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I am following the notes of Gathmann to learn myself about commutative algebra. I have the following exercise written in them (without solutions at the end):

Exercise 1.13. Show that the equation of ideals $$(x^3-x^2,x^2y-x^2,xy-y,y^2-y) = (x^2,y)\cap(x-1,y-1)$$ holds in the polynomial ring $\mathbb{C}[x,y]$. Is this a radical ideal? What is its zero locus in $\mathbb{A}^2_{\mathbb{C}}$?

While the two last questions are easy to solve: it is not radical because $x(x-1)$ is not in the ideal but $(x(x-1))^2 = x^2(x-1)^2 = (x-1)(x^3 - x^2)$, so is in the ideal, and its zero locus is obviously $\{(0,0),(1,1)\}$.

The inclusion $\subseteq$ is easy too, because the product of ideals is included in the intersection. But of the $\supseteq$ part, I suppose that I have an element of the intersection, so a poynomial $p$ such that there exists $p_1,p_2,p_3,p_4 \in \mathbb{C}[x,y]$ such that: $$p = p_1x^2 + p_2y$$ $$p = p_3(x-1) + p_4(y-1)$$

I tried to add them or but them equal, without success. The only thing I can prove is that $p^2$ is in the ideal, by multiplying the two writings, but the ideal is not radical, so it doesn't work either!

Someone can give me a hint or help me please?

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  • $\begingroup$ Consider the ideal $I=(zx^2,zy, (1-z)(x-1), (1-z)(y-1))$. The intersection of $(x^2,y)$ and $(x-1,y-1)$ are the elements of $I$ that don't depend on $z$. Assume that any term containing $z$ is larger than any term non containing it. Get a new set of generators of $I$ that is a Groebner basis in that monomial order. Then the generators not depending on $z$ generate the intersection. $\endgroup$ – logarithm May 6 at 19:25
  • $\begingroup$ That solution seems good, but even if I know about Groebner basis from a class I followed this semester, I have not see them in the notes I am reading, so there is probably an easier solution (yours looks like elimination theory, but the only thing I know about it is the Wikipedia page so I would like not use it...) $\endgroup$ – eti902 May 6 at 19:33
  • $\begingroup$ It is just polynomial long division and the basis of computing all these operations between ideals. You can just compute without ever saying the name. $\endgroup$ – logarithm May 6 at 19:37
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    $\begingroup$ Hint. Use this: if $I+J=R$, then $I\cap J=IJ$. $\endgroup$ – user26857 May 6 at 20:15
  • $\begingroup$ Yes that's perfect, thank you! $\endgroup$ – eti902 May 7 at 3:22

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