An affine variety $X$ over a field $k$ is irreducible if and only if its defining ideal $I(X)$ is prime (in this post we use the convention that varieties are not necessarily irreducible). Hence, it is useful to be able to determine whether or not a given ideal $I\subset k[x_1,\ldots,x_n]$ is prime.
Suppose I can explicitly write down all of the equations defining $X$, that is $X=Z(f_1,\ldots,f_m)=Z(J)$, so that $I(X)=\sqrt J$. I'm interested in learning different techniques for determining if $I(X)$ is prime, or equivalently, if $X$ is irreducible. I'm also interested in different sufficient conditions on the $f_i$ which imply that $I(X)$ is prime, so that the problem is reduced to checking something about the $f_i$. If possible, references to the literature would be great. Here are some first techniques of which I'm already aware:
- Show that if $ab\in I$, then either $a\in I$ or $b\in I$. This is just the definition of a prime ideal, but perhaps we can reduce the list of pairs $(a,b)$ we must check to some finite list involving the $f_i$ and their divisors.
- Use a computer algebra system such as Magma or GAP. This is troublesome as $m$ and $n$ grow.
- Construct a surjective morphism of varieties $\varphi:Y\to X$ such that $Y$ is irreducible.
How else might we show that $I(X)$ is prime?
Localization is a powerful tool when it comes to proving an ideal is prime. Namely, if $x \in R$ is a nonzero divisor then $R$ is an integral domain if and only if $R_x$ is an integral domain, where $R_x$ is the localization of $R$ with respect to $\{1, x, x^2, \dots\}$. By being able to assume $x$ is invertible, one can often simplify the relations defining the quotient.
The interaction of dimension theory with properties of rings such as complete intersection and Cohen-Macaulay are also important. For example, in a local CM ring, a sequence $\{a_1, \dots, a_n\}$ is a regular sequence if and only if $\dim R/(a_1, \dots a_i) = \dim R - i$. This reduces checking whether certain quotients are integral to dimension counting. Cohen-Macaulay can be difficult to check unless you have a complete intersection in which case it is automatic.