Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 2 down vote accepted

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.