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March 5

Problem 64 (Ascending Chains of Ideals In a PID Are Finite)

Suppose that $R$ is a ring, and that $A_1\subseteq A_2\subseteq A_3\subseteq \cdots$ is nested sequence of ideals of $R$.

1. Prove that the union $\ds \bigcup_{n=1}^{\infty} A_n$ is an ideal of $R$.
2. If $R$ is a principle ideal domain, prove that there exists an integer $k$ so that $A_k=A_n$ for every $n\geq k$. In other words, prove that every ascending chain of ideals in a principle ideal domain contain finitely many different ideals.
3. Challenge: Give an example of a ring $R$ and an infinite ascending chain of ideals so that no two ideals are equal.

You can present in class even if you don't get the third part.

Problem 65 (Every Element is a Product of Irreducibles in a PID)

Suppose that $D$ is a principle ideal domain with $a\in D$ where $a\neq 0$ is not a unit. Prove that $a$ can be written as the product of irreducibles.

Problem 66 (Every Polynomial Has A Root In Some Extension Field)

Let $F$ be a field and $f(x)\in F[x]$ be nonzero and not a unit. Prove that there exists an extension field $E$ in which $f(x)$ has a zero.

Definition.The Derivative Of A Polynomial Over a Ring

Let $R$ be a ring and $f(x)\in R[x]$. If $f(x)=0$, then we define the derivative of $f(x)$ to be zero. Otherwise, we define the derivative of $f(x) = a_nx^n+\cdots +a_1x+a_0$ where $a_n\neq 0$ to be $$f'(x) = n\cdot a_nx^{n-1}+(n-1)a_{n-1}x^{n-2}+\cdots + 2a_2 x + a_1.$$

Problem 67 (The Sum and Product Rule for Derivatives)

Suppose that $F$ is a field and that $f(x),g(x)\in F[x]$.

1. Prove the sum rule, namely that $(f+g)'(x)=f'(x)+g'(x)$.
2. Prove the product rule, namely that $(fg)'(x)=f'(x)g(x)+f(x)g'(x)$.
3. What properties of the field $F$ were needed to complete your proofs? State a stronger theorem that requires less assumptions.

Problem 68 (Multiple Zeros and The Derivative Proof)

Suppose that $F$ is field and $f(x)\in F[x]$. Prove that $f(x)$ has a zero of multiplicity greater than one in some extension field $E$ if and only if $f$ and $f'$ have a common factor in $F[x]$.

Defintion.Field extensions generated by subsets and Simple Extensions

Suppose that $E$ is a field extension of $F$ and $S\subset E$.

1. We'll use the notation $F(S)$ to represent the smallest subfield of $E$ that contains both $F$ and $S$.
2. If $S=\{a\}$ then we call $F(a)$ a simple extension.

Problem 69 (Showing The Existence of a Field Extension Generated by a subset)

Suppose that $E$ is an extension field of the field $F$. Let $S$ be a subset of $E$.

1. Prove that there is a subfield of $E$ that contains both $F$ and $S$.
2. Prove that the intersection of all subfields of $E$ that contain both $F$ and $S$ is a subfield of $E$.

Problem 70 (Factoring By An Irreducible Gives A Simple Extension)

Let $F$ be a field and $p(x)\in F[x]$ be irreducible over $F$. If $a$ is a zero of $p(x)$ in some extension $E$ of $F$, then $F(a)$ is isomorphic to $F[x]/\left<p(x)\right>$. Furthermore, if $\deg p(x) = n$, then every member of $F(a)$ can be uniquely expressed in the form $c_{n-1}a^{n-1} + c_{n-2}a^{n-2} + \cdots + c_{1}a + c_{0}$ where $c_i\in F$. (In other words, the set $\left\{1, a, a^2, \ldots, a^{n-1}\right\}$ is a basis for $F(a)$ over $F$).

Problem 71 (Simple Extensions For The Same Polynomial Are Isomorphic)

Let $F$ be a field and let $p(x)\in F[x]$ be irreducible over $F$. If $a$ is a zero of $p(x)$ in some extension $E$ of $F$ and $b$ is a zero of $p(x)$ in some extension $E'$ of $F$, then the fields $F(a)$ and $F(b)$ are isomorphic.

March 7

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In our proof that every nonzero, nonunit, in a PID can be written as a product of irreducibles, there was a part of our proof that all of you took for granted. The following problem has you complete the details.

Problem 72 (Ideals Generated By A Reducible Are Properly Contained By Non Unit Factors)

Prove that if $d$ is reducible and $d=ab$ where $a$ and $b$ are non units, then $\left<d\right>$ is a proper subset of $\left<a\right>$.

Problem 74 (Existence Of A Splitting Field)

Let $F$ be a field and $f(x)\in F[x]$ have positive degree. We have already shown that there exists an extension field $E$ of $F$ in which $f(x)$ has a zero $a$, in other words we can write $f(x) = (x-a)g(x)$ where $a,g(x)\in E[x]$ (by the Factor Theorem). Prove that there exists an extension field $E'$ of $F$ in which $f(x)$ can be written as a product of linear factors, i.e. we can write $f(x) = (x-a_1)(x-a_2)\cdots (x-a_n)g$ where $a_1,a_2,\ldots, a_n,g\in E'$ and also $g$ is a unit.

Definition.Splitting Field of $f(x)$ over $F$

Look this on up in the text

Problem 73 (Unique Factorization In A PID)

Suppose that $D$ is a principal ideal domain. Suppose that $f=p_1p_2\cdots p_m = q_1q_2\cdots q_n,$ where each $p_i$ and $q_i$ is irreducible over $D$. Prove that $n=m$ and after rearranging, for each $i$ we have $p_i=u_iq_i$ for some unit $i$. In other words, prove that $D$ uniquely factors as a product of irreducibles.

We have now shown that in every principle ideal domain, every nonzero nonunit can be written as a product of irreducibles (we'll call this a factorization), and that any two factorization are essentially the same (after rearrangement and multiplying by units). Any integral domain that satisfies this property is called a unique factorization domain.

Definition (Unique Factorization Domain)

Let $D$ be an integral domain. We say that $D$ is a unique factorization domain if the following two properties hold.

1. If $d\in D$ is not a unit, and not zero, then we can write $d$ as a product of irreducibles over $D$.
2. If we have written $d$ as a product of irreducibles over $D$ in two ways, say $d=p_1p_2\cdots p_n$ and $d=q_1q_2\cdots q_m$, then $n=m$ and after rearranging we have for each $i$ that $p_i=u_iq_i$ for some unit $u_i$.

The next problem should follow immediately from a fact that we have already shown.

Problem 75 (Polynomial Rings Over A Field Are Unique Factorization Domains)

Prove that $F[x]$ is a unique factorization domain if $F$ is a field.

What was the key that made this all work? What is the key to proving that $F[x]$ is a principle ideal domain? The key is the Division Algorithm, which is also called the Euclidean Algorithm. If an integral domain has something similar to the division algorithm, then we'll call it a Euclidean domain.

Definition.Euclidean Domain

Please look this one up the book.

Take a look at the proof we used to show that $F[x]$ is a PID whenever $F$ is a field. The key to this proof was the division algorithm. You should be able to generalize this proof to show that any Euclidean domain is a principle ideal domain.

Problem 76 (Euclidean Domains Are PIDs)

Prove that a Euclidean domain is a principle ideal domain. Prove also that a Euclidean domain is a unique factorization domain.

Problem 77 (Both Z And FX Are Euclidean Domains)

Prove that $Z$ is a Euclidean domain with function $d(a)=|a|$. Prove that $F[x]$ is a Euclidean domain for a field $F$ with $d(f(x))=\deg f(x)$.

Problem 78 (The Gaussian Integers Is A Euclidean Domain)

Prove that $Z[i]$ is a Euclidean domain, using the function $d(a+bi)=a^2+b^2$.

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March 19

Problem 79 (In A Euclidean Domain All Units Have The Same Measure)

Suppose $D$ is a Euclidean domain with measure $d$. Let $u$ be a unit. Prove that that $d(u)=d(1)$, in other words prove that all units must have the exact same measure.

Problem 80 (In A Euclidean Domain Associates Have The Same Measure)

Suppose $D$ is a Euclidean domain with measure $d$. Suppose that $a$ and $b$ are nonzero associates, in other words assume that $a=ub$ for some unit $u$. Prove that $d(a)=d(b)$.

Problem 81 (Describing the Extension Field $\mathbb{Q}(\pi)$ of $\mathbb{Q}$)

Consider the field $\mathbb{Q}$ and the element $\pi$. It is known that the element $\pi$ is not the zero of any polynomial in $\mathbb{Q}[x]$. However, we also know that $\pi\in \mathbb{R}$, which is an extension field of $\mathbb{Q}$. This means we can talk about $\mathbb{Q}(\pi)$, the smallest subfield of $\mathbb{R}$ that contains both $\mathbb{Q}$ and $\pi$. Describe the elements of $\mathbb{Q}$ in a constructive way. In other words, give a way to obtain all elements of $\mathbb{Q}(\pi)$ by giving a way to describe any element of this field.

The derivative of a polynomial can help us understand the next two problems.

Problem 82 (Irreducibles Over A Field Of Characteristic Zero Have No Repeated Zeros)

Let $F$ be a field of characteristic zero and $f(x)\in F[x]$ be irreducible over $F$. Suppose that $E$ is a splitting field for $f(x)$, so that we can write $f(x) = c(x-a_1)(x-a_2)\cdots(x-a_n)$ for $c\in F$ and $a_1, \ldots, a_n\in E$. Show that $a_i\neq a_j$ if $i\neq j$, in other words show that $f$ has no zero of multiplicity greater than one.

Problem 83 (Irreducibles Over A Field Of Characteristic $p$ May Have Repeated Zeros)

Let $F$ be a field of characteristic $p$ and $f(x)\in F[x]$ be irreducible over $F$. Suppose that $E$ is a splitting field for $f(x)$, so that we can write $f(x) = c(x-a_1)(x-a_2)\cdots(x-a_n)$ for $c\in F$ and $a_1, \ldots, a_n\in E$.

1. Show that $f(x)$ might have a zero with multiplicity greater than one (give an example of such a polynomial).
2. If $f(x)$ has a multiple zero, then what can we say about the coefficients of $f(x)$?

Definition.

Problem.

Problem.

March 21

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March 26

We discussed what problems to work on in class on Monday. Sorry I did not have time to type them up here. It's been a busy week and stats has been consuming all my time. I have typed up quite a few problems on Friday's list, that we will be focusing on for the next week.

March 28

Problem

Prove that the dimension of a vector space is well defined. In other words, prove that if $A =\{a_1,\ldots,a_n\}$ and $B=\{b_1,\ldots,b_m\}$ are two bases for a vector space $V$ over $F$, then we must have $m=n$.

Problem

Let $E$ be an extension field of $F$. Let $a,b\in E$. Prove that $F(a)(b)=F(a,b)$.

Definition.Degree Of A Field Extension

We defined $ [E:F] $ to be the dimension of $E$ as a vector space over $F$. When we see $ [E:F]=n $, we'll say, "The degree of $E$ over $F$ is $n$." We say that $E$ is a finite field extension of $F$ if $ [E:F] $ is finite.

Problem

Suppose that $K$ is a finite field extension of the field $E$, and $E$ is a finite field extension of the field $F$. Prove that $ [K:F] = [K:E] [E:F] $.

Definition.Splitting Field

Let $E$ be an extension field of $F$ and $f(x)\in F[x]$. We say that $F[x]$ splits in $E$ if $f(x)$ can be factored as a product of linear factors in $E[x]$. A splitting field for $f(x)$ over $F$ is a field extension $E$ of $F$ in which $f(x)$ splits in $E$ but $f(x)$ does not split in any other proper subfield of $E$.

Problem

Suppose $f(x)\in F[x]$ where $F$ is a field. Let $E$ and $E'$ be two splitting fields for $f(x)$ over $F$. Prove that $E$ and $E'$ are isomorphic.

Defintion.Perfect Field

We say that $F$ is a perfect field if every irreducible polynomial $p(x)$ over $F$ has no zeros of multiplicity greater than one.

Problem

Show that a field $F$ is perfect if and only if either (1) we know $F$ has characteristic zero or (2) we know $F$ has characteristic $p$ and $F=F^p$, where $F^p=\{a^p\mid a\in F\}$. In other words, a field is perfect if an only if it has characteristic zero or every element has a $p$th root where $p$ is the characteristic of $F$.

Problem.Finite Fields Are Perfect

Suppose that $F$ is a finite field with characteristic $p$. Prove that $F$ is a perfect field.

Problem.

Let $p(x)$ be irreducible over a field $F$, and let $E$ be a splitting field for $p(x)$. Suppose that $a$ is zero of $p(x)$ of multiplicity $k$. Prove that if $b$ is a zero of $p(x)$, then $b$ must have multiplicity $k$ as well. In other words, we can write $$p(x) = (x-a_1)^k(x-a_2)^k\cdots (x-a_j)^k$$ where $a_1,\ldots,a_j$ are the zeros of $p(x)$ and $k$ is the common multiplicity.

Problem.A Polynomial Ring over a UFD is a UFD

If $D$ is a unique factorization domain, prove that $D[x]$ is a unique factorization domain.

Definition.Algebraic and Transcendental Extensions

Let $E$ be an extension field of the field $F$. Let $a\in E$.

• We say that $a$ is algebraic over $F$ if $a$ is the zero of some nonzero polynomial in $F[x]$. Otherwise we say $a$ is transcendental over $F$.
• An extension $E$ is called an algebraic extension of $F$ if every element of $E$ is algebraic over $F$. Otherwise we say $E$ is a transcendental extension of $F$.

Problem

Suppose that $E$ is a field extension of $F$, and let $a\in E$.

1. If $a$ is algebraic over $F$, prove that $F(a)\approx F[x]/\left<p(x)\right>$ where $p(x)$ is a polynomial in $F[x]$ of minimal degree such that $p(a)=0$.
2. If $a$ is transcendental over $F$, prove that $F(a)\approx F(x)$, where $F(x)$ is the field of fraction of the integral domain $F[x]$.

You may assume in your work that the map $\phi:F(x)\to F(a)$ defined by $\phi(f(x))=f(a)$ is a ring homomorphism.

Problem

Suppose that $a$ is algebraic over a field $F$. Prove that there exists a unique monic polynomial $p(x)\in F[x]$ of minimal degree such that $p(a)=0$, and prove that $p(x)$ is irreducible over $F$.

We call the polynomial $p(x)$ above the minimal polynomial for $a$ over $F$. The degree of an algebraic element $a$ over $F$ is the degree of the minimal polynomial for $a$ over $F$.

Problem

Suppose that $a$ is algebraic over $F$ and the minimal polynomial for $a$ over $F$ is $p(x)$. Suppose that $f(x)\in F[x]$ has the property that $f(a)=0$. Prove that $f(x)=p(x)q(x)$ for some $q(x)\in F[x]$, in other words prove that the minimal polynomial for $a$ over $F$ divides every polynomial in $F[x]$ that has $a$ as a zero.

Problem.Finite Extensions Are Algebraic

Suppose that $E$ is a field extension of $F$ and that $ [E:F]=n$ is finite. Prove that $E$ is an algebraic extension of $F$.

Problem.Not Every Algebraic Extension is Finite

Construct an example of an algebraic extension $E$ over a field $F$ that is not a finite extension, and prove your claims. See exercise 3 on page 378 if you need a jump start (it gives you an example and lets you prove the claims).

Problem

Show that $\mathbb{Q}(\sqrt2,\sqrt3) = \mathbb{Q}(\sqrt2+\sqrt3)$. In particular, explain how to obtain the minimal polynomial for $(\sqrt2+\sqrt3)$ over $\mathbb{Q}$.

Problem.Every Finite Extension Is A Simple Extension When The Characteristic Is Zero

Suppose that $F$ a field of characteristic 0, and let $a$ and $b$ be algebraic over $F$. Prove that there exists $c\in F(a,b)$ such that $F(a,b)=F(c)$.

Problem

Suppose that $ [E:\mathbb{Q}]=2$. Prove that there exists an integer $d$ such that $\mathbb{Q}(\sqrt{d})=E$ and $d$ is not divisible by the square of any prime.

Problem

Find the degree and a basis for $E=Q(\sqrt 2, \sqrt[3]{2},\sqrt[4]{2})$ over $\mathbb{Q}$. Then find an element $c\in E$ so that $Q(c) = E$. What is the minimal polynomial for the $c$ you chose?

Problem.An Algebraic Extension of An Algebraic Extension Is Algebraic

Suppose that $K$ is an algebraic extension of $E$, and suppose that $E$ is an algebraic extension of $F$. Prove that $K$ is an algebraic extension of $F$.

Problem.

Suppose that $K$ is an extension field of $F$. Let $E$ be the set of all elements of $K$ that are algebraic over $F$ which means we must have $F\subseteq E\subseteq K$. Prove that $E$ is a field. In other words, we are showing that the set of elements that are algebraic over $F$ is a field extension of $F$.

Definition.Algebraically Closed

We say a field $F$ is algebraically if the field has no proper algebraic extensions.

Problem

Prove that a field is algebraically closed if and only if every polynomial $f(x)\in F[x]$ splits in $F$. In other words, a field is algebraically closed if and only if it contains the zeros of every polynomial in $F[x]$.

Problem

Prove that $\mathbb{C}$ is algebraically closed.

Problem.Existence Of A Field Of Order $p^n$

Pick a prime $p$ and a positive integer $n$ and consider the polynomial $f(x)= x^{p^n}-x\in \mathbb{Z}_p[x]$. Prove that the splitting field for $f(x)$ over $\mathbb{Z}_p$ is precisely the set of zeros of $f(x)$, which means $E$ has $p^n$ elements. Here are some hints.

• Show that $f(x)$ has no multiple zeros by looking at $f'$.
• Show that the set of zeros of $f(x)$ is closed under addition, subtraction, multiplication, and division by nonzero elements.

Problem.Uniqueness Of A Field Of Order $p^n$

Suppose that $K$ is a field of order $p^n$. Prove that $K$ is isomorphic to the splitting field for $f(x) = x^{p^n}-x$ over $\mathbb{Z}_p$.

We have now shown that any finite field must have order $p^n$ and must be isomorphic to the splitting field for $f(x) = x^{p^n}-x$ over $\mathbb{Z}_p$. This field is called the Galois Field of order $p^n$ and written $GF(p^n)$.

Problem.The Multiplicative Group Of Finite Field Is Cyclic

Suppose that $F = GF(p^n)$. Prove that the multiplicative group $F^*$ is cyclic.

Problem.A Finite Extension Of A Finite Field Is Simple

Suppose $F$ is a finite field and suppose that $E$ is a finite extension of $F$. Prove that $E=F(a)$ for some $a\in E$.

If $F$ is a perfect field, then is it true that any finite extension is always a simple extension? We know this is true if $F$ has characteristic zero, or if $F$ is finite with characteristic $p$. What if $F$ is an infinite perfect field with characteristic $p$? If you are interested in reading more on this topic, do a google search on separable extensions.

Problem

Let $F$ be a field. Prove that there exists an algebraic extension $E$ of $F$ that is algebraically closed.

March 31

For more problems, see AllProblems