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

I have a question about the following proof in Atiyah-Macdonald:

enter image description here

1:Why is $\Omega$ infinite? Are all algebraically closed fields infinite?

2: How does the existence of $\xi$ follow from $\Omega$ being infinite?


share|cite|improve this question
1.) If $F$ is a finite field, $p(X) = 1+\prod_{a \in F} (X-a)$ is a polynomial without root. 2.) A polynomial of degree $n$ can have at most $n$ roots, but there are more than $n$ elements in $\Omega$. – martini Jul 18 '12 at 10:48
@martini Oh, of course! Thank you! If you make this truly helpful comment into an answer I will upvote and accept it! – Rudy the Reindeer Jul 18 '12 at 11:02
Ok. Will do. ${}{}$ – martini Jul 18 '12 at 11:05
What is "AM"? Please don't use abbreviations in question titles. – Henning Makholm Jul 18 '12 at 11:22
@HenningMakholm Atiyah-Macdonald. Sorry. – Rudy the Reindeer Jul 18 '12 at 12:47
up vote 6 down vote accepted

1.) Yes, algebraically closed fields are infinite: For, if $\mathbb F$ is a finite field, we can consider the polynomial $p(X) = 1 + \prod_{a \in \mathbb F}(X-a)$, which has no zeros as $p(\alpha) = 1 + 0 = 1$ for each $\alpha \in \mathbb F$, so $\mathbb F$ is not algebraically closed.

2.) $\xi$ is choosed such that $\xi$ is not a zero of $q(X) = \sum_{i=0}^n f(a_i)X^{n-i}$, as $f(a_0) \ne 0$, $q$ has degree $n$ and hence at most $n$ different zeros. As $\Omega$ has, by 1.), infinitely many elements, there is a $\xi \in \Omega$ with $0 \ne q(\xi) = \sum_{i=0}^n f(a_i)\xi^{n-i}$.

share|cite|improve this answer
Thank you! ${}{}$ – Rudy the Reindeer Jul 18 '12 at 11:11

Hint $\rm\,(1)$ follows by a polynomial analog of Euclid's proof of infinitely many primes. If $\rm\:D\:$ is a domain, then Euclid's method yields infinitely many nonassociate irreducible polynomials $\rm\in D[x].\,$ When $\rm\,D\,$ is an algebraically closed field, irreducibles are linear, associate to a monic prime $\rm\:x-a.\:$ Thus infinitely many nonassociate primes $\rm\:x-a_i$ $\Rightarrow\,$ infinitely many $\rm\:a_i$ $\Rightarrow$ $\rm\:D$ infinite.

For (2) recall a ring $\rm\,D\,$ is a domain iff every polynomial $\rm\,f\ne 0\,$ over $\rm\,D\,$ has at most $\rm\,deg\ f\,$ roots.

share|cite|improve this answer
Thank you! ${}{}$ – Rudy the Reindeer Jul 19 '12 at 5:30
@ClarkKent Note that $(1)$ in Martini's answer is essentially the above Euclid's method expressed as a proof by contradiction, just as is often done in the integer case. – Bill Dubuque Jul 19 '12 at 5:37
Dear Bill, yes, thank you! – Rudy the Reindeer Jul 19 '12 at 5:45

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.