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Let $F$ be a field and let $f(x) \in F[x]$ be irreducible of prime degree $p$. Let $E/F$ be a finite extension. Prove: If $f(x)$ is not irreducible in $E[x]$, then $p \mid [E:F]$. (Hint: Consider a field $L$ with $E \subseteq L$ and $L$ as a root of $\alpha$ of $f(x)$.)

Proof: Let $F$ be a field and let $f(x) \in F[x]$ be irreducible of prime degree $p$. Let $E/F$ be a finite extension. By Proposition 20 there exists a field $L$ containing $F$ with $[L:F] = p$ in which $f(x)$ has a root $\alpha$. If $E \subseteq L$ then we can consider $p = [L : F] = [L : E][E : F]$. It must follow that $[E:F] = p$ since if $[L:E] = p$, then $[E:F]=1$ which would contradict $f(x)$ being irreducible in $F$. Hence $[E:F] = p$ and immediately $p \mid [E:F]$. Now if $L \subseteq E$ then we consider $[E:F] = [E:L][L:F] = [E:L]p$ and thus $p \mid [E:F]$.

Proposition 20: Let $F$ be a field. Let $p(x) \in F[x]$ be irreducible of degree $n$. Then there exists a field $K$ containing $F$ with $[K:F] = n$ in which $p(x)$ has a root.

Is this proof correct? If not, what is wrong and how would I fix it. If its right, alternate proofs/approaches welcomed.

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Is your question really about finite fields? That is, is $F$ necessarily a finite set? If not, then your tags are a bit off. –  Jyrki Lahtonen Feb 28 '13 at 12:50
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I agree with Jyrki, and think that there is some misunderstanding of the terminology here: a finite extension of fields is not a finite field, the latter of which is a field with finitely many elements. Thanks. –  awllower Feb 28 '13 at 17:12

3 Answers 3

up vote 3 down vote accepted

Your proof technique doesn't quite work. There's no reason that $L$ needs to be comparable to $E$. The hint I believe is suggesting the following approach. If $f(x)$ takes a root $\alpha \in E$ we are done (why?). Otherwise let $g(x)$ be an irreducible factor of $f(x)$ over $E$. Now apply Proposition 20 to $E$ and $g(x)$. A bit more clever degree arithmetic should finish the argument now.

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What does it mean for $L$ to be comparable to $E$? –  Robert Feb 28 '13 at 8:44
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@Robert I mean that there's no reason that $L \subset E$ or $E \subset L$ needs to happen. For instance if $f(x)$ doesn't take a root over $E$ and is of degree $p$ and $L=F(\alpha)$ where $\alpha$ is a root of $f$. Then $E \cap L=F$. –  JSchlather Feb 28 '13 at 9:19

Let $L$ be the splitting field of $f(x)$ over $\mathbb F$, and $K$ the splitting field of $f(x)$ over $E$. Then $L\subseteq K$, and $p\mid [L:F]$, so $p\mid[K:F]$. But $f$ is reducible over $E$, this means that $p$ does not divide $[K:E]$. Hence $p\mid[E:F]$. Q.E.D.

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Your notation for fields is very inconsistent. Either stick to blackboard or stick to regular math font. Personally I don't see the need for blackboard except for exceptional fields. –  JSchlather Feb 28 '13 at 18:32
    
@JSchlather Thanks for pointing it out. –  awllower Mar 1 '13 at 0:15

Let $F$ be a field, $f(x) \in F[x]$ irreducible of prime degree $p$. Let $E/F$ be a finite extension. Suppose $f(x)$ is reducible in $E[x]$. Let $\alpha$ be a root of $f(x)$. Consider $F(\alpha)$. It follows that $[F(\alpha) : F] = p = \deg \alpha$ since $f(x)$ is irreducible in $F$ of degree $p$. Now observe that $[E(\alpha) : E] < p$ since $f(x)$ is reducible in $E$ and $f(x)$ is not the minimal polynomial for $\alpha$ over $E$, but $m_{\alpha, E}(x) \mid f(x)$ so $[E(\alpha) : E] = \deg m_{\alpha, E}(x) < \deg f(x) = p$. Now \begin{align*} [E(\alpha) : F] &= [E(\alpha) : F(\alpha)][F(\alpha) : F] = p[E(\alpha) : F(\alpha)]\\ [E(\alpha) : F] &= [E(\alpha) : E][E : F] \end{align*} So $ [E(\alpha) : E][E : F] = p[E(\alpha) : F(\alpha)]$. It follows that $p \mid [E(\alpha) : E][E : F]$. It immediately follows that $p \mid [E(\alpha) : E]$ or $p \mid [E : F] $. Conclude that $p \mid [E:F]$.

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