3
$\begingroup$

An exercise in Lang's algebra book is: let $k$ an infinite field, and $E$ an algebraic extension of $k$. Then $E$ has the same cardinality as $k$. How can one can prove this?

$\endgroup$
  • 1
    $\begingroup$ Might I inquire as to why you unaccepted my answer? $\endgroup$ – Asaf Karagila Jun 29 '14 at 18:13
11
$\begingroup$

HINT: First prove that $k[x]$ and $k$ have the same cardinality, you can do that by showing that $k[x]\cong\bigcup k^n$, and by induction $k^n$ and $k$ have the same cardinality, so $|k[x]|=\aleph_0\cdot|k|=|k|$; then show there is a surjection from $k[x]$ onto $E$, and an injection from $k$ into $E$. Conclude the wanted equality.

(Note that this makes a heavy use of the axiom of choice, and indeed without the axiom of choice one might have a counterexample of an algebraic closure of the rational numbers which is not countable.)

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Classic and clean! I know it's possible for the closure not to exist or for multiple nonisomorphic versions to exist, but is it known for sure that the algebraic closure of the rationals can be larger without choice? $\endgroup$ – Steven Stadnicki Dec 12 '13 at 22:51
  • $\begingroup$ Steven, the most famous result is due to Lauchli's, where he shows the consistency of an uncountable closure of the rational. $\endgroup$ – Asaf Karagila Dec 12 '13 at 22:55
  • 2
    $\begingroup$ To read more about the choice connection, math.stackexchange.com/questions/114978/… should be just fine. $\endgroup$ – Asaf Karagila Dec 12 '13 at 23:12
  • 1
    $\begingroup$ @000: The axiom of choice is indeed used to show that $k[x]$ and $k$ have the same cardinality, as well to show that injection+surjection implies a bijection. $\endgroup$ – Asaf Karagila Dec 14 '13 at 21:07
  • 1
    $\begingroup$ @infintedimensional the roots of $p$ are exactly the roots of $p^n$ for all $n>1$. $\endgroup$ – Asaf Karagila Aug 19 at 11:07
6
$\begingroup$

There is a canonical map $\operatorname {minpol}:E\to k[X]$ sending an element $e\in E$ to its minimal polynomial $\operatorname {minpol}_e(X)\in k[X]$ over $k$.
Since the image $\operatorname {minpol}(E)\subset k[X]$ contains all the affine polynomials $X-q \;(q\in k)$, that image $\operatorname {minpol}(E)$ has the same cardinality as $k$ (since $k$ and $k[X]$ have same cardinality, as already remarked by Asaf).
Since the fibers of $\operatorname {minpol}: E\to \operatorname {minpol}(E)$ are finite and non empty, the set $E$ also has the cardinality of $k$, as desired.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Yeah, but there's nothing wrong with that. :-) $\endgroup$ – Asaf Karagila Dec 12 '13 at 23:09
  • 3
    $\begingroup$ Dear 000, thanks for the kind words, but don't worry about acceptance: I'm happy just knowing that my answer was of some help to you. $\endgroup$ – Georges Elencwajg Dec 13 '13 at 16:37
  • $\begingroup$ Some bravely anonymous users have downvoted this answer. I don't care a fig but want to reassure readers that the answer is perfectly correct and completete: if anybody thinks otherwise, let him comment here. But I'm not holding my breath... $\endgroup$ – Georges Elencwajg Feb 21 '15 at 9:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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