Exercise 8.2.4

Proof.

(a)

$F\subset L\subset M$, and $F\subset M$ is a Galois extension.

The Theorem of the Primitive Element implies that $L=F(\alpha )$ for some $\alpha \in L$.

Since $F\subset M$ is a Galois extension, the minimal polynomial $h$ of $\alpha$ over $F$ is separable and splits completely over $M$, say $h(x)=(x-{\alpha }_1)\cdots (x-{\alpha }_r)$, where ${\alpha }_1=\alpha$.

$K=F({\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_r)$ is a Galois extension of $F$ containing $L$, since $K$ is the splitting field of a separable polynomial $h\in F[x]$.

We show that $F\subset K$ is the Galois closure of $F\subset L$.

$\text{∙}$ $F\subset K$ is a Galois extension (as previously proved) and $K\supset L$.

$\text{∙}$ Suppose that $L\subset K^{\prime }$ is another extension such that $K^{\prime }$ is Galois over $F$.

As $F\subset K^{\prime }$ is normal, and $\alpha \in L\subset K^{\prime }$, the polynomial $h\in F[x]$ with a root in $K^{\prime }$ splits completely over $K^{\prime }$:

$h(x)=(x-{\beta }_1)\cdots (x-{\beta }_r)$, where ${\beta }_1=\alpha$, and ${\beta }_i\in K^{\prime },1\le i\le r$.

Let $K^{″}=F({\beta }_1,\cdots ,{\beta }_r)$.

$K$ and $K^{″}$ are both splitting fields of $h$ over $F$. By the Theorem of Unicity of the splitting field, there exist an isomorphism $\phi :K\to K^{″}$ which is the identity on $F$. As $K^{″}\subset K^{\prime }$, $\phi$ gives a field homomorphism of $K$ in $K^{\prime }$ which is the identity on $F$.

By definition of a Galois closure, $F\subset K$ is a Galois closure of $F\subset L$.

Note: in Exercise 7.3.13, we have seen that there exists a unique sub-extension of $F\subset M$ which is a Galois closure of $F\subset L$, so $K$ is the unique Galois closure of $F\subset L$ included in $M$, the smallest subfield $K$ of $M$ such that $F\subset L\subset K\subset M$ which is Galois over $F$. If $\alpha \in K^{\prime }$, $L\subset K^{\prime }$, and $F\subset K^{\prime }$ is a Galois extension, then $\alpha ={\alpha }_1\in K^{\prime }$ implies that ${\alpha }_1,\dots ,{\alpha }_r\in K^{\prime }$, so $K=F({\alpha }_1,\dots ,{\alpha }_r)\subset K^{\prime }$.

(b)

If $L^{\prime }$ is a conjugate field of $L$ in $M$, there exists a $F$-automorphism $\sigma$ of the field $M$ such that $L^{\prime }=\sigma (L)$, $\sigma \in \mathrm{Gal}(M∕F)$.

As $\sigma$ sends $\alpha$ on a conjugate of $\alpha$, $\sigma (\alpha )={\alpha }_i,1\le i\le r$, and

$$L^{\prime }=\sigma (L)=\sigma (F(\alpha ))=F({\alpha }_i).$$

Indeed, an element $u\in L$ is of the form $u=g(\alpha ),g\in F[x]$, so $\sigma (u)=g({\alpha }_i)\in F({\alpha }_i)$, therefore $\sigma (L)\subset F({\alpha }_i)$.

Conversely, an element $v\in F({\alpha }_i)$ is of the form $v=g({\alpha }_i)$. So $v$ is the image of $u=g(\alpha )\in L$ by $\sigma$, so $L^{\prime }=\sigma (L)=F({\alpha }_i)$, therefore the conjugates of $L$ are among the $F({\alpha }_i),1\le i\le r$.

Moreover, if $L^{″}=F({\alpha }_i)$, for any $i=1,\dots ,r$, we know that there exists $\sigma \in \mathrm{Gal}(M∕F)$ such that $\sigma (\alpha )={\alpha }_i$ (Prop.5.1.8). As previously proved, $L^{″}=F({\alpha }_i)=\sigma (F(\alpha ))=\sigma (L)$.

The conjugate fields of $L$ in $M$ are the $r$ subfields $F({\alpha }_i),1\le i\le r$.