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Suppose we have an algebraically closed field $F$ and $n+1$ variables $X_0, \dots, X_n$, where $n > 1$. Does there exist an irreducible homogeneous polynomial in these variables of degree $d$ for any positive integer $d > 1$? In other words, does there always exist an irreducible hypersurface of arbitrary degree?

Of course, I am also interested in constructions of these polynomials.

Thank you.

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2 Answers 2

up vote 9 down vote accepted

Yes, this is straightforward. First note that a homogeneous polynomial $f(X_0, ... X_n)$ which is not divisible by $X_0$ is irreducible if and only if $f(1, ... X_n)$ is irreducible, so the problem reduces to constructing irreducible polynomials in $k[x_1, ... x_n]$ of degree $d$. To do this we can take $x_1^2 - x_2 h(x_3, ... x_n)$ for any polynomial $h$ of degree $d-1$. (If $n = 2$ then instead use, for example, $x_1^2 - (x_2 + x_2^d)$. The important thing is that whatever comes after $x_1^2$ should not itself be a square.)

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Proposition: The Fermat polynomial $f = X_0^d + \cdots + X_n^d$ is irreducible for $n\ge 2$ in characteristic zero.

Proof: By induction on $n$ using Eisenstein's criterion (e.g. in Lang's Algebra, Part One, Theorem IV.3.1). For $n=2$, we view $X_0^d+X_1^d+X_2^d$ as an element of $k[X_1,X_2][X_0]$. Now, $X_1^d+X_2^d = (X_1-e_1 X_2)\cdots(X_1-e_d X_2)$, where $e_1,\dots,e_d$ are the roots of the polynomial $\xi^d+1$. Since $k$ is not necessarily algebraically closed, the $e_i$ will lie in general in the algebraic closure $\bar{k}$ of $k$. Second, all $e_i$ are distinct. This implies that there is a factor $f \in k[\xi]$ of the polynomial $\xi^d+1$ (possibly the entire polynomial itself), that is irreducible over $k$ and $\xi^d+1 \not\in (f^2)$. Now let $\mathfrak{P}$ be the prime ideal generated by $f$. Then $X_1^d+X_2^d \in \mathfrak{P}-\mathfrak{P}^2$ and so Eisenstein's criterion gives that $X_0^d+X_1^d+X_2^d$ is irreducible. Next, for $n >2$ we view $X_0^d + X_1^d+\cdots+X_n^d$ as an element of $k[X_1,\dots,X_n][X_0]$. By induction hypothesis $X_1^d+\cdots+X_n^d$ is irreducible and we can take the prime ideal $\mathfrak{P}$ of Eisenstein's criterion to be the ideal generated by $X_1^d+\cdots+X_n^d$.

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