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Context :

Definition : We say $f(x) \in F[x]$ splits over the field extension $E/F$ if $$ f(x) = c (x-\alpha_1)\cdots(x-\alpha_n) $$ for some $c, \alpha_1, \ldots, \alpha_n \in E$. A splitting field of $f$ is a minimal extension $E$ such that $f$ splits over $E/F$.

Theorem: Every polynomial $f(x) \in F[x]$ has a unique splitting field up to isomorphism.

Now, the unicity part of the Theorem is a particular case of the more general result below.

Lemma: Let $\sigma : F \to F'$ be a field isomorphism. For $f(x) = \displaystyle \sum_{i=0}^n a_i x^i \in F[x]$, define $\sigma(f)(x) := \displaystyle \sum_{i=0}^n \sigma(a_i) x^i \in F'[x]$. Let $E$ (respectively $E'$) be a splitting field of $f$ (respectively of $\sigma(f)$). Then, $\sigma$ extends to an isomorphism $\widetilde{\sigma} : E \to E'$.

Indeed, taking $F' = F$ and $\sigma = id_F$ shows that any two splitting fields of $f$ must be isomorphic, whatever $f$ is in $F[x]$.

Now, I struggle with the proof of this Lemma. Basically it goes like this.

Proof of the Lemma: We induct over $n$, the number of roots of $f$ in the splitting field $E$ (there are $\deg f$ such roots, counting multiplicity) that are not in $F$.

If $n=0$, then $E = F$ and since the images of the roots of $f$ are roots of $\sigma(f)$, we also have $E' = F'$. Hence. there is no need to extend $\sigma$.

If $n>0$, let $\alpha \in E \backslash F$ be a root of $f$ and let $m(x)$ be its minimal polynomial over $F$ ($m$ exists since $f(\alpha) = 0$ and $f \in F[x]$). Then we have the following sequence of isomorphisms $$ F(\alpha) \to F[x]/(m(x)) \to F'[x]/(\sigma(m)(x)) \to F'(\alpha'), $$ given by $$ \alpha \mapsto \overline{x} \mapsto \overline{x} \mapsto \alpha', $$ where $\alpha'$ is any root of $\sigma(m)$ in $E' \backslash F'$. Hence by composition we get an isomorphism $\sigma_1 : F(\alpha) \to F'(\alpha')$ extending $\sigma$. Since $f$ has less than $n$ roots in $E \backslash F(\alpha)$, induction completes the proof.

Question: What exactly are the three isomorphisms ? I know that since $\alpha$ is algebraic over $F$, we have $F(\alpha) = F[\alpha]$ and so every element in $F(\alpha)$ can be written $f(\alpha)$ for a certain polynomial $f(x) \in F[x]$, therefore knowing the image of $\alpha$ gives a lot of information about the isomorphism. But what are the elements of $F$ mapped to in the first isomorphism, for example ?

Also how can we see instantly that these are isomorphisms ?

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The isomorphism $F(a) \rightarrow F[x]/(m(x))$ just maps $a$ to $x$ (and $F$ to itself).

It's actually easier to show the map the other way is an isomorphism. The map $F[x] \rightarrow F(a)$ mapping $x$ to $a$ is well-defined, and surjective because $F(a) = F[a]$. Its kernel consists of polynomials $p(x) \in F[x]$ such that $p(a)=0$. These are exactly the polynomials that the minimal polynomial $m(x)$ of $a$ divide (by the usual argument of showing the remainder after division is zero by minimality). So the kernel is $(m(x))$ and $F[x]/(m(x)) \rightarrow F(a), x \mapsto a$ is an isomorphism. Its inverse is clearly the map sending $a$ back to $x$.

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  • $\begingroup$ @Prometeheus I understand how you use the First Isomorphism Theorem. However, trying to prove directly that $g : \alpha \mapsto \overline{x} : F(\alpha) \to F[x]/(m(x))$ is an isomorphism, I don't see how the fact that $\alpha$ is a root of $m(x)$ is to be used. It seems to me that the way we define the map, i.e. $g(\displaystyle \sum_{i=0}^n a_i \alpha^i) := \displaystyle \sum_{i=0}^n a_i \overline{x}^i$, implies that we'll have $g(ab)=g(a)g(b)$, $g(a+b)=g(a)+g(b)$, $g(1)=1$ and surjectivity of $g$, for any $\alpha$ and $m(x)$. Hence it seems to me $g$ would be an isomorphism if $m(x)$ is.. $\endgroup$ – Amateur Sep 11 '14 at 3:30
  • $\begingroup$ .. taken to be irreducible and $\alpha$ not necessarily chosen to be a root of $m(x)$. However, in that case, $g(m(\alpha)) = \overline{m(x)} = 0$ seems to contradict injectivity ? $\endgroup$ – Amateur Sep 11 '14 at 3:33
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    $\begingroup$ The map $\alpha \mapsto \overline{x}$ would not be well-defined unless $\alpha$ is a root of $m(x)$. That's because relations satisfied in the domain must also be satisfied in the codomain. You have $m(\alpha)$ mapping to $\overline{m(x)}$, so if $m(\alpha)$ is zero, $\overline{m(x)}$ better be zero as well. $\endgroup$ – Prometheus Sep 11 '14 at 3:48
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    $\begingroup$ So whatever $\alpha$ maps to, must satisfy $p(x)$, otherwise we have $\phi(0)=0$ on the one hand from the fact that $\phi$ must a homomorphism, and on the other hand we'd have $\phi(0)=\phi(\alpha)\neq 0$. That means the map is not well-defined. Well-definedness is not really a technical term. Whenever you claim to define something and it turns out not to make sense, the thing is not well-defined. $\endgroup$ – Prometheus Sep 11 '14 at 4:08
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    $\begingroup$ I'm not sure what you are confused about then. You are claiming that $m(\alpha)=0$ is not used at all in the construction, but you are trying to show that by assuming it's not true, then deriving a contradiction. That just shows it must be true. $\endgroup$ – Prometheus Sep 11 '14 at 4:10

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