I am trying to understand the following:

Theorem I. If the polynomial $p(x)$ is irreducible in $F[x]$ and if $a$ is a root of $p(x),$ then $F(a) \cong F'(b)$ where $b$ is a root of $p'(t) \in F'[t].$ Moreover, this isomorphism $\Phi$ can be chosen so that $\Phi(a) = b$ and $\Phi(\alpha) = \beta \in F'$ for every $\alpha \in F.$

Corollary. If $p(x) \in F[x]$ and $a,b$ are both roots of $p(x),$ then $F(a) \cong F(b)$ by an isomorphism defined by $\phi:F(a) \to F(b), \hspace{1mm} a \mapsto b$ and leaves every element of $F$ fixed.

If I'm understanding this correctly, the preceding corollary states that if the roots $a$ and $b$ contained in the adjoined extensions $F(a)$ and $F(b)$ are from the same polynomial $p(x)$ over a field $F$, then we may restrict the mapping $\Phi$ in theorem I so that the adjoined roots $a,b$ are mapped to one another but $F$ is mapped to itself.

Theorem II. Let $\tau: F \to F', \hspace{1mm} \alpha \mapsto \beta$ be a field isomorphism, let $f(x) \in F[x], \hspace{1mm} \deg f(x) \geqslant 1$ and let $K$ be a splitting field of $f(x)$ over $F$ and $K'$ a splitting field of $f'(t)$ over $F'.$ Then $\tau: F \to F', \hspace{1mm} \tau(\alpha) = \beta$ extends to the field isomorphism $\psi:K \to K', \hspace{1mm} \alpha \mapsto \psi(\alpha) = \beta.$ In particular (proven in next theorem), any two splitting fields $K_1,K_2$ of the same polynomial over $F$ satisfy the property that $K_1 \cong K_2$ and that $\psi(\alpha) = \alpha$ for every $\alpha \in F.$

This theorem is similar to theorem $I$ except that the isomorphism $\Phi$ is between splitting fields instead of between fields with adjoined roots. So, $\Phi$ not only works between two extensions that extend the base fields $F,F'$ of two irreducible polynomials $p(x)$ and $p'(t)$ by a single root, but it also works between the smallest two fields that contain all the roots of the two distinct polynomials $f(x)$ and $f'(x)$. The "in particular" part of the statement follows from this theorem because if we select $\tau: F \to F$ to be the identity map such that $\tau(\alpha) = \alpha$ for every $\alpha \in F,$ then $\tau$ can be extended to the isomorphism $\psi:K \to K'$ between $F$ and $F'.$

Is this reasoning correct?

  • 1
    $\begingroup$ Essentially correct. There is no restriction in the corollary. It says "$\phi:F(a)\rightarrow F'(b)$ is an isomorphism where $\phi|_F:F\rightarrow F'$ and $a\mapsto b$." For example, taking $F(X)\rightarrow F(a)$ where $F(X)$ are rational functions in $X$ and the map is simply $X\mapsto a$ or, evaluation of a rational function at $X=a$. However, it is also much deeper than this example and I encourage you to explore with not so weird an example: $X^2+1$ over $\mathbb{Q}$. $\endgroup$ – Eoin Jun 8 '15 at 3:07
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    $\begingroup$ The polynomials $f(x)$ and $f'(t)$ are not "distinct" in the usual sense. $f'(t)$ is the image of $f(x)$ under an isomorphism given by $\tau: F\rightarrow F'$ and $x\mapsto t$. So they are distinct polynomials but, they are still "isomorphic" in some sense (probably close to a very very loose sense). It is correct however, to take the identity map and extend this to a mapping which does not have to be an identity $\psi: K\rightarrow K'$. It is simply an isomorphism. For example, the mapping of $\mathbb{Q}[i]\rightarrow \mathbb{Q}[i]$ sending $i\mapsto -i$ would be such a map. $\endgroup$ – Eoin Jun 8 '15 at 3:15
  • $\begingroup$ @Eoin Thanks for the help. $\endgroup$ – St Vincent Jun 8 '15 at 4:40

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