# Fundamental theorem of algebra for finite fields

1,435

Multivariate polynomials over an infinite field can have infinitely many roots, as pointed out by other users.

As for the univariate case, the answer is yes: if $f$ is a (univariate) polynomial over a field $K$ and $a\in K$ is a root, then we can use the division algorithm to show that $f(x)=(x-a)q(x)$ for some polynomial $q$ over $K$.

Note that this isn't the fundamental theorem of algebra, which says that every complex (univariate) polynomial of degree $n$ has exactly $n$ roots, counting multiplicity, in the complex numbers.

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### Jessica

PhD student in computer science

Updated on August 01, 2022

• Jessica 10 months

Over the real numbers, a polynomial with degree $n$ has at most $n$ roots, regardless of whether it's a single variate or a multivariate polynomial. Is the same true when the polynomial is over a finite field like $\mathbb{Z}_p$?

• Multivariable polynomials of degree $n$ don't have at most $n$ roots. They generally have infinitely many roots.
• Dear eeeeeeeeee, The polynomial $x^2 + y ^2 - 1$ is a $2$-variable polynomial of degree $2$, which certainly has more than $2$ roots over $\mathbb R$. So I think there is an error in your assertion. Regards,
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