# Abstract Algebra/Splitting Fields and Algebraic Closures
## Splitting Fields

Let F be a field and p(x) be a nonconstant polynomial in F(x). We already know that we can find a field extension of F that contains a root of p(x). However, we would like to know whether an extension E of F containing all of the roots of p(x) exists. In other words, can we find a field extension of F such that p(x) factors into a product of linear polynomials? What is the "smallest" extension containing all the roots of p(x)?
Let F be a field and $p(x)=\alpha _{0}+\alpha _{1}x+\cdots +\alpha _{n}x^{n}$ be a nonconstant polynomial in F[x]. An extension field E of F is a splitting field of p(x) if there exist elements $\alpha _{1},\cdots ,\alpha _{n}$ in E such that $E=F(\alpha _{1},\cdots ,\alpha _{n})$ and
$p(x)=(x-\alpha _{1})(x-\alpha _{2})\cdots (x-\alpha _{n})$ in E[x].
A polynomial $p(x)\in F[x]$ splits in E if it is the product of linear factors in E[x].
Example 1: Let $p(x)=x^{4}+2x^{2}-8$ be in $\mathbb {Q} [x]$. Then p(x) has irreducible factors $x^{2}-2$ and $x^{2}+4$. Therefore, the field $\mathbb {Q} ({\sqrt {2}},i)$ is a splitting field for p(x).
Example 2: Let $p(x)=x^{3}-3$ be in $\mathbb {Q} [x]$. Then p(x) has a root in the field $\mathbb {Q} ({\sqrt[{3}]{3}})$. However, this field is not a splitting field for p(x) since the complex cube roots of 3, ${\frac {-{\sqrt[{3}]{3}}\pm ({\sqrt[{6}]{3}})^{5}i}{2}}$ are not in $\mathbb {Q} ({\sqrt[{3}]{3}})$.
Theorem Let p(x) $\in$ F(x) be a nonconstant polynomial. Then there exists a splitting field E for p(x).
Proof. We will use mathematical induction on the degree of p(x). If $deg(p(x))=1$, then p(x) is a linear polynomial and $E=F$. Assume that the theorem is true for all polynomials of degree k with $1\leq k<n$ and let $deg(p(x))=n$. We can assume that p(x) is irreducible; otherwise, by our induction hypothesis, we are done. There exists a field K such that p(x) has a zero $\alpha _{1}$ in K. Hence, $p(x)=(x-\alpha _{1})q(x)$, where $q(x)\in K(x)$. Since $deg(q(x))=n-1$, there exists a splitting field $E\supset K$ of q(x) that contains the zeros $\alpha _{2},\cdots ,\alpha _{n}$ of p(x) by our induction hypothesis. Consequently,
$E=K(\alpha _{2},\cdots ,\alpha _{n})=F(\alpha _{1},\cdots ,\alpha _{n})$
is a splitting field of p(x).
The question of uniqueness now arises for splitting fields. This question is answered in the affirmative. Given two splitting fields K and L of a polynomial $p(x)\in F(x)$, there exists a field isomorphism $\phi :K\to L$ that preserves F. In order to prove this result, we must first prove a lemma.
Lemma Theorem Let $\phi :E\to F$ be an isomorphism of fields. Let K be an extension field of E and $\alpha \in K$ be algebraic over E with minimal polynomial p(x). Suppose that L is an extension field of F such that $\beta$ is root of the polynomial in F[x] obtained from p(x) under the image of $\phi$. Then $\phi$ extends to a unique isomorphism $\psi :E(\alpha )\to F(\beta )$ such that $\psi (\alpha )=\beta$ and $\psi$ agrees with $\phi$ on E.
Lemma Proof. If p(x) has degree n, then we can write any element in $E(\alpha )$ as a linear combination of $1,\alpha ,\cdots ,\alpha ^{n-1}$. Therefore, the isomorphism that we are seeking must be
$\phi (a_{0}+a_{1}\alpha +\cdots +a_{n-1}\alpha ^{n-1})=\psi (a_{0})+\psi (a_{1})\beta +\cdots +\psi (a_{n-1})\beta ^{n-1}$,
where
$a_{0}+a_{1}\alpha +\cdots +a_{n-1}\alpha ^{n-1}$
is an element in $E(\alpha )$. The fact that $\phi$ is an isomorphism could be checked by direct computation; however, it is easier to observe that $\phi$ is a composition of maps that we already know to be isomorphisms.
We can extend $\psi$ to be an isomorphism from E[x] to F[x], which we will also denote by $\psi$, by letting
$\psi (a_{0}+a_{1}x+\cdots +a_{n}x^{n})=\psi (a_{1})x+\cdots +\psi (a_{n})x^{n}$.
This extension agrees with the original isomorphism $\psi :E\to F$, since constant polynomials get mapped to constant polynomials. By assumption, $\psi (p(x))=q(x)$; hence, $\psi$ maps $\left\langle p(x)\right\rangle$ onto $\left\langle q(x)\right\rangle$. Consequently, we have an isomorphism ${\overline {\psi }}:E[x]/\left\langle p(x)\right\rangle \to F[x]/\left\langle q(x)\right\rangle$. We have isomorphisms $\sigma :E[x]/\left\langle p(x)\right\rangle \to F(\alpha )$ and $\tau :F[x]/\left\langle q(x)\right\rangle \to F(\beta )$, defined by evaluation at $\alpha$ and $\beta$, respectively. Therefore, $\psi =\tau ^{-1}{\overline {\phi }}\sigma$ is the required isomorphism.
Now write $p(x)=(x-\alpha )f(x)$ and $q(x)=(x-\beta )g(x)$, where the degrees of f(x) and g(x) are less than the degrees of p(x) and q(x), respectively. The field extension K is a splitting field for f(x) over E(α), and L is a splitting field for g(x) over F(β). By our induction hypotheses there exists an isomorphism $\psi :K\to L$ such that $\psi$ agrees with ${\bar {\phi }}$ on E(α). Hence, there exists an isomorphism $\psi :K\to L$ such that $\psi$ agrees with $\psi$ on E.
Corollary Let p(x) be a polynomial in F[x]. Then there exists a splitting field K of p(x) that is unique up to isomorphism.

## Algebraic Closures

Given a field F, the question arises as to whether or not we can find a field E such that every polynomial p(x) has a root in E. This leads us to the following theorem.
Theorem 21.11 Let E be an extension field of F. The set of elements in E that are algebraic over F form a field.
Proof. Let $\alpha ,\beta \in E$ be algebraic over F. Then $F(\alpha ,\beta )$ is a finite extension of F. Since every element of $F(\alpha ,\beta )$ is algebraic over $F,\alpha \pm \beta ,\alpha \beta$, and $\alpha /\beta {\text{ }}(\beta \neq 0)$ are all algebraic over F. Consequently, the set of elements in E that are algebraic over F forms a field.
Corollary 21.12 The set of all algebraic numbers forms a field; that is, the set of all complex numbers that are algebraic over $\mathbb {Q}$ makes up a field.
Let E be a field extension of a field F. We define the algebraic closure of a field F in E to be the field consisting of all elements in E that are algebraic over F. A field F is algebraically closed if every nonconstant polynomial in F[x] has a root in F.
Theorem 21.13 A field F is algebraically closed if and only if every nonconstant polynomial in F[x] factors into linear factors over F[x].
Proof. Let F be an algebraically closed field. If $p(x)\in F[x]$ is a nonconstant polynomial, then p(x) has a zero in F, say α. Therefore, $x-\alpha$ must be a factor of p(x) and so $p(x)=(x-\alpha )q_{1}(x)$, where $deg(q_{1}(x))=deg(p(x))-1$. Continue this process with $q_{1}(x)$ to find a factorization
$p(x)=(x-\alpha )(x-\beta )q_{2}(x)$
where $deg(q_{2}(x))=deg(p(x))-2$. The process must eventually stop since the degree of p(x) is finite.
Conversely, suppose that every nonconstant polynomial p(x) in F[x] factors into linear factors. Let $ax-b$ be such a factor. Then $p(b/a)=0$. Consequently, F is algebraically closed.
Corollary 21.14 An algebraically closed field F has no proper algebraic extension E.
Proof. Let E be an algebraic extension of F; then $F\subset E$. For $\alpha \in E$, the minimal polynomial of α is $x-\alpha$. Therefore, $\alpha \in F$ and $F=E$.
Theorem 21.15 Every field F has a unique algebraic closure.
It is a nontrivial fact that every field has a unique algebraic closure. The proof is not extremely difficult, but requires some rather sophisticated set theory. We refer the reader to [3], [4], or [8] for a proof of this result.
We now state the Fundamental Theorem of Algebra, first proven by Gauss at the age of 22 in his doctoral thesis. This theorem states that every polynomial with coefficients in the complex numbers has a root in the complex numbers. The proof of this theorem will be given in Abstract Algebra/Galois Theory.
Theorem 21.16 (Fundamental Theorem of Algebra) The field of complex numbers is algebraically closed.