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my lecturer used this theorem in the title without proof. I didn't find a proper proof by myself. Could anyone help me out, please?

user26857
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Tobi92sr
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2 Answers2

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Let's suppose that $ K $ is finite and write $ K=\{\alpha_{1}, \ldots , \alpha_{n}\}$. Now take the polynomial $ p (x)=(x-\alpha_{1})\ldots (x-\alpha_{n}) +1\in K[x]$. It's easy to see that $ p (x) $ doesn't have any roots in $ K $. Hence, $ K $ is not algebraically closed.

Xam
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  • Thanks for the fast response. – Tobi92sr Dec 01 '16 at 21:15
  • @Tobi92sr you're welcome. – Xam Dec 01 '16 at 21:20
  • This can be presented nicely via a polynomial analog of Euclid's proof that there are infinitely many primes - see this answer. – Bill Dubuque Dec 01 '16 at 21:51
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    Does this proof work? A field $K$ is algebraically closed if every non-constant polynomial in $K\left[x\right]$ has a root. You appear to have constructed a constant polynomial with no root. That is, $p\left(x\right)=1$ for all $x\in K$, so $p$ is constant, right? – Joshua Tilley Jul 23 '18 at 14:35
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    @JoshuaTilley it seems to me you're confusing a polynomial with a polynomial function. The only constant polynomials are those of degree 0. This one is thus clearly not constant, although everywhere you evaluate it you get the same answer (i.e. although it's constant as a function). – Janou Glaeser Jul 30 '20 at 23:23
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A different approach is to consider the fundamental theorem of Algebra. Since $F$ is algebraically closed, it will contain every root of every polynomial in $F[x]$, even those with degree larger than $|F|$.

Of course you should now proof that there exists such a polynomial with $m > n := |F|$ distinct roots, whereas @Xammm solution constructs a contradiction right away, so this is not a practical proof but it may give you another point of view.

cronos2
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    What does the fundamental theorem of algebra (which speaks of $\mathbb C$) have to do with anything? You are just using the definition of being algebraically closed. – Wojowu Dec 01 '16 at 21:24