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Let $A_{0}=k[x,y]$, $\mathfrak{m}=(x,y)$. Let $A=(A_{0})_{\mathfrak{m}}$. We wish to compute the characteristic polynomial, $\chi_{\mathfrak{q}}$, of the $\mathfrak{m}$-primary ideals (i) $(x,y)$, (ii) $(x,y^2)$, (iii) $(x,y)^2$ and finally check that the degrees of each polynomial are equal.

I am only really familiar with char. poly. in linear algebraic terms, i.e., Jordan canonical form computations. I read through Atiyah-Macdonald, but I did not find it terribly helpful in direct computation. How might I do this problem? Any help would be appreciated!

user 3462
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Well, it seems that $\chi_{\mathfrak q}(n)=l(A/\mathfrak q^n)$ for $n\gg0$.

In general, it is not an easy task to compute these lengths! That's why the theory reduces the finding of the degree from the general case to the most simple one when $\mathfrak q=\mathfrak m$. (However I won't exclude the possibility to perform such computation in your particular cases.) In your case this is easy to find since you know that $\mathfrak m^n/\mathfrak m^{n+1}$ is a $A/\mathfrak m\cong k$-vector space having a basis with $n+1$ elements, so $\chi_{\mathfrak m}(n)=n(n+1)/2$ is a polynomial of degree $2$ for $n\gg 0$. (But, of course, you know this from Krull-Chevalley-Samuel theorem which says that the degree of the characteristic polynomial equals the Krull dimension of $A$.)

  • Ah, we can also see that the Hilbert polynomials for the three ideals are equivalent. Namely, let $\mathfrak{m}=(x,y)$, $\mathfrak{g}=(x,y^2)$, $\mathfrak{h}=(x,y)^2$. Then we have $l(\mathfrak{g_{1}})=1$, $l(\mathfrak{g_{n}})=l(\mathfrak{m_{n}})$ for $n \geq 2$ and $l(\mathfrak{h_{1}})=0$, $l(\mathfrak{h_{2}})=2$, $l(\mathfrak{h_{n}})=l(\mathfrak{m_{n}})$ for $n \geq 3$. Then our ideals have the same Hilbert polynomial, i.e., $H(n)=n+1$ and the degrees of our char. polys. should be equal. – user 3462 Dec 04 '13 at 19:10