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Show that an algebraic closed field must have infinite many elements.

Let's suppose that an algebraic closed field $K$ has finite many elements.

But how could I get a contradiction??

Git Gud
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Mary Star
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1 Answers1

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Suppose $K=\{a_1, \ldots ,a_n\}$, for some natural number $n$.

Consider the polynomial $f(x)=(x-a_1)(x-a_2)\ldots(x-a_n)+1$.

Since you assumed that $K$ is algebraically closed, there is a root $a$ of $f(x)$ in $K$, that is $f(a_i)=0$, for some $i\in \{1, \ldots ,n\}$.

Try to get a contradiction.

Git Gud
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  • So, since $f(a_i)=1 \neq 0$ do we conclude that $K$ contains a finite set of elements?? – Mary Star Nov 07 '14 at 00:36
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    @MaryStar I don't get your comment. On the one hand one has $f(a_i)=1$, on the other $f(a_i)=0$. – Git Gud Nov 07 '14 at 00:37
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    Perhaps it is clearer to express this as: no finite field can be algebraically closed. – lhf Nov 07 '14 at 00:50
  • Ok... Why do we write the polynomial $f(x)$ in that form?? – Mary Star Nov 07 '14 at 00:58
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    @MaryStar Are you asking the motivation behind choosing that polynomial? One reason is it works :P For something a little more enlightening recall that a field is also a ring and in rings there are such things as 'prime elements' and polynomials of degree $1$ are prime elements. Now try to make a parallel with the prime numbers in the ring $\mathbb Z$ and Euclid's proof for the infinitude of prime numbers. – Git Gud Nov 07 '14 at 01:16
  • I see... What can we conclude from that for the fields $\mathbb{Z}_p, p=\text{ prime number }$?? Do we conclude that the fields $\mathbb{Z}_p, p=\text{ prime number }$ are not algebraic closed, since they have a finite number of elements. Is this correct?? – Mary Star Nov 07 '14 at 01:29
  • @MaryStar In the ring $\mathbb Z_p$, $x-a_i$ are prime elements. This is what I meant. And it's just motivation for the polynomial, then one checks it works. – Git Gud Nov 07 '14 at 10:11
  • @MaryStar Also note that you can easily transform this proof by contradiction in a proof by contraposition. – Git Gud Nov 08 '14 at 13:20
  • Could you explain me further what we conclude for the fields $\mathbb{Z}_p$ where $p$ is a prime ?? $$$$ To transform ths proof by contradiction in a proof by contraposition do we suppose that $K$ is not algebraically closed and we conclude that $K$ has a finite set of elements?? – Mary Star Nov 08 '14 at 13:42
  • The contrapositive is simply that any finite field is not algebraically closed. So you take a finite field, provide a polynomial which doesn't decompose in the field and you're done. I think you're missing the point regarding that other thing. Nonetheless I'm unable to explain better unless you ask a specific question. – Git Gud Nov 08 '14 at 13:45