Originally Posted by

**aliceinwonderland** This is the sketch of the proof.

Let $\displaystyle f \in K[x]$ and use induction on $\displaystyle n = [\Sigma: K]$.

For $\displaystyle n=1,\Sigma = K$ and f splits over K, which imples that $\displaystyle i(f)$ splits over $\displaystyle K'$. Thus $\displaystyle \Sigma'=K'$.

If n > 1, assume by induction that $\displaystyle f \in K[x]$ of degree less than n holds the given isomorphism extension property. Then f must have an irreducible factor g of degree greater than 1. Let $\displaystyle \alpha$ be a root of g in $\displaystyle \Sigma$ such that $\displaystyle f=gf_1$, where $\displaystyle f_1$ has degree n-1. Verify that

$\displaystyle K(\alpha) \cong \frac{K[x]}{(g)} \cong \frac{K'[x]}{(g')} \cong K'(\alpha')$, where i sends $\displaystyle \alpha$ to $\displaystyle \alpha'$ and g to g' with coefficients.

Since $\displaystyle \Sigma$ is a splitting field of f over $\displaystyle K(\alpha)$ and $\displaystyle \Sigma'$ is a splitting field of i(f) over $\displaystyle K'(\alpha')$, the induction hypothesis implies that j extends to an isomorphism $\displaystyle \Sigma \cong \Sigma'$.