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Here's who we'll have present today.

  1. Carla (second)
  2. Nick (last)
  3. Caley (third)
  4. Kimberly (fourth)
  5. Sam (fifth)
  6. Darrel (first)

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)

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

  1. (Reflexive) For every $A\in S$, we know that $A\cong A$ (so the ordered pair $(A,A)$ is always in $\mathscr{C}$).
  2. (Symmetric) If $A\cong B$, then $B\cong A.$
  3. (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 $(a,b)\cong(c,d)$.

  1. Prove that the relation above is an equivalence relation.
  2. 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 $$ [a/b]+[c/d]=[(ad+bc)/(bd)]\quad \text{and}\quad [a/b]\cdot[c/d]=[ac/bd]. $$

  1. 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.
  2. Prove that $(F,+,\cdot)$ is a field.
  3. 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 $n$.

  1. Find all solutions to the equation $x^2-5x+6=0$ as a polynomial in $\mathbb{Z}_5[x]$.
  2. 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.
  3. 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.
  4. 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.
  5. 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.

For more problems, see AllProblems