Extending field homomorphisms to automorphisms

I have $L/K$ a finite field extension and an irreducible polynomial which has two roots in $L$, $\alpha$ and $\beta$. I’m trying to show there is an automorphism of $L$ that fixes $K$ and switches $\alpha$ and $\beta$.

My thoughts so far are to use the fact that the inclusion $K \to L$ can be extended to an isomorphism $K(\alpha) \to K(\beta)$. Now I know that $L$ is normal, so it is the splitting field of some polynomial. I would like to show that $L$ is the splitting field of $f$ considered as a polynomial in say $K(\alpha)$ and then the theorem on uniqueness of splitting fields we would have that the isomorphism $K(\alpha)$ to $K(\beta)$ would then extend to an automorphism of $L$. However I am having trouble showing that $L$ is the splitting field.

I know certainly that $f \in K(\alpha)[x]$ splits in $L$ but I’m not sure how to see there is no intermediate field between $K(\alpha)$ and $L$ where $f$ might split.

I’m aware that the splitting fields of f over $K(\alpha)$ and $K(\beta)$ are unique, but I’m then unsure how to extend that isomorphism to an automorphism of L


Solutions Collecting From Web of "Extending field homomorphisms to automorphisms"

Let’s see the general case. Then your question follows as a Corollary.

First consider a isomorphism $$\sigma: K \to K’$$

Now write $h(x) = \sum_{j=0}^{n} a_j X^j$ irreducible in $K[X]$. If $\alpha$ is a root of $h$ and $\beta$ a root of $h^{\sigma}(x) = \sum_{j=0}^{n} \sigma(a_j) X^j$, then there exists $\hat{\sigma}: K[\alpha] \to K'[\beta]$. In fact,

Take the following isomorphism

$$\begin{align} \hat{\sigma}:K[\alpha] &\to K'[\beta]\\ \sum_{j=0}^{n} x_j \alpha^j&\mapsto\sum_{j=0}^{n} \sigma(x_j) \beta^j \end{align}$$

it is well-defined and the only isomorphism such that $\hat{\sigma}|_K = \sigma$ and $\hat{\sigma}(\alpha) = \beta $.

Now if $f = f_{1}^{e_1}f_{2}^{e_2}\ldots f_{k}^{e_k}$ where $f_{j} \in K[X]$ are distinct and irreducible in $K[X]$ and $\alpha_1, \ldots , \alpha _ k$ arre all distinct roots of $f$, then $$f^{\sigma} = (f_{1}^{\sigma})^{e_1}\ldots (f^{\sigma}_{k})^{e_k}$$

where $f^{\sigma} \in K'[X]$ are distinct and irreducible in $K'[X]$.

The number $r$ of roots of $f$ is the sum of the degrees of $f_1,…,f_k$ then it is the number of roots of $f^{\sigma}$.

If we write $K_{1} = K[\alpha_1], K_{2} = K[\alpha_2], \ldots , K_{r} = K[\alpha_r] = L(f,K)$ then you apply the result above to each extension $K_j[\alpha_{j+1}]/K_j$ to get

$$\hat{\sigma}: L(f,K) \to L(f^{\sigma}, K’)$$

such that $\hat{\sigma}|_K = \sigma$ and $\sigma(\alpha_j), j = 1,…,r$ are the roots of $f^{\sigma}$.