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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?

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2 Answers 2

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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.

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  • $\begingroup$ Thank you. This is helpful. Usually I'm not working with complete intersection, nor CM, but the localization technique could be helpful. It seems that for many equations in many variables, the hard part may be to prove a certain element $x$ is a nonzero divisor. $\endgroup$
    – Jared
    Commented Mar 31, 2014 at 17:50
  • $\begingroup$ The second paragraph can used to prove (or at least, attempt a proof) that an element is not a zero divisor, when it applies. In my experience a great deal of algebraic sets turn out to be complete intersections (or at least, locally complete intersections), which makes me curious as to what you are working with. $\endgroup$
    – RghtHndSd
    Commented Mar 31, 2014 at 18:02
  • $\begingroup$ Currently I'm studying the variety of triples of pairwise commuting matrices. Even in the case of pairs of commuting matrices, it's not known if the ideal defined by the equations $XY-YX$ is radical. $\endgroup$
    – Jared
    Commented Mar 31, 2014 at 18:19
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You could take the quotient by I(X) and see if it gives an integral domain.

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    $\begingroup$ Isn't this rather tautological? Could you give an example where it is easier to check that the quotient is an integral domain than it is to check that the ideal is prime? $\endgroup$
    – RghtHndSd
    Commented Mar 31, 2014 at 15:41
  • $\begingroup$ Sure, consider the ideal $(x^i - y^j)$ in $R[x,y]$ with $R$ an integral domain and $i,j$ relatively prime integers. Then $R[x,y] / (x^i - y^j) \approx R[t]$ and $R$ an integral domain implies $R[t]$ is an integral domain and so $(x^i - y^j)$ is prime. $\endgroup$
    – aegbert
    Commented Mar 31, 2014 at 15:46
  • $\begingroup$ Taking $R = k$, $i = 3$, and $j = 2$ you've just proved that the cuspidal cubic curve is nonsingular. But of course it has a singularity at the origin. $\endgroup$
    – RghtHndSd
    Commented Mar 31, 2014 at 15:48
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    $\begingroup$ Oops, your right, I meant isomorphic to a subring of R[t] $\endgroup$
    – aegbert
    Commented Mar 31, 2014 at 15:51
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    $\begingroup$ Ah, of course. I certainly don't see any easy way of doing this without first taking the quotient. Thanks for the example. $\endgroup$
    – RghtHndSd
    Commented Mar 31, 2014 at 15:54

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