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I'm stuck on a exercise from my class notes of Commutative Algebra.Exercise goes as follows:

What are the irreducible components of the algebraic set $V(y^2-xz,z^2-x^2y)$ in $\mathbb{A}^3_K$? Here I'm just letting $K$ be an algebraically closed field.

I've tried to solve this problem and I've found that one irreducible component will be $V(y,z)$ but I'm unable to solve it completely. I guess this problem requires a t"trick" and I'm unable to catch that trick. I solve such problems in the following way:

Normally, what I do is take the equations determining an algebraic set $V(I)$, and usually one of them factors so that $V(I)$ decomposes as $V(J_1)\cup V(J_2)\cup\cdots$ or something. After breaking things down enough, I can eventually find that $K[x,y,z]/J_i$ is an integral domain, so $J_i$ is prime, and $V(J_i)$ is irreducible. Any hints/ideas?

user26857
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Arpit Kansal
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    This should be helpful: http://math.stackexchange.com/questions/1023753/what-are-the-irreducible-components-of-vxy-z3-xz-y3-in-mathbba3-k?rq=1 – user26857 Oct 27 '15 at 13:03

2 Answers2

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In general , you can use primary decomposition in Macaulay2.

In your case,
The primary decomposition of your ideal is $(y^2 - xz, x^2 y - z^2 , x^3 - yz)$ and $(z, y)$ . This means that their intersection is your ideal $(y^2-xz,z^2-x^2y)$ and both ideals are primary.

Then you can take radical of both ideal to find the irreducible components. The radical of both ideals remain the same so the above two ideals are prime and their the prime ideals of the irreducible components.

Ben
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  • Thank you,I'm interested in finding primary decomposition of $(y^2-xz,z^2-x^2y)$ without using Macaulay2.Thats why I was finding irreducible components.Regards, – Arpit Kansal Oct 27 '15 at 15:05
  • By taking radical, you can find irreducible component after finding its priamary decomposition. – Ben Oct 27 '15 at 15:39
  • But I don't see any "good" way of finding primary decomposition. – Arpit Kansal Oct 28 '15 at 02:54
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Okay, so you have already found out that $I_1 = (y,z)$ is one irreducible component of $I=(y^2-xz,z^2-x^2y)$. Then to get rid of $I_1$ in $I$, one can compute $I:(g) = \{ f \in k[x,y,z] \mid rg \subset I \}$ for various $g$. It is a fact that all irreducible components of $I$ occurs in this fashion.

So let's compute $I:(z)$.

By nothing that $x^2(y^2-xz)+y(z^2-x^2y)=yz^2-x^3z=z(yz-x^3)$ is in $I$, we see that $yz- x^3 \in I:(z)$. It is also clear that $I \subset I:(z)$. In fact it is true that that $I:(z) = I+(yz-x^3)=(y^2-xz,z^2-x^2y,yz-x^3)$, but we don't need this.

We have found a candidate for another component if $I$. So we need to check that this is in fact a prime ideal. I'm not completely sure how to do this without a computer, we can note that the ideal $I:(z)$ is given by the minors of $$ \begin{bmatrix} z & y & x \\ x^2 & z & y \end{bmatrix} $$ And such ideals are very often prime.

Fredrik Meyer
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  • Sorry..I'm confused.I know that all irreducible components of $I$ are the prime ideals of the form $I:g$ for $g \in R$.But I don't understand why do you say we don't need $I:(z)$ ? – Arpit Kansal Oct 28 '15 at 03:00
  • One more question: I've never seen the representation of prime ideals in the form of minors.Can you please explain a little more(or some reference please)? – Arpit Kansal Oct 28 '15 at 03:03
  • I know that $I+(yz-x^3)$ is a prime ideal,because it's kernel of a ring morphism.Have a look here:http://math.stackexchange.com/questions/1487657/neat-way-to-find-the-kernel-of-the-ring-homomorphism – Arpit Kansal Oct 28 '15 at 03:08
  • @Arpit Sorry, the answer was a bit hastily written. To summerize: you had already found one component $I_1=(y,z)$. By luck I guessed that $I:(z)$ would be another. Instead of formally computing $I:(z)$ (because I'm not sure how), I guessed that it the form of $I_2$ (the other prime ideal). If $I_2$ was a prime ideal, we would be done, because we have found two prime ideals whose intersection is $I$. (and you seem to know that $I_2$ is a prime ideal, so we're done) – Fredrik Meyer Oct 28 '15 at 09:04
  • Sorry,I'm still confused.How did you compute/guessed the form of $I:(z)$ ? I can of course see that $I+(yz-x^3) \subset I:(z)$ but not the reverse inequality? – Arpit Kansal Oct 28 '15 at 14:55