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I was reading over my notes from complex analysis and saw the fundamental theorem of algebra which states that:

A polynomial of positive degree over a field $\mathbb{C}$ of complex numbers has a root in $\mathbb{C}.$

Now, I've noticed that the proof of this requires the use of (complex) analysis. I'm wondering whether anyone knows any proofs of this outside the realm of (complex) analysis?

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The wiki page on the theorem lists some proofs. –  Arthur Jul 10 at 21:22
    
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There are lots of proofs for this one. I suspect if someone compiled a list of theorems ranked by the number of published proofs, this one would rank pretty high. But not as high as the Pythagorean theorem or the law of quadratic reciprocity. –  Michael Hardy Jul 10 at 21:55

1 Answer 1

this is only a line of argument, and is only for a special case. i wonder if it can be tightened up?

let the coefficients of $p(z) = a_0+a_1z+\dots+a_mz^m$ be Gaussian integers $\in \mathbb{Z}+i\mathbb{Z}$ and $a_0=1$.

then by the division algorithm we may find a non-terminating series $q(z)=\sum_{j=0}^{\infty} b_jz^j$ (with $b_0=1$) such that formally $p(z)q(z) = 1$. since the coefficients of $q(z)$ are integers its radius of convergence cannot exceed unity. i want to deduce that $q$ must have a pole in the closed unit disc, and that therefore $p$ must have a root there.

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