several presenters.

Note: If you decide to type up this for your problem, you can pick three of them and prove just them.

January 27

The two facts from the problem Obtaining A New Ideal By Adding One Element will help us characterize the properties of an ideal $A$ that will result in $R/A$ being a field. Remember, the following:

We need to make sure that if $b\notin A$, then there exists $c\in R$ such that $bc\in 1+A$, or equivalently $1\in bc+A$.
So given any element $b\notin A$, if we can show that the ideal $B=\{bc+a\mid c\in R, a\in A\}$ must contain 1, which means $B=R$, then we'd win.
This is basically the same as showing that given any ideal $B$ trapped between $A$ and $R$, so $A\subset B\subset R$, then if $B$ is not equal to $A$, it must equal $R$. We call an ideal such as $A$ a maximal ideal.

The next problem has you carefully show that $R/A$ is a field if and only if $A$ is a maximal ideal.

Problem 31 ($R/A$ Is A Field Iff $A$ Is Maximal)

Let $R$ be a commutative ring with unity, and let $A$ be a proper ideal of $R$. Prove that the following are equivalent.

$R/A$ is a field.
Whenever $B$ is an ideal of $R$ and $A\subseteq B\subseteq R$, then either $B=A$ or $B=R$. (We say that $A$ is a maximal ideal.)

Definition (Prime Ideal And Maximal Ideal)

A prime ideal $A$ of a commutative ring $R$ is a proper ideal of $R$ such that $a,b\in R$ and $ab\in A$ imply $a\in A$ or $b\in A$.
A maximal ideal $A$ of a commutative ring $R$ is a proper ideal of $R$ such that, whenever $B$ is an ideal of $R$ and $A\subseteq B\subseteq R$, then $B=A$ or $B=R$.

We'll have Nick, Laura, and Joe share their results related to last time.

We'll introduce the following problems, together with many related ideas.

Show that $\mathbb{Z}/\left<n\right>\approx \mathbb{Z}_n$ using the first isomorphism theorem. Make sure you prove that the kernel is what you claim it is. The division algorithm should be useful.
Practice long division.
Prove that $\mathbb{Z}[x]/\left<x^2+1\right>\approx \mathbb{Z}[i]$, by using the first isomorphism theorem.
Prove that $\mathbb{Z}[x]/\left<x^2+1\right>$ is an integral domain using linear algebra.
Why is it enough to show that no two polynomials can multiply together to give $(x^2+1)$, instead of showing that no two polynomials can multiply together to give $(x^2+1)p(x)$ for some polynomial $p(x)$?

January 29

We've seen half of this proof in class. We still need to see the other half. Please come ready to explain why if $B$ is a maximal ideal, then $R/A$ is a field.

Problem 31 ($R/A$ Is A Field Iff $A$ Is Maximal)

Let $R$ be a commutative ring with unity, and let $A$ be a proper ideal of $R$. Prove that the following are equivalent.

$R/A$ is a field.
Whenever $B$ is an ideal of $R$ and $A\subseteq B\subseteq R$, then either $B=A$ or $B=R$. (We say that $A$ is a maximal ideal.)

We won't be going back to prove all the properties of ring homomorphisms, as their proofs are practically identical to the group theoretic versions. If there is one you want to see in class, please let me know. If you are struggling with the material, I strongly suggest you take the time to prove each of these properties. We'll take time in class to prove the first isomorphism theorem, which is a quick application of these properties.

Theorem (First Isomorphism Theorem For Rings)

Let $\phi$ be a ring homomorphism from $R$ to $S$. The mapping from $R/\ker\phi$ to $\phi(R)$, given by $r+\ker\phi \to \phi(r)$ is an isomorphism. In symbols, we have $R/\ker\phi\cong \phi(R).$

Problem 38 (First Isomorphism Theorem For Rings Proof)

Prove the first isomorphism theorem for rings. You'll need to first prove that the map defined in the statement of the theorem is well-defined, then you can use the properties of ring homomorphisms to show that the map is an isomorphism.

The next few definitions are things you already know, I am just including them here for completeness. Make sure you read the definition of the degree of a polynomial.

Definition (Polynomial Ring With Coefficients In R)

Let $R$ be commutative ring. The set of formal symbols $$R[x] = \{a_nx^n+\cdots+a_1x+a_0|a_i\in R, n\in Z^+\}$$ is called the ring of polynomials over $R$ in the indeterminate $x$. Two elements are considered equal if and only if the have the same coefficients. Addition is defined component wise, and multiplication is defined using regular polynomial distribution.

Definition (Degree Of A Polynomial)

If $f(x) = a_nx^n+\cdots+a_1x+a_0$ with $a_n\neq 0$ then we say the degree of $f(x)$ is $n$. The polynomial $f(x)=0$ has no degree (it is not degree zero).

Let's now prove that given any polynomials $f(x)$ and $g(x)$, we can always perform long division and write $f(x)=q(x)g(x)+r(x)$ where the degree of $r$ is less than the degree of $g$, or $r=0$. To figure out a proof, just pretend like you are going to do long division. What would your first step be? You should be able to reduce the degree of $f(x)$, and then use induction on the degree of $f(x)$.

Theorem (Division Algorithm For Polynomials)

Let $F$ be a field and let $f(x)$ and $g(x)\in F[x]$ with $g(x)\neq 0$. Then there exist unique polynomials $q(x)$ and $r(x)$ such that $f(x)=g(x)q(x)+r(x)$ and either $r(x)=0$ or $\deg r(x) < \deg g(x)$.

Problem 40 (Division Algorithm For Polynomials Proof)

Prove Theorem Division Algorithm For Polynomials.

A lot of our work has been focused on determining when a ring must be a field. If it's a finite ring, there's a really easy way to immediately throw out the possibility that the ring is a field, by just counting elements. If the order of the ring is not a power of a prime, then it can't be a field. The next problem has you show this.

Problem 41 (The Order Of A Finite Field)

Suppose that $F$ is a finite field of characteristic $p$. Show that the order of $F$ must be $p^n$ for some integer $n$.

Click for a hint.

What's the additive order of every element in $F$? Use the fundamental theorem of finite Abelian groups, and just pay attention to the additive group, ignoring completely the multiplicative part of $F$.

Suppose we know that $D$ is an integral domain. If $D$ is finite, we've already shown it must be a field. However, if $D$ is not finite, can we somehow obtain a field from $D$? As an example, we already know that we can obtain the rationals $\mathbb{Q}$ from $\mathbb{Z}$ by defining division, in that we say $a/b=c/d$ if and only $ad=bc$ where $b\neq 0$ and $d\neq 0$. We'll show that this approach allows us to take any integral domain and from it create a field (called the field of quotients).

To obtain this field of fractions, let's first review what an equivalence relation is, as we use equivalence relations to define division.

Definition (Equivalence Relation)

Let $S$ be set. Let $\cong$ be a relation on $S$, meaning $\cong$ is a collection $\mathscr{C}$ of ordered pairs of $S$. We way that $A\cong B$ if and only if the ordered pair $(A,B)$ is an element of $\mathscr{C}$. We say that $\cong$ is an equivalence relation if and only if

(Reflexive) For every $A\in S$, we know that $A\cong A$ (so the ordered pair $(A,A)$ is always in $\mathscr{C}$).
(Symmetric) If $A\cong B$, then $B\cong A.$
(Transitive) If $A\cong B$ and $B\cong C$, then $A\cong C$.

Problem 42 (The Rational Are Obtained From The Integers From An Equivalence Relation)

Consider the set of ordered pairs $S=\{(a,b)\mid a,b\in\mathbb{Z},b\neq0\}$. Define a relation on $S$ by saying that $(a,b)\cong(c,d)$ if and only if $ad=bc$. We'll generally write $(a/b)=(c/d)$ to mean that $(a,b)\cong(c,d)$.

Prove that the relation above is an equivalence relation.
If we replace $\mathbb{Z}$ with any integral domain $D$, prove that we still obtain an equivalence relation on $S$. (If your work on part 1 didn't refer to the integers specifically, then this part should automatically follow.)

Problem 43 (The Field Of Quotients Of An Integral Domain)

Consider again the set of ordered pairs $S=\{(a,b)\mid a,b\in\mathbb{Z},b\neq0\}$, together with the equivalence relation $(a,b)\cong(c,d)$ if and only if $ad=bc$. Let $F$ be the set of equivalence classes of $S$ under the relation $\cong$. We can write an element in $F$ as $ [a/b] $ where the brackets remind us that there are infinitely many ways to represent elements in $F$ (for example we know $ [2/4]=[3/6] $ because $2\cdot 6=3\cdot 4$). On the set $F$, define the operations $$ [a/b]+[c/d]=[(ad+bc)/(bd)]\quad \text{and}\quad [a/b]\cdot[c/d]=[ac/bd]. $$

Prove $+$ and $\cdot$ are binary operations on $F$. This will require you show that if $ [a/b] \cong [a'/b']$ and $ [c/d] \cong [c'/d']$ then we must have $ [(ad+bc)/(bd)] \cong [(a'd'+b'c')/(b'd')] $, and a similar fact for multiplication.
Prove that $(F,+,\cdot)$ is a field.
If we replace $\mathbb{Z}$ with any integral domain $D$, prove that $(F,+,\cdot)$ is a field.

January 31

We won't be going back to prove all the properties of ring homomorphisms, as their proofs are practically identical to the group theoretic versions. If there is one you want to see in class, please let me know. If you are struggling with the material, I strongly suggest you take the time to prove each of these properties (they are a good review). We'll take time in class to prove the first isomorphism theorem, which is a quick application of these properties, and any others that you want to see.

Theorem (First Isomorphism Theorem For Rings)

Let $\phi$ be a ring homomorphism from $R$ to $S$. The mapping from $R/\ker\phi$ to $\phi(R)$, given by $r+\ker\phi \to \phi(r)$ is an isomorphism. In symbols, we have $R/\ker\phi\cong \phi(R).$

Problem 38 (First Isomorphism Theorem For Rings Proof)

The next few definitions are things you already know, I am just including them here for completeness. Make sure you read the definition of the degree of a polynomial.

Definition (Polynomial Ring With Coefficients In R)

Definition (Degree Of A Polynomial)

If $f(x) = a_nx^n+\cdots+a_1x+a_0$ with $a_n\neq 0$ then we say the degree of $f(x)$ is $n$. The polynomial $f(x)=0$ has no degree (it is not degree zero).

Theorem (Division Algorithm For Polynomials)

Problem 40 (Division Algorithm For Polynomials Proof)

Prove Theorem Division Algorithm For Polynomials.

Problem 41 (The Order Of A Finite Field)

Suppose that $F$ is a finite field of characteristic $p$. Show that the order of $F$ must be $p^n$ for some integer $n$.

Click for a hint.

To obtain this field of fractions, let's first review what an equivalence relation is, as we use equivalence relations to define division.

Definition (Equivalence Relation)

(Reflexive) For every $A\in S$, we know that $A\cong A$ (so the ordered pair $(A,A)$ is always in $\mathscr{C}$).
(Symmetric) If $A\cong B$, then $B\cong A.$
(Transitive) If $A\cong B$ and $B\cong C$, then $A\cong C$.

Problem 42 (The Rational Are Obtained From The Integers From An Equivalence Relation)

Prove that the relation above is an equivalence relation.
If we replace $\mathbb{Z}$ with any integral domain $D$, prove that we still obtain an equivalence relation on $S$. (If your work on part 1 didn't refer to the integers specifically, then this part should automatically follow.)

Problem 43 (The Field Of Quotients Of An Integral Domain)

Prove $+$ and $\cdot$ are binary operations on $F$. This will require you show that if $ [a/b] \cong [a'/b']$ and $ [c/d] \cong [c'/d']$ then we must have $ [(ad+bc)/(bd)] \cong [(a'd'+b'c')/(b'd')] $, and a similar fact for multiplication.
Prove that $(F,+,\cdot)$ is a field.
If we replace $\mathbb{Z}$ with any integral domain $D$, prove that $(F,+,\cdot)$ is a field.

Problem 34 (Why Are Zero Divisors Such A Problem)

Consider the polynomial $p(x)=x^2-5x+6 = (x-2)(x-3)$. We can view this as a polynomial in $\mathbb{Z}_n[x]$ for any given $n$.

Find all solutions to the equation $x^2-5x+6=0$ as a polynomial in $\mathbb{Z}_5[x]$.
Show that the equation $x^2-5x+6=0$ as a polynomial in $\mathbb{Z}_6[x]$ has 4 solutions. Then show that $x^2-5x+6$ can be factored as both $(x+4)(x+3)$ and $(x)(x+1)$. As a suggestion, since there are only 6 numbers in $\mathbb{Z}_6$, you can quickly determine which is a zero by just plugging it into the equation and then seeing if you get 0 mod 6.
Find all solutions to $x^2-5x+6=0$ as a polynomial in $\mathbb{Z}_8[x]$. Then factor this polynomial using only coefficients in $\mathbb{Z}_8$. Again, there are only 8 numbers to test to see if they are zeros. Just plug them all in. You might want to use Excel or write a computer program to make this fast.
Find all solutions to $x^2-5x+6=0$ as a polynomial in $\mathbb{Z}_{14}[x]$. Factor this polynomial using only coefficients in $\mathbb{Z}_{14}$. Then factor it in a different way.
Find all solutions to $x^2-5x+6=0$ as a polynomial in $\mathbb{Z}_{30}[x]$. (You should get more than 4 answers.) One way to factor the polynomial is of course $(x-2)(x-3)=(x+28)(x+27)$. Find one other way.

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