20140328

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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.

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