## Splitting Fields (Normal Extensions)

### Definition

Definition: Let $f(x)\in F[x]$ be a polynomial and let $K/F$ be an extension field. $K$ is called a splitting field for $f(x)$ if

• $f$ splits into linear factors in $K$
• $f$ does not split into linear factors over any proper subfield of $K$.

### Splitting fields exist

Proposition: Any polynomial $f(x)\in F[x]$ has a splitting field.

Proof: If all irreducible factors of $f(x)$ have degree 1 then $F$ is a splitting field. Otherwise, let $\alpha$ be a root of an irreducible factor of $f$ of degree greater than $1$ and let $F_1=F(\alpha)$. Write $f(x)=(x-\alpha)f_1(x)$ and, by induction, let $E$ be a splitting field for $f_1(x)$ over $F(\alpha)$. Then all the roots of $f(x)$ belong to $E$. Let $K$ be the subfield of $E$ generated over $F$ by the roots of $f(x)$. This is your splitting field.

Remark: Some books say that if $K/F$ is the splitting field over $F$ for a polynomial, then $K$ is called a normal extension.

### Degrees of splitting fields

Proposition: If $f(x)\in F[x]$ has degree $n$ then its splitting field has degree at most $n!$.

Proof: It can be obtained by adjoining roots successively of polynomials of degree $n$, $n-1$, \ldots.

## Examples

1. $f(x)=(x^2-2)(x^2-3).$ Splitting field is $\Q(\sqrt{2},\sqrt{3})$ which has degree 4.
2. $f(x)=x^3-2$ which is irreducible by Eisenstein. Three roots are $\sqrt[3]{2},\omega\sqrt[3]{2},\omega^2\sqrt[3]{2}$ where $\omega=e^{2\pi i/3}$ is a cube root of one. Since $$\omega = \frac{-1+\sqrt{-3}}{2}$$ this field has degree six and contains $\sqrt{-3}$.
3. $x^4+4$ “looks irreducible” but it isn’t. It factors as $(x^2+2x+2)(x^2-2x+2)$. It splits over the field $\Q(i)$ because $(\pm 1 \pm i)^2=\pm 2i$ so $(\pm 1 \pm i)^4=-4.$
4. The splitting field of $x^n-1$ is called the $n^{th}$ cyclotomic field and is generated by $e^{2\pi a/n}$ where $a$ is an integer relatively prime to $n$. If $n$ is prime, then $x^p-1$ then it factors as $(x-1)(1+x+\cdots+x^{p-1})$; the second factor is irreducible so that field has degree $p-1$.
5. The splitting field of $x^p-2$ has degree $p(p-1)$.

## Uniqueness of splitting fields

### Extensions of isomorphisms

Theorem: (DF Theorem 27 p. 541) Let $\phi: F\to F’$ be a field isomorphism. Let $f(x)\in F[x]$ and let $f’(x)\in F’[x]$ be the polynomial obtained from $f$ by applying $\phi$ to its coefficients. Let $E/F$ be a splitting field of $f$ and let $E’/F’$ be a splitting field of $f’$. Then there is an isomorphism $\sigma:E\to E’$ which makes the following diagram commutative (the vertical arrows are the inclusion maps):

$\begin{xy}\xymatrix{ E \ar[r]^{\sigma} & E' \\ F\ar[u]\ar[r]^{\phi} & F'\ar[u] \\ } \end{xy}$

Corollary: Any two splitting fields for $f(x)$ are isomorphic via an isomorphism that is the identity on $F$.

### More on extensions

The extension theorem can seem a little mysterious. Let’s look more closely at an application.

Let $f(x)=x^3-2$ and let $E/\Q$ be its splitting field (which has degree 6 over $\Q$). Inside this field there are three isomorphic cubic extensions: $L_1=\Q(\sqrt[3]{2})$, $L_2=\Q(\omega\sqrt[3]{2})$, and $L_3=\Q(\omega^2\sqrt[3]{2})$ where $\omega=e^{2\pi i/3}$ is a cube root of unity.

$\begin{xy} \xymatrix{ & E & \\ L_1\ar[ur] & L_2\ar[u] & L_3\ar[ul] \\ & \Q\ar[ur]\ar[u]\ar[ul] &\\ } \end{xy}$

Now $E$ is a splitting field for $f(x)$ over each of $L_1$, $L_2$, and $L_3$.

### Still more on extensions

We can apply the theorem to (for example) the diagram

$\begin{xy}\xymatrix{ E \ar[r]^{\sigma} & E \\ \Q(\sqrt[3]{2})\ar[u]\ar[r]^{\phi} & \Q(\omega\sqrt[3]{2})\ar[u] \\ } \end{xy}$

where $\phi$ is the isomorphism that sends $\sqrt[3]{2}\to\omega\sqrt[3]{2}$ and fixes $\Q$. It follows that there is an automorphism $\sigma$ of the splitting field that extends $\phi$.

### Automorphisms of splitting fields of irreducibles

In general, if $f(x)$ is an irreducible polynomial over $F$, and $\alpha$ and $\beta$ are two roots of $f(x)$ in its splitting field $E/F$, then there is an automorphism $E\to E$ fixing $F$ sending $\alpha$ to $\beta$. In particular the automorphism group of $E$ fixing $F$ permutes the roots of $f(x)$ transitively.

### Proof of the extension theorem

The proof is by induction. If all roots of $f(x)$ belong to $F$, then all roots of $f’(x)$ belong to $F’$, and $E=F$ and $E’=F’$ so the identity map works. Now suppose we know the result for all $f$ of degree less than $n$ and suppose that $f$ is of degree $n$. Choose an irreducible factor $p(x)$ of $f(x)$ of degree at least $2$, and the corresponding factor $p’(x)$ of $f’(x)$. Since $F[x]/p(x)$ is isomorphic to $F’[x]/p’(x)$, we have an isomorphism $\phi’: F(\alpha)\to F’(\beta)$ that restricts to $\phi:F\to F’$.

Let $f(x)=(x-\alpha)f_1(x)$ and $f’(x)=(x-\beta)f’_1(x)$. Now $E$ (resp. $E’$) is a splitting field for $f_1$ (resp $f_1’$) and by induction we have an isomorphism $\sigma: E\to E’$ that restricts to $\phi’:F(\alpha)\to F’(\beta)$. This $\sigma$ also restricts to $\phi:F\to F’$ (since $\phi’$ does).

### Another property of splitting fields

Proposition: Let $K/F$ be the splitting field of a polynomial. Then if $g(x)\in F[x]$ is any irreducible polynomial over $F$, and $\alpha\in K$ is a root of $g(x)$, then all roots of $g(x)$ belong to $K$. (In other words, if $K/F$ is a splitting field for some polynomial, then any polynomial in $F[x]$ is either irreducible or splits into linear factors over $K$.)

Proof: Suppose that $K$ is the splitting field of $f(x)\in F[x]$. Suppose that $\alpha\in K$ and let $\beta$ be another root of $g(x)$ and consider the field $K(\beta)$. Then $K(\beta)$ is the splitting field of $f(x)$ over $F(\beta)$. ($K$ contains all the roots of $f(x)$, and it must contain $\beta$ if it contains $F(\beta)$. ) But then we have the diagram:

$\begin{xy} \xymatrix{ K\ar[r] & K(\beta) \\ F(\alpha)\ar[u]\ar[r] & F(\beta)\ar[u]\\ } \end{xy}$

The extension theorem tells us that there is an isomorphism from $K$ to $K(\beta)$ carrying $F(\alpha)$ to $F(\beta)$ and fixing the field $F$. Therefore $[K:F]=[K(\beta):F]$. But then $$[K(\beta):F]=[K(\beta):K][K:F].$$ This forces $[K(\beta):K]=1$ so $\beta\in K$.

## Algebraic Closures

### Algebraic closure

Definition: A field $F$ is algebraically closed if it has no nontrivial algebraic extensions; in other words, if every irreducible polynomial over $F$ has degree $1$.

Definition: If $F$ is a field, then $\overline{F}$ is an algebraic closure of $F$ if $\overline{F}/F$ is algebraic and every polynomial in $F[x]$ splits completely in $\overline{F}$.

So notice that the complex numbers are algebraically closed, but they are not an algebraic closure of $\Q$, because they contain transcendental elements.

### Algebraic closures are algebraically closed.

Lemma: If $\overline{F}$ is an algebraic closure of $F$, then $\overline{F}$ is algebraically closed.

This lemma says that if every polynomial with coefficients in $F$ has a root in $\overline{F}$, then every polynomial with coefficients in $\overline{F}$ has a root in $F$.

To prove this, let $f(x)\in\overline{F}[x]$. Let $F_1/F$ be the extension of $F$ generated by the coefficients of $f$. Since $F_1$ is generated by finitely many algebraic elements, $F_1/F$ is finite and a root $\alpha$ of $f(x)\in F_1[x]$ is finite over $F_1$. Therefore $f$ has a root in a finite extension of $F$, which is therefore in $\overline{F}$.

### Every field has an algebraic closure

Theorem: Given a field $F$, there exists an algebraically closed field containing $F$.

Proof: See Proposition 30 in DF on p. 544.

Theorem: If $K/F$ is algebraically closed, then the collection of elements of $K$ that are algebraic over $F$ is an algebraic closure of $F$.

Since $\C$ is algebraically closed, the set of algebraic numbers inside $\C$ is an algebraic closure of $\Q$. The construction of $\R$ and $\C$ is primarily by analysis, and the proof that $\C$ is algebraically closed is also analytic – at least, the usual proof.

## Separability

Separability is a phenomenon that is important when studying polynomials over fields of characteristic $p$.

Definition: A polynomial is separable if it has distinct roots, and inseparable if it has repeated roots.

Proposition: An irreducible polynomial over a field with characteristic $0$ is separable. It is inseparable over a field with characteristic $p$ if and only if its derivative is zero.

Proof: If $\alpha$ is a repeated root of a polynomial $f(x)$, then $f’(\alpha)=0$ where $f’$ is the “formal derivative” of $f$. Conversely, if $\alpha$ is a common root of $f(x)$ and $f’(x)$, then $\alpha$ is a multiple root of $f(x)$. This is because of the product rule; on the one hand:

$\frac{d}{dx}((x-a)^r g(x))=r(x-a)^{r-1}g(x)+(x-a)^r g(x)$

so if $a$ is a multiple root, then it is a root of $f’(x)$. On the other hand, if $a$ is a common root of $f(x)$ and $f’(x)$, write

$f(x)=(x-a)g(x)$

so

$f'(x)=(x-a)g'(x)+g(x).$

Since $f’(a)=0$, we have $g(a)=0$ so $g(x)$ is divisible by $(x-a)$.

Now if $f(x)$ is irreducible, then since $f’(x)$ has degree less than $f(x)$, if it is nonzero it is relatively prime to $f(x)$. In characteristic $0$, it is automatically nonzero. In characteristic $p$, it could be zero. For example the derivative of $x^p-a$ is zero.

Notice that if a polynomial has derivative zero (over a field of characteristic $p$) it must be a polynomial in $x^p$. From this one can see that any irreducible polynomial $f(x)$ over a field with characteristic $p$ is of the form $f_0(x^{p^k})$ for some power of $p$, and $f_0(x)$ is a separable polynomial.

## The Frobenius map

If $F$ is a field of characteristic $p$, then the map $\phi: F\to F$ given by $\phi(x)=x^p$ is a field endomorphism called the Frobenius map or the Frobenius endomorphism.

If the Frobenius map is surjective, then evey irreducible polynomial over $F$ is separable. Such a field is called perfect.