Decomposing an affine algebraic set into irreducible ones

Question

Let $X\subset\mathbb C^4$ be given by the system \begin{align} x_1x_4 - x_2x_3 &=0\\ x_1x_3 - x_2^2 &=0 \end{align} I need to decompose this into a union of irreducible sets. The obvious approach is to try and find the set by obtaining relations on the variables that are easily visualizable. Once that is done it might be clear, what the decomposition is. This worked on the previous system, where I was able to get rid of the mixed term via a linear transformation of coordinates: something like $x_1 = y_1 - y_3$ and $x_3 = y_1+y_3$. This doesn't seem to work here, since I've got both $x_1x_4$ and $x_1x_3$, for example. The change I found that does work is degenerate.

My question is: does one need to obtain clear relations on the variables in order to decompose the set, and if so, how to do it?

Thank you.

Answer 1

Call your ideal $I$.

If you want to do these things by hand, the best thing to do is to stare at the equation. Then you notice that the ideal $I_1=(x_1,x_2)$ lies in your ideal, and it is certainly prime.

Thus, we want to remove $I_1$ from the primary decomposition of $I$. This can be done by calculating the ideal quotient $I:(x_1x_2)$. Now, to do this, you better work with a Gröbner basis, and the two generators here is not a Gröbner basis. So we compute an $S$-pair, with respect to the lexicographic order (call your two generators $f_1$ and $f_2$): $$S(f_1,f_2)= \frac{lcm(in(f_1),in(f_2))}{f_1}f_1 - \frac{lcm(in(f_1),in(f_2))}{f_2} f_2 = \frac{x_1x_3x_4}{x_1x_4}(x_1x_4-x_2x_3)-\frac{x_1x_3x_4}{x_1x_3}(x_1x_3-x_2^2)$$

This equals $S(f_1,f_2)=x_3x_2^2-x_2x_3^2$. Call this polynomial $f_3$. You can check that this polynomial does not reduce to zero modulo the two others, and that the set $\{ f_1,f_2,f_3 \}$ does not produce any more $S$-pairs. Thus $\{f_1,f_2,f_3\}$ is a Gröbner basis for $I$. But having a Gröbner basis, it is easy to calcualat ethe ideal quotient!

You just quotient each of the elements in the Gröbner basis! So quotening each by $x_1x_2$, we see that $I:(x_1x_2)$ is $$(x_1x_4-x_2x_3,x_1x_3-x_2^2,x_3x_2-x_3^2)$$

This is a prime ideal of dimension $2$. So your ideal is the intersection a plane and a curve lying on a quadric.

If you don't know Gröbner bases, I recommend the section in Eisenbud's "Commutative Algebra with a View Towards Algebraic Geometry".