LEC ALG3 20

Splitting fields

Definition 342.1(Splitting field).

Let $k$ be a field and $f(x)\in k[x]$. A field extension $K/k$ such that $f$ splits into linear factors in $K$, and such that $K=k({\alpha }_1,\dots ,{\alpha }_n)$ is generated by all the roots of $f$, is called a splitting field of $f$.

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Theorem 342.2(Theorem).

Let $k$ be a field. Then any $f(x)\in k[x]$ of positive degree has a splitting field.

Proof.

Here’s the easy way out: Let $k^{\text{a}}$ be the algebraic closure of $k$. $f(x)$ splits completely in $k^{\text{a}}$. Let ${\alpha }_1,\dots ,{\alpha }_n$ be the roots of $f$ in $k^{\text{a}}$. Then, $k({\alpha }_1,\dots ,{\alpha }_n)$ is a splitting field of $f$.

Here’s what Clare did: Let $f(x)$ have degree $d$. By Prp 338.5, there exists an extension $E_1/k$ in which $f(x)$ has a root. Suppose $f(x)=(x-{\alpha }_1)f_1(x)$ with $f_1(x)\in E_1[x]$. Let $E_2/E_1$ be an extension in which $f_1(x)$ has a root; suppose $f_1(x)=(x-{\alpha }_2)f_2(x)$ with $f_2(x)\in E_2[x]$. Keep going to obtain an extension $E_{d-1}/\dots /E_1/k$ in which $f$ splits completely.□

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Exercise 342.3(Exercise).

$f(x)=x^3-2\in \mathbb{Q}[x]$. Construct a splitting field abstractly, without appealing to $\mathbb{C}$.

$x^3-2$ is irreducible by Thm 118.2. Let $k_1:=\mathbb{Q}[x]/(x^3-2)$, and let $\overline{x}$ be the image of $x$ in $k_1$. We will identify $\mathbb{Q}$ with its embedding in $k_1$. As we know, $\overline{x}$ is a root of $y^3-2\in k_1[y]$. Use polynomial division to obtain $y^3-2=(y-\overline{x})(y^2+y\overline{x}+{\overline{x}}^2)$.

We now need to determine if $g(y):=y^2+y\overline{x}+{\overline{x}}^2$ is irreducible in $k_1[x]$. The discriminant of $g(y)$ is $\Delta ={\overline{x}}^2-4{\overline{x}}^2=-3{\overline{x}}^2$. A quadratic over a field is reducible iff its discriminant is a square in that field. Since $\overline{x}$ is nonzero, $-3{\overline{x}}^2$ is a square in $k_1$ iff $-3$ is a square in $k_1$. Thus, reducibility forces the existence of $\alpha \in k_1$ such that ${\alpha }^2=-3$. The minimal polynomial of $\alpha$ over $\mathbb{Q}$ is $x^2-3$, so the degree of the extension $\mathbb{Q}(\alpha )/\mathbb{Q}$ is $2$. We now have the tower $k_1/\mathbb{Q}(\alpha )/\mathbb{Q}$, and Prp 322.2 tells us that $[\mathbb{Q}(\alpha ):\mathbb{Q}]=2$ divides $[k_1:\mathbb{Q}]=3$ - impossible. Thus, $g(y)$ is irreducible in $k_1$.

Let $k_2=k_1[y]/(y^2+y\overline{x}+{\overline{x}}^2)$, and let $\overline{y}$ be the image of $y$ in $k_2$. We will identify $k_1$ with its embedding in $k_2$. Again, $\overline{y}$ is a root of $z^2+z\overline{x}+{\overline{x}}^2\in k_2[z]$. Perform polynomial long division again to obtain $z^2+z\overline{x}+{\overline{x}}^2=(z-\overline{y})(z+(\overline{x}+\overline{y}))$.

Thus, $k_2$ is a splitting field of $x^3-2\in \mathbb{Q}[x]$, and $z^3-2\in k_2[z]$ splits as

$$z^3-2=(z-\overline{x})(z-\overline{y})(z+(\overline{x}+\overline{y})).$$

$[k_2:\mathbb{Q}]=[k_2:k_1][k_1:\mathbb{Q}]=2\cdot 3=6$.

Theorem 342.4(Theorem).

Let $K$ be a splitting field of the polynomial $f(x)\in k[x]$. If $E$ is another splitting field of $f$, then there exists an isomorphism $\sigma :E\to K$ inducing the identity on $k$. If $k\subseteq K\subseteq k^{\text{a}}$, where $k^{\text{a}}$ is an algebraic closure of $k$, then any embedding of $E$ in $k^{\text{a}}$ inducing the identity on $k$ must be an isomorphism of $E$ onto $K$.

Proof.

Let $K^{\text{a}}$ be an algebraic closure of $K$. By Cor 119.20, $K^{\text{a}}$ is algebraic over $k$, hence is an algebraic closure of $k$. By Thm 338.13, there exists an embedding $\tau :E\to K^{\text{a}}$ inducing the identity on $k$. In $E[x]$, we have the factorization

$$f(x)=c(x-{\beta }_1)\dots (x-{\beta }_n),$$

where $c\in k$ and ${\beta }_i\in E$. Since $\tau {|}_k$ is the identity, we have¹

$$f(x)=f^{\tau }(x)=c(x-{\beta }_1^{\tau })\dots (x-{\beta }_n^{\tau }).$$

Since $f(x)$ already has a factorization

$$f(x)=c(x-{\alpha }_1)\dots (x-{\alpha }_n)$$

in $K[x]$, and since its factorization must be unique in $K^{\text{a}}$, we must have that $\tau$ is a bijection from $\{{\beta }_1,\dots ,{\beta }_n\}$ onto $\{{\alpha }_1,\dots ,{\alpha }_n\}$. Thus, we have (see Rmk 119.8)

$$K=k({\alpha }_1,\dots ,{\alpha }_n)=k({\beta }_{i_1}^{\tau },\dots ,{\beta }_{i_n}^{\tau })=\tau (k({\beta }_1,\dots ,{\beta }_n))=\tau E.$$ □

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Notes on gcd

Lemma 342.5(Lemma).

Let $f,g\in F[x]$ and $K/F$. Then the gcd of $f$ and $g$ in $F[x]$ is equal to their gcd in $K[x]$.

Proof.

Let$d$ and $d^{\prime }$ be the gcd of $f$ and $g$ in $F[x]$ and $K[x]$ respectively. We have to show that $d|d^{\prime }$ and $d^{\prime }|d$. The former is clear.

Since $F[x]$ is a PID, we can write $d=af+bg$ for some $a,b\in F[x]$. This equation holds true in $K[x]$. Since $d^{\prime }|f$ and $d^{\prime }|g$ in $K[x]$, $d\in (d^{\prime })$, which implies $d^{\prime }|d$.□

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Proposition 342.6(Proposition).

Let $K/F$ be a field extension. If $f,g\in F[x]$ have a common root in $K$, then they are not relatively prime in $F[x]$. Conversely, if $f$ and $g$ are not relatively prime in $F[x]$, then there exists an extension field where $f$ and $g$ have a common root.

Proof.

If $\alpha$ is a common root of $f$ and $g$ in $K[x]$, the gcd of $f$ and $g$ in $K[x]$ is not a unit. By Lem 5, it follows that the gcd of $f$ and $g$ in $F[x]$ is not a unit.

Conversely, suppose $f$ and $g$ are not relatively prime. Since $F[x]$ is a PID, $(f,g)=(d)$, where $d$ is the gcd of $f$ and $g$ in $F[x]$. Split $d$ into irreducible factors. Let $h$ be an irreducible factor of $g$. There exists, by Prp 338.5, an extension field in which $h$ has a root, say $\alpha$. Let $K^{\prime }=F(\alpha )$. Then $f$ and $g$ have a root in $K^{\prime }$.□

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Corollary 342.7(Corollary).

Let $f(x)\in F[x]$. Then there exists $K/F$ where $f$ has a multiple root $⟺$ $f$ and $f^{\prime }$ are not relatively prime in $F[x]$.

Proof.

Using Exr 333.2 and Lem 5, there exists $K/F$ where $f$ has multiple root $⟺$ there exists $a\in K$ such that $f(a)=0$ and $f^{\prime }(a)=0$ $⟺$ $f$ and $f^{\prime }$ are not relatively prime in $K[x]$ $⟺$ $f$ and $f^{\prime }$ are not relatively prime in $F[x]$.□

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Normal extensions

If $p(x)$ is irreducible in $k[x]$, $k[x]/(p(x))$ does not have to be the splitting field of $p(x)$. For example, take $k=\mathbb{Q}$ and $p(x)=x^3-2$ . Since $p(x)=\text{Irr}(\sqrt[3]{2};\mathbb{Q})$, we have

$$\frac{\mathbb{Q}[x]}{(x^3-2)}\cong \mathbb{Q}[\sqrt[3]{2}]$$

by Prp 119.9. Thus, $\mathbb{Q}[x]/(x^3-2)$ contains the real root $\sqrt[3]{2}$ but does not contain the roots $\omega \sqrt[3]{2},{\omega }^2\sqrt[3]{2}$.

Theorem 342.8(Theorem).

Let $K$ be an algebraic extension of $k$, contained in an algebraic closure $k^{\text{a}}$ of $k$. Then the following conditions are equivalent:

1. Every embedding of $K$ in $k^{\text{a}}$ induces an automorphism of $K$.
2. $K$ is the splitting field of a family of polynomials in $k[x]$.
3. Every irreducible polynomial of $k[x]$ which has a root in $K$ splits completely into linear factors in $K$.
   An extension satisfying one of these properties is called a normal extension

Footnotes

1. When you multiply $c(x-{\beta }_1^{\tau })\dots (x-{\beta }_n^{\tau })$ out, you just get $f(x)$ with each of its coefficients wrapped in $\tau$, which is just $f(x)$. ↩