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Relevant notations and results

  • Let $K / F$ be a field extension and let $\alpha \in K$ be algebraic over $F$. We denote the minimal polynomial of $\alpha$ over $F$ by $\min(F,\alpha)$.
  • If $\sigma : F \to F'$ is a field homomorphism, then there is an induced ring homomorphism $F[x] \to F'[x]$, which we also denote by $\sigma$, given by $\sigma\left( \sum a_i x^i \right) = \sum \sigma(a_i) x^i$.
  • Lemma 3.17. Let $\sigma : F \to F'$ be a field isomorphism. Let $f(x) \in F[x]$ be irreducible, let $\alpha$ be a root of $f$ in some extension field $K$ of $F$, and let $\alpha'$ be a root of $\sigma(f)$ in some extension $K'$ of $F'$. Then there is an isomorphism $\tau : F(\alpha) \to F'(\alpha')$ with $\tau(\alpha) = \alpha'$ and $\tau |_F = \sigma$.

I am reading Patrick Morandi's Field and Galois Theory, and on page 34, he states and proves the following theorem, which is a special case of the Isomorphism Extension Theorem.

Theorem 3.19. Let $\sigma : F \to F'$ be a field isomorphism, let $f(x) \in F[x]$ and let $\sigma(f) \in F'[x]$ be the corresponding polynomial over $F'$. Let $K$ be a splitting field of $f$ over $F$, and let $K'$ be a splitting field of $\sigma(f)$ over $F'$. Then there is an isomorphism $\tau : K \to K'$ with $\tau|_F = \sigma$. Furthermore, if $\alpha \in K$ and if $\alpha'$ is any root of $\sigma(\min(F,\alpha))$ in $K'$, then $\tau$ can be chosen so that $\tau(\alpha) = \alpha'$.

Proof: We prove this by induction on $n = [K:F]$. If $n = 1$, then $f$ splits over $F$, and the result is trivial in this case. So, suppose that $n > 1$ and that the result is true for splitting fields of degree less than $n$. $\color{red}{\text{If $f$ splits over $F$, then the result is clear.}}$ If not, let $p(x)$ be a nonlinear irreducible factor of $f(x)$, $\color{blue}{\text{let $\alpha$ be a root of $p$,}}$ and let $\alpha'$ be a root of $\sigma(p)$. Set $L = F(\alpha)$ and $L' = F'(\alpha')$. Then $[L:F] > 1$, so $[K:L] < n$. By Lemma 3.17, there is a field isomorphism $\rho : L \to L'$ with $\rho(\alpha) = \alpha'$. Since $K$ is a splitting field over $L$ for $f(x)$ and $K'$ is a splitting field over $L'$ for $\sigma(f)$, by induction the isomorphism $\rho$ extends to an isomorphism $\tau : K \to K'$. The isomorphism $\tau$ is then an extension of $\sigma$ (and $\rho$), and $\tau(\alpha) = \rho(\alpha) = \alpha'$. $$\tag*{$\blacksquare$}$$


My doubts are related to the highlighted parts in the proof:

  1. Upon assuming that $n = [K:F] > 1$, it is certain that $f$ cannot split over $F$, because then we would have $K = F \implies [K:F] = 1$. So, isn't the statement highlighted in red redundant?
  2. In the proof, $\alpha$ is taken to be a root of (an irreducible factor of) $f$, and $\alpha'$ a root of the corresponding polynomial. But in the statement of the theorem $\alpha$ is an arbitrary element of $K$. It doesn't seem to be obvious how to use the given proof to construct an isomorphism $K \to K'$ which sends this arbitrary $\alpha$ to an $\alpha'$. What am I missing here?
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The proof is a little jumbled up. It is true that the sentence highlighted in red is redundant. Moreover, in the proof the author seems to be showing the existence of an isomorphism $\tau : K \to K'$ with $\tau |_F = \sigma$, which only incidentally maps a root of an irreducible factor $p$ of $f$ to a root of $\sigma(p)$. The latter half of the theorem still needs proof.

To correct/complete the proof, we can work in the following manner.


If $n = 1$, then $f$ splits over $F$, and the result is trivial in this case. So, suppose that $n > 1$ and that the result is true for splitting fields of degree less than $n$.

Let $\alpha \in K$ and $p = \min(F,\alpha)$. Suppose that $\deg(p) > 1$. Let $\alpha' \in K'$ be a root of $\sigma(p)$. Set $L = F(\alpha)$ and $L' = F'(\alpha')$. Then, $[L:F] > 1$, so $[K:L] < n$. By Lemma 3.17, there is a field isomorphism $\rho : L \to L'$ with $\rho(\alpha) = \alpha'$ and $\rho |_F = \sigma$. Since $K$ is a splitting field over $L$ for $f(x)$ and $K'$ is a splitting field over $L'$ for $\sigma(f)$, by induction the isomorphism $\rho$ extends to an isomorphism $\tau : K \to K'$. The isomorphism $\tau$ is then an extension of $\sigma$ (and $\rho$), and $\tau(\alpha) = \rho(\alpha) = \alpha'$.

And, what if $\deg(p) = 1$? Then, $\alpha \in F$, $p = x - \alpha$ and $\sigma(p) = x - \sigma(\alpha)$, so $\alpha' = \sigma(\alpha)$. So, any isomorphism $\tau : K \to K'$ with $\tau |_F = \sigma$ will satisfy $\tau(\alpha) = \alpha'$. So, if $\alpha \in K$ is such that $\deg(\min(F,\alpha)) = 1$, then we only need to show the existence of an isomorphism $\tau : K \to K'$ with $\tau |_F = \sigma$ to complete the proof. But this is the content of the proof as given in the textbook, so we are done.


Incidentally, the proof of Theorem 3.20 (Isomorphism Extension Theorem), given in the textbook on pages 34-35, also suffers from the same flaw. There, too, the author constructs an isomorphism $\tau : K \to K'$ of splitting fields that extends the isomorphism $\sigma : F \to F'$ of the base fields, but neglects to prove that $\tau$ can be chosen so that $\tau(\alpha) = \alpha'$, where $\alpha \in K$ and $\alpha'$ is any root of $\sigma(\min(F,\alpha))$. A similar argument as above can complete that proof as well.

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