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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?

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  • $\begingroup$ $\tau(f_i)=\sigma(f_i)$ 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 $\tau(f_i)=\sigma(f_i)$ $\endgroup$ – reuns Aug 18 at 23:12
  • $\begingroup$ 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. $\endgroup$ – northcity4 Aug 19 at 6:39
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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 $F'(\tau(X)).$

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  • $\begingroup$ My understanding is that F(X)=F[X] if X contains a single algebraic element. Is there a generalization of this? $\endgroup$ – northcity4 Aug 17 at 18:59
  • $\begingroup$ I see. The polynomial argument only works if X is finite. $\endgroup$ – northcity4 Aug 17 at 20:05

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