## Section9.1Splitting Fields

Let $F$ be a field and $p(x)$ be a nonconstant polynomial in $F[x]\text{.}$ We already know that we can find a field extension of $F$ that contains a root of $p(x)\text{.}$ 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 $p(x)\text{?}$

Let $F$ be a field and $p(x) = a_0 + a_1 x + \cdots + a_n x^n$ be a nonconstant polynomial in $F[x]\text{.}$ An extension field $E$ of $F$ is a splitting field of $p(x)$ if there exist elements $\alpha_1, \ldots, \alpha_n$ in $E$ such that $E = F( \alpha_1, \ldots, \alpha_n )$ and

\begin{equation*} p(x) = ( x - \alpha_1 )(x - \alpha_2) \cdots (x - \alpha_n)\text{.} \end{equation*}

A polynomial $p(x) \in F[x]$ splits in $E$ if it is the product of linear factors in $E[x]\text{.}$

###### Example9.1.

Let $p(x) = x^4 + 2x^2 - 8$ be in ${\mathbb Q}[x]\text{.}$ Then $p(x)$ has irreducible factors $x^2 -2$ and $x^2 + 4\text{.}$ Therefore, the field ${\mathbb Q}( \sqrt{2}, i )$ is a splitting field for $p(x)\text{.}$

###### Example9.2.

Let $p(x) = x^3 - 3$ be in ${\mathbb Q}[x]\text{.}$ Then $p(x)$ has a root in the field ${\mathbb Q}( \sqrt[3]{3}\, )\text{.}$ However, this field is not a splitting field for $p(x)$ since the complex cube roots of 3,

\begin{equation*} \frac{ -\sqrt[3]{3} \pm (\sqrt[6]{3}\, )^5 i }{2}\text{,} \end{equation*}

are not in ${\mathbb Q}( \sqrt[3]{3}\, )\text{.}$

###### Proof.

We will use mathematical induction on the degree of $p(x)\text{.}$ If $\deg p(x) = 1\text{,}$ then $p(x)$ is a linear polynomial and $E = F\text{.}$ Assume that the theorem is true for all polynomials of degree $k$ with $1 \leq k \lt n$ and let $\deg p(x) = n\text{.}$ We can assume that $p(x)$ is irreducible; otherwise, by our induction hypothesis, we are done. By Theorem 8.10, there exists a field $K$ such that $p(x)$ has a zero $\alpha_1$ in $K\text{.}$ Hence, $p(x) = (x - \alpha_1)q(x)\text{,}$ where $q(x) \in K[x]\text{.}$ Since $\deg q(x) = n -1\text{,}$ there exists a splitting field $E \supset K$ of $q(x)$ that contains the zeros $\alpha_2, \ldots, \alpha_n$ of $p(x)$ by our induction hypothesis. Consequently,

\begin{equation*} E = K(\alpha_2, \ldots, \alpha_n) = F(\alpha_1, \ldots, \alpha_n) \end{equation*}

is a splitting field of $p(x)\text{.}$

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 $p(x) \in F[x]\text{,}$ there exists a field isomorphism $\phi : K \rightarrow L$ that preserves $F\text{.}$ In order to prove this result, we must first prove a lemma.

###### Proof.

If $p(x)$ has degree $n\text{,}$ then by Theorem 8.18 we can write any element in $E( \alpha )$ as a linear combination of $1, \alpha, \ldots, \alpha^{n - 1}\text{.}$ Therefore, the isomorphism that we are seeking must be

\begin{equation*} \overline{\phi}( a_0 + a_1 \alpha + \cdots + a_{n - 1} \alpha^{n - 1}) = \phi(a_0) + \phi(a_1) \beta + \cdots + \phi(a_{n - 1}) \beta^{n - 1}\text{,} \end{equation*}

where

\begin{equation*} a_0 + a_1 \alpha + \cdots + a_{n - 1} \alpha^{n - 1} \end{equation*}

is an element in $E(\alpha)\text{.}$ The fact that $\overline{\phi}$ is an isomorphism could be checked by direct computation; however, it is easier to observe that $\overline{\phi}$ is a composition of maps that we already know to be isomorphisms.

We can extend $\phi$ to be an isomorphism from $E[x]$ to $F[x]\text{,}$ which we will also denote by $\phi\text{,}$ by letting

\begin{equation*} \phi( a_0 + a_1 x + \cdots + a_n x^n ) = \phi( a_0 ) + \phi(a_1) x + \cdots + \phi(a_n) x^n\text{.} \end{equation*}

This extension agrees with the original isomorphism $\phi : E \rightarrow F\text{,}$ since constant polynomials get mapped to constant polynomials. By assumption, $\phi(p(x)) = q(x)\text{;}$ hence, $\phi$ maps $\langle p(x) \rangle$ onto $\langle q(x) \rangle\text{.}$ Consequently, we have an isomorphism $\psi : E[x] / \langle p(x) \rangle \rightarrow F[x]/\langle q(x) \rangle\text{.}$ By Proposition 8.17, we have isomorphisms $\sigma: E[x]/\langle p(x) \rangle \rightarrow E(\alpha)$ and $\tau : F[x]/\langle q(x) \rangle \rightarrow F( \beta )\text{,}$ defined by evaluation at $\alpha$ and $\beta\text{,}$ respectively. Therefore, $\overline{\phi} = \tau \psi \sigma^{-1}$ is the required isomorphism.

We leave the proof of uniqueness as a exercise.

###### Proof.

We will use mathematical induction on the degree of $p(x)\text{.}$ We can assume that $p(x)$ is irreducible over $E\text{.}$ Therefore, $q(x)$ is also irreducible over $F\text{.}$ If $\deg p(x) = 1\text{,}$ then by the definition of a splitting field, $K = E$ and $L = F$ and there is nothing to prove.

Assume that the theorem holds for all polynomials of degree less than $n\text{.}$ Since $K$ is a splitting field of $p(x)\text{,}$ all of the roots of $p(x)$ are in $K\text{.}$ Choose one of these roots, say $\alpha\text{,}$ such that $E \subset E( \alpha ) \subset K\text{.}$ Similarly, we can find a root $\beta$ of $q(x)$ in $L$ such that $F \subset F( \beta) \subset L\text{.}$ By Lemma 9.4, there exists an isomorphism $\overline{\phi} : E(\alpha ) \rightarrow F( \beta)$ such that $\overline{\phi}( \alpha ) = \beta$ and $\overline{\phi}$ agrees with $\phi$ on $E\text{.}$

Now write $p(x) = (x - \alpha ) f(x)$ and $q(x) = ( x - \beta) g(x)\text{,}$ where the degrees of $f(x)$ and $g(x)$ are less than the degrees of $p(x)$ and $q(x)\text{,}$ respectively. The field extension $K$ is a splitting field for $f(x)$ over $E( \alpha)\text{,}$ and $L$ is a splitting field for $g(x)$ over $F( \beta )\text{.}$ By our induction hypothesis there exists an isomorphism $\psi : K \rightarrow L$ such that $\psi$ agrees with $\overline{\phi}$ on $E( \alpha)\text{.}$ Hence, there exists an isomorphism $\psi : K \rightarrow L$ such that $\psi$ agrees with $\phi$ on $E\text{.}$

### Subsection9.1.1Separable Extensions

Many of the results that we will soon prove depend on the fact that a polynomial $f(x)$ in $F[x]$ has no repeated roots in its splitting field. Thus we need to know exactly when a polynomial factors into distinct linear factors in its splitting field. Let $E$ be the splitting field of a polynomial $f(x)$ in $F[x]\text{.}$ Suppose that $f(x)$ factors over $E$ as

\begin{equation*} f(x) = (x - \alpha_1)^{n_1} (x - \alpha_2)^{n_2} \cdots (x - \alpha_r)^{n_r} = \prod_{i = 1}^{r} (x - \alpha_i)^{n_i}\text{.} \end{equation*}

We define the multiplicity of a root $\alpha_i$ of $f(x)$ to be $n_i\text{.}$ A root with multiplicity 1 is called a simple root. Recall that a polynomial $f(x) \in F[x]$ of degree $n$ is separable if it has $n$ distinct roots in its splitting field $E\text{.}$ Equivalently, $f(x)$ is separable if it factors into distinct linear factors over $E[x]\text{.}$ An extension $E$ of $F$ is a separable extension of $F$ if every element in $E$ is the root of a separable polynomial in $F[x]\text{.}$

###### Proof.

First assume that $\chr F = 0\text{.}$ Since $\deg f'(x) \lt \deg f(x)$ and $f(x)$ is irreducible, the only way $\gcd( f(x), f'(x)) \neq 1$ is if $f'(x)$ is the zero polynomial; however, this is impossible in a field of characteristic zero. If $\chr F = p\text{,}$ then $f'(x)$ can be the zero polynomial if every coefficient of $f'(x)$ is a multiple of $p\text{.}$ This can happen only if we have a polynomial of the form $f(x) = a_0 + a_1 x^p + a_2 x^{2p} + \cdots + a_n x^{np}\text{.}$

Certainly extensions of a field $F$ of the form $F(\alpha)$ are some of the easiest to study and understand. Given a field extension $E$ of $F\text{,}$ the obvious question to ask is when it is possible to find an element $\alpha \in E$ such that $E = F( \alpha )\text{.}$ In this case, $\alpha$ is called a primitive element. We already know that primitive elements exist for certain extensions. For example,

\begin{equation*} {\mathbb Q}( \sqrt{3}, \sqrt{5}\, ) = {\mathbb Q}( \sqrt{3} + \sqrt{5}\, ) \end{equation*}

and

\begin{equation*} {\mathbb Q}( \sqrt[3]{5}, \sqrt{5}\, i ) = {\mathbb Q}( \sqrt[6]{5}\, i )\text{.} \end{equation*}

The next theorem tells us that we can often find a primitive element.

###### Proof.

Suppose that $E$ is a finite extension of an infinite field (it turns out that there is no problem if $F$ is a finite field, although we will not consider this case now anyway). We will prove the result for $F(\alpha, \beta)\text{.}$ The general case easily follows when we use mathematical induction. Let $f(x)$ and $g(x)$ be the minimal polynomials of $\alpha$ and $\beta\text{,}$ respectively. Let $K$ be the field in which both $f(x)$ and $g(x)$ split. Suppose that $f(x)$ has zeros $\alpha = \alpha_1, \ldots, \alpha_n$ in $K$ and $g(x)$ has zeros $\beta = \beta_1, \ldots, \beta_m$ in $K\text{.}$ All of these zeros have multiplicity $1\text{,}$ since $E$ is separable over $F\text{.}$ Since $F$ is infinite, we can find an $a$ in $F$ such that

\begin{equation*} a \neq \frac{\alpha_i - \alpha}{\beta - \beta_j} \end{equation*}

for all $i$ and $j$ with $j \neq 1\text{.}$ Therefore, $a( \beta - \beta_j ) \neq \alpha_i - \alpha\text{.}$ Let $\gamma = \alpha + a \beta\text{.}$ Then

\begin{equation*} \gamma = \alpha + a \beta \neq \alpha_i + a \beta_j; \end{equation*}

hence, $\gamma - a \beta_j \neq \alpha_i$ for all $i, j$ with $j \neq 1\text{.}$ Define $h(x) \in F( \gamma )[x]$ by $h(x) = f( \gamma - ax)\text{.}$ Then $h( \beta ) = f( \alpha ) = 0\text{.}$ However, $h( \beta_j ) \neq 0$ for $j \neq 1\text{.}$ Hence, $h(x)$ and $g(x)$ have a single common factor in $F( \gamma )[x]\text{;}$ that is, the minimal polynomial of $\beta$ over $F( \gamma )$ must be linear, since $\beta$ is the only zero common to both $g(x)$ and $h(x)\text{.}$ So $\beta \in F( \gamma )$ and $\alpha = \gamma - a \beta$ is in $F( \gamma )\text{.}$ Hence, $F( \alpha, \beta ) = F( \gamma )\text{.}$

### Exercises9.1.2Collected Homework

###### 1.

Consider the field $\Q(\sqrt[4]{3}, i)\text{.}$ Note this is the same field as $\Q(\sqrt[4]{3}, \sqrt[4]{3}i)\text{.}$

###### (a)

Is this a splitting field for some polynomial in $\Q[x]\text{?}$ If so, what is that polynomial, and what is its degree?

Hint

Guess a polynomial for one of these roots and then try to factor it completely over the complex numbers to see if that polynomial splits in the field.

###### (b)

What is the $[Q(\sqrt[4]{3}, i):\Q]\text{?}$ Explain how you know.

###### (c)

It turns out that there are exactly 8 different fields between $\Q$ and $\Q(\sqrt[4]{3}, i)\text{.}$ Find at least four of these (more if you can) and draw a large tower diagram showing how these are related (with degrees marked appropriately).

Hint

The next part of this problem might be helpful (and you can assume it is true for this part).

###### (d)

Prove that the fields $\Q(\sqrt[4]{3})$ and $\Q(\sqrt[4]{3}i)$ are isomorphic, but not equal.

###### 2.

Bonus: The field $\Q(\sqrt[3]{7}, \sqrt{2}i)$ can also be written as $\Q(\gamma)$ for a primitive element $\gamma\text{.}$ Find such a $\gamma$ by working through the proof of Theorem 9.8 with this example. Explain how you found the number $a$ from the theorem.