# Proof Explanation *Field and Galois Theory*

The following is with regards to Lemma 3.18 from Field and Galois Theory by Patrick Morandi.

Let $$\sigma : F \to F'$$ be a field isomorphism, $$K$$ be a field extension of F, and $$K'$$ be a field extension of $$F'$$. Suppose that $$K$$ is a splitting field of $$\{f_i\}$$ over F and that $$\tau: K \to K'$$ is a homomorphism with $$\tau|_F = \sigma$$. If $$f_i':=\sigma(f_i)$$, then $$\tau(K)$$ is a splitting field of $$\{f_i'\}$$ over $$F'$$.

To clarify, since $$\sigma$$ and $$\tau$$ are field homomorphisms, there are natural induced ring homomorphisms $$\sigma':F[x]\to F'[x]$$ and $$\tau':K[x]\to K'[x]$$. So when he writes $$\sigma(f)$$ or $$\tau(f)$$, he really means $$\sigma'(f)$$ or $$\tau'(f)$$.

I have two questions regarding the proof. The first is that Morandi assumes that $$\tau(K)$$ is a field. Now if we assume that ring homomorphisms must map $$1_F\to 1_{F'}$$, then it will be the case $$\tau$$ is injective and thus, $$\tau(K)$$ is a field. Should this be the case?

For my second question, let $$X$$ be the set of all the roots of all the $$f_i$$ in $$K$$. Then $$K=F(X)$$. Morandi claims that $$\tau(F(X))=F'(\tau(X)).$$ I was able to show that $$F'(\tau(X))\subseteq \tau(F(X))$$, but I am not convinced of the other inclusion. Could someone kindly explain this?

• $\tau(f_i)=\sigma(f_i)$ means applying $\sigma$ to the coefficients of $f_i$, that it splits completely implies $\tau(f_i)$ splits completely, and that $K$ is generated over $F$ by the roots of the $f_i$ means $\tau(K)$ is generated over $\tau(F)=\sigma(F)$ by the roots of the $\tau(f_i)=\sigma(f_i)$ – reuns Aug 18 at 23:12
• This was the point of my question (which has been answered). Really the question is what do we mean by "generated." Generally we mean elements in $K$ can be written as polynomials in terms of the generators, but this is only true (the only case I am aware of) if the extension is finite. I guess if $X$ is a set of algebraic elements and is finite, then $[F(X):F]$ is finite. Furthermore, it is true your comment shows $\tau(K)\subseteq F'(\tau(X))$, but I figured it was also worth mentioning why the other inclusion was trivial. – northcity4 Aug 19 at 6:39

To see why $$\tau(K)$$ is a field, you are correct in saying that $$\tau$$ is injective. Furthermore, it is surjective. Hence, they are isomorphic as rings so in particular it has a field structure. To see the other inclusion, note that elements in $$\tau(F(X))$$ is simply $$\tau$$ of polynomials with coefficients $$F$$ over the elements of $$X.$$ This is because the extension is algebraic. Take any monomial of the form $$f \alpha_1^{e_1} \ldots \alpha_k^{e_k}$$ with $$f \in F$$ and $$\alpha_i \in X.$$ Then $$\tau(f\alpha_1^{e_1} \ldots \alpha_k^{e_k})$$ is just $$\tau(f)\tau(\alpha_1)^{e_1}\ldots \tau(\alpha_k)^{e_k}$$ which is an element in $$F'(\tau(X)).$$