# Fundamental theorem of algebra for finite fields

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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 6 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 $\mathbb{Z}_p$?

• Seth over 8 years
Multivariable polynomials of degree $n$ don't have at most $n$ roots. They generally have infinitely many roots.
• Matt E over 8 years
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,
• Jessica over 8 years
Aw, shucks. Thanks for the correction.
• rschwieb over 8 years
But finite fields do have algebraic closures, where any polynomial will split into linear factors.
• Jessica over 8 years
@rschwieb even multivariable polynomials?
• rschwieb over 8 years
@eeeeeeeeee No, not multivariable polynomials. The statement of the fundamental theorem of algebra has nothing to do with multivariable polynomials, so it did not occur to me to mention them.