Primary decomposition of an ideal

Question

Suppose $I$ is the ideal $(xy, yz, zx)$ in $R =\Bbb R[x, y, z]$. I want to compute the primary decomposition of $I$.

I have viewed many post on this topic, as I suspect, the primary decomposition of $I$ would be:

$$ (xy, yz, zx)= (x,y)\cap(x,z)\cap(y,z),$$

but I cannot show how.

@JosephCurwen am I wrong, please more details. I could not understand the topic properly? — Alvaro, Jan 06 '14 at 23:50

score 1 · Accepted Answer · edited Jan 07 '14 at 08:29

1

Your ideal is radical and monomial, for the monomial ideals holds the formula $$\sqrt{(A,BC)}=\sqrt{(A,B)}\cap \sqrt{(A,C)}$$ with $A,B,C \in k[x,y,z]$.

Now you have $$(xy,yz,zx)=\sqrt{(xy,yz,zx)}=\sqrt{(xy,yz,z)}\cap \sqrt{(xy,yz,x)}=\sqrt{(xy,z)}\cap \sqrt{(yz,x)}=$$$$=\sqrt{(x,z)}\cap \sqrt{(y,z)}\cap \sqrt{(y,x)}\cap \sqrt{(z,x)}=(x,z)\cap (y,z) \cap (y,x)$$ and then your primary decomposition.

edited Jan 07 '14 at 08:29

answered Jan 06 '14 at 23:57

Sabino Di Trani

5,101

thanks a lot for the answer, I am stacked in step 2. – Alvaro Jan 07 '14 at 00:01
What is not clear? – Sabino Di Trani Jan 07 '14 at 00:04
How to know that $I$ is radical and monomial ?? – Alvaro Jan 07 '14 at 01:40
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I suppose that $A$, $B$, $C$ are monomial ideals, and not elements in $k[x,y,z]$. Furthermore, in this case a simpler formula works: $I = ((a) + J) \cap ((b) + J)$ with $I,J$ monomial ideals and $a,b$ coprime monomials. – Jan 07 '14 at 08:37
@Alvaro $I$ is monomial because is generated by monomials and radical because this monomial generators are squarefree – Sabino Di Trani Jan 07 '14 at 09:50
If you need more explanations about radical of a monomial ideal I think this could be useful http://math.stackexchange.com/questions/125907/the-radical-of-a-monomial-ideal-is-also-monomial – Sabino Di Trani Jan 07 '14 at 09:53
@JosephCurwen Thanks for explanation. – Alvaro Jan 07 '14 at 12:19

Primary decomposition of an ideal

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