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

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

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

Thanks.

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5  
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! –  Matt N. 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. –  Matt N. Jul 18 '12 at 12:47

2 Answers 2

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 $0 \ne q(\xi) = \sum_{i=0}^n f(a_i)\xi^{n-i}$.

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Thank you! ${}{}$ –  Matt N. 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.

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Thank you! ${}{}$ –  Matt N. 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! –  Matt N. Jul 19 '12 at 5:45

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