I was looking at this post which explains why finite fields are not algebraically closed. But the comment of Joshua Tilley made me doubt. If $K=\{\alpha_1,\ldots,\alpha_n\}$ is a finite field then the polynomial $p(x)=(x-\alpha_1)\ldots (x-\alpha_n)+1$ is just constant no? But the definition of algebraically closed is that every non-constant polynomial has a root. So how does this prove anything? I know there is something I don't get since it is not the first time I see this proof.
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2$p(x)$ is constant as a function $K \to K$ not as a polynomial. The two rings (of polynomials in $n$ variables and of polynomial functions $K^n \to K$) are the same iff the field is not finite – reuns Jun 13 '19 at 21:04
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Oh, I didn't knew that. Why $K^n\to K$? – roi_saumon Jun 13 '19 at 21:05
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@bof But $\, x^p-1 = (x-1)^p\ $ over a field of characteristic $p.\ \ $ – Bill Dubuque Jun 13 '19 at 21:21
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2See [Equality of polynomials: formal vs. functional](https://math.stackexchange.com/q/25024/242) and [Why are polynomials defined to be “formal”?](https://math.stackexchange.com/q/98345/242) and also [Do we really need polynomials (In contrast to polynomial functions)?](https://math.stackexchange.com/q/390260/242) – Bill Dubuque Jun 13 '19 at 21:24
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@BillDubuque Oops. Thanks for the correction. I guess what I said would be right if $n\not\equiv 0\pmod p$ where $p$ is the characteristic. – bof Jun 13 '19 at 22:14
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Note that a polynomial over a field is not a function. It is a formal expression of the type $\sum_{i=0}^n a_ix^i$ where $n\in\mathbb{N}\cup\{0\}$, $a_i$ are elements in the field. Two polynomials are equal if and only if all the corresponding coefficients are the same. For example, over $\mathbb{F_2}$ the polynomials $x+1$ and $x^2+1$ are the same as functions but they are different polynomials. So no, the fact that the polynomial you defined is equal to $1$ at every point does not make it constant. It is a polynomial of degree $n$.
Mark
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