How to decompose into irreducible components

Question

I only know how to find the irreducible components when I know what the image is, but there are lots of equations that are hard to figure out their images, is there any systematic way to find the irreducible components?

Answer 1

The general question "how to find irreducible components?" has a long answer. Here is an extract of the book "Ideals, varieties and algorithms" by Cox, Little, O'Shea, which points you to the literature where algorithms are presented.

It would be nice to have more general methods that could be applied to any ideal. Theorems 2,4,5, and 6 tell us that certain decompositions exist, but the proofs give no indication how to find them. The problem is that the proofs rely on the Hilbert Basis Theorem, which is intrinsically nonconstructive. Based on what we have seen in §§5 and 6, the following questions arise naturally:

$\bullet$ (Primality) Is there an algorithm for deciding if a given ideal is prime?

$\bullet$ (Irreducibility) Is there an algorithm for deciding if a given affine variety is irreducible?

$\bullet$ (Decomposition) Is there an algorithm for finding the minimal decomposition of a given variety or radical ideal?

The answer to all three questions is yes, and descriptions of the algorithms can be found in the works of HERMANN (1926), MINES, RICHMAN, and RUITENBERG (1988), and SEIDENBERG (1974, 1984). As in §2, the algorithms in these articles are not very practical. However, the work of GIANNI, TRAGER, and ZACHARIAS (1988) has recently led to algorithms implemented in AXIOM and REDUCE that answer the above questions. See also Chapter 8 of BECKER and WEISPFENNING (1993) and, for the primality algorithm, §4.4 of ADAMS and LOUSTAUNAU (1994). A different algorithm for studying these questions, based on ideas of EISENBUD, HUNEKE and VASCONCELOS (1992), has been partially implemented in Macaulay.

In the example, however, we may use the following calculation:

$(x^2+y^2+z^2,x^2-y^2-z^2+1)\\=(x^2+y^2+z^2,(-y^2-z^2)-y^2-z^2+1)\\=(x^2+y^2+z^2,y^2+z^2 - 1/2)\\=(x^2 + 1/2,y^2+z^2 - 1/2)\\=(x^2-c,y^2+z^2+c)$

where $c = -1/2$. What follows, works for any $c \in k^*$. Assuming that $k$ is algebraically closed, we factor $x^2-c=(x+\sqrt{c})(x-\sqrt{c})$ and the ideals $(x+\sqrt{c}),(x-\sqrt{c})$ are coprime. Hence, the ideal becomes $(x+\sqrt{c},y^2+z^2+c) \cap (x-\sqrt{c},y^2+z^2+c)$ and both these ideals are prime, because $k[x,y,z]$ modulo them is (in each case) $k[y,z]/(y^2+z^2+c)$, which is an integral domain since $y^2+z^2+c$ is irreducible (apply Eisenstein's criterion with $z + \sqrt{-c}$).