# Irreducibility of Hypersurface

Can any one help me, how to show that a hypersurface in $\mathbb{A}^n$ is irreducible iff the defining equation F is a power of an irreducible polynomial G(i.e G can not be written as a product of two non constant polynomial). thank you. here I want to inform that I don't know what is irreducible hypersurface.

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– Blah Apr 5 '12 at 17:46
Blah's link says "An algebraic variety is called irreducible if it cannot be written as the union of nonempty algebraic varieties" . That definition is hilariously false. Here is a correct definition : a variety $X$ is irreducible if it cannot be written as the union of two closed subvarieties $Y,Y'\subsetneq X$ strictly included in the given variety $X$. – Georges Elencwajg Apr 5 '12 at 21:41
I find it quite extraordinary that you do not ask what an irreducible hypersurface is... It should be obvious to you that it is simply impossible to prove what you want without knowing what an irreducible hypersurface is! Out of curiosity: why are you trying to prove a characterization of a notion that you don't know? – Mariano Suárez-Alvarez Apr 5 '12 at 22:39

For $k$ an algebraically closed field, a hypersurface $H\subset \mathbb A^n_k$ is a closed subset of the form $H=V(F)$, where $F\in k[T_1,...,T_n]\setminus k \;$ is some non-constant polynomial.
The easy irreducibility criterion for $H$ (like for any closed subset of $\mathbb A^n_k$) is that the ideal of polynomials vanishing on it, $I(H)\subset k[T_1,...,Tn]$, be prime.
$$I(H)=I(V(F))=rad (F)$$ If the factorization of $F\in k[T_1,...,T_n]$ into irreducible polynomials is $F=G_1^{n_1}\cdot G_2^{n_2} \cdot \ldots \cdot G_r^{n_r}$, we have $$rad (F)= (G_1\cdot G_2 \cdot \ldots \cdot G_r )$$ and this will be a prime ideal iff $r=1$ i.e. finally iff $F=G^r$ with $G$ an irreducible polynomial .