# Countable Field Extension of a Countable Field

Okay, first question on this site, apologies in advance for any mistakes I may make.

Question: So I need to show that an algebraic field extension $E:F$, with $F$ being countable, is countable.

My Thoughts: As far as I am aware, $E$ being algebraic means that every element in $E$ is algebraic (root of some polynomial in $F[x]$ ) and being countable means that there is a natural mapping of the integers onto $E$.

At the moment I am thinking that as every element in $E$ is algebraic then it has some corresponding polynomial in $F[x]$, that it is a root of, and then because $F$ is countable I need a bijection (or possibly just a surjection) of $F$ onto $E$ and that that may be achievable through the connection between an element of $E$ and it's polynomial in $F[x]$. I just can't see how to make this rigorous.

Note: I did try to find something resembling this question but mostly found analysis based questions. This may be down to my unfamiliarity with the website though.

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Let $L/K$ be an algebraic extension. We have a natural map $\phi: L \rightarrow K[t]$ given by $\phi(\alpha)=\mathrm{min}_K(\alpha,t)$. Sadly this map need not be injective, for instance if $K=\mathbb Q$ and $L=\mathbb Q(\sqrt{2})$ then $\phi(\sqrt{2})=\phi(-\sqrt{2})$. So we have to work a bit harder. If $p(t) \in K[t]$ then hat can we say about the cardinality of $\phi^{-1}(p(t))$.

Now work in your original setting and recall that a countable union of finite sets is countable.

Edit: The argument concludes by noting that for each $p(t) \in K[t]$ we have that $|\phi^{-1}(p(t)) | \leq \deg p(t)$ so the fibers of $\phi$ are finite. In particular

$$L=\bigcup_{p(t) \in K[t]} \phi^{-1}(p(t))$$ so $L$ is a countable union of finite sets and thereby countable. This proof shows in general that any algebraic extension of an infinite field has the same cardinality as the base field.

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Is your notation L/K the same as L:K? Also, I assume by the way you say it need not be injective that we do in fact need a bijective map for the countability? –  Nytram12 Feb 17 '13 at 8:31
@Nytram12 Yes $L/K$ means that $K \subset L$ is an extension of fields. Finding an explicition bijection to a countable set is one way to show countability but it's not the only way. –  JSchlather Feb 17 '13 at 8:36
Oh? What other ways are there? –  Nytram12 Feb 17 '13 at 8:38
@Nytram12 Well countable sets are closed under countable unions and finite products for instance. So if you can write your field $E$ as a countable union of countable sets then you've shown its countable. –  JSchlather Feb 17 '13 at 8:52
Fair enough. I suspect I am meant to go with the bijection route here. Thanks for your help –  Nytram12 Feb 17 '13 at 18:25