Prove that $\mu ((X,Y)^n (X+1, Y+1)^m)=(n+1)(m+1)$

Question

Let $k$ be a field (if it helps, we can think $k$ algebraically closed and/or of characteristic $0$), and consider the polynomial ring $k[X,Y]$. For an ideal $I$ of $k[X,Y]$ I denote by $$\mu(I) = \min \{ n \ge 1 : \exists x_1, \dots x_n \mbox{ generators of } I\}$$ the minimum cardinality of systems of generators of $I$.

I am interested in proving for all $n,m \ge 1$ the equality $$\mu ((X,Y)^n (X+1, Y+1)^m)=(n+1)(m+1)$$

What I know:

1. $\mu((X,Y)^n) = n+1$ : I localize $k[X,Y]_{(X,Y)}$ and find a lower bound $\mu((X,Y)^n) \ge n+1$ using Nakayama's lemma. On the other hand $\{X^n, X^{n-1}Y, \dots,Y^n \}$ is a system of generators of $(X,Y)^n$ formed by $n+1$ elements, so $\mu((X,Y)^n) \le n+1$.

2. Similarly I have shown $\mu((X+1, Y+1)^m)=m+1$

3. Putting together 1. and 2. I have $$\mu ((X,Y)^n (X+1, Y+1)^m) \le (n+1)(m+1)$$

4. Using Macaulay2 (although I know very little about this software), I found out that the formula is correct for large values of $n,m$, (if I didn't make mistakes), so I believe that it is correct.

What I don't know: everything concerning Groebner bases and stuff like that.

Please can you give me a way to prove this equality?

Answer 1

The equality is wrong. In your situation the correct number is $\max\{m,n\}+1$. The quickest proof is by appealing to Serre's codimension two trick. If $J\subset k[x,y]=R$ is a height two ideal, then $\mu(J)=\mu(\mathrm{Ext}^1(J,R))+1$ which in turn is same as $\mu(\mathrm{Ext}^2(R/J,R))+1$. In your case, since $(x,y)$ and $(x+1,y+1)$ are comaximal, Chinese remainder theorem will give us the estimate I said.

Just to illustrate this, let me look at the case for simplicity, where $n=m$. Then $(x,y)=(x-y,y)$ and $(x+1,y+1)=(x-y, y+1)$. Thus, $(x,y)^n(x+1,y+1)^n=((x-y, y)(x-y,y+1) )^n=(x-y, y(y+1))^n$, which is $n+1$ generated.

This is to answer Georges comment. I will assume and it is clear that $\mu(\mathrm{Ext}^1(J,R))=\mu(\mathrm{Ext}^2(R/J,R))$. If this number is $d$, then easy homological algebra says that there exists an element $\eta\in \mathrm{Ext}^1(J,R^d)$ such that the corresponding extension, $0\to R^d\to P\to J\to 0$ has the property, when we apply $\mathrm{Hom}(*,R)$, we get a surjection $\mathrm{Hom}(R^d,R)\to \mathrm{Ext}^1(J,R)$. This implies by ext calculations that $P$ is a projective module of rank $d+1$ and since it is free (by Seshadri's theorem), $\mu(J)=d+1$.

If $J=IL$, where $I,L$ are height two comaximal ideals as in the above question, then $R/J=R/I\oplus R/L$ by Chinese remainder theorem. Thus $\mathrm{Ext}^2(R/J,R)=\mathrm{Ext}^2(R/I,R)\oplus \mathrm{Ext}^2(R/L,R)$. Since $I,L$ are comaximal, easy to check that $\mu(\mathrm{Ext}^2(R/J,R))=\max\{ \mu(\mathrm{Ext}^2(R/I,R)),\mu(\mathrm{Ext}^2(R/L,R))\}$.

Finally, for the above question, writing $u=x-y$, I claim the ideal in question is generated by (assuming say $n\geq m$) the following. $u^ky^{n-k}$ for $n\geq k\geq m$ and $u^ky^{n-k}(y+1)^{m-k}$ for $0\leq k<m$. These are contained in the given ideal is obvious and it is equal can be checked by inverting $y$ and $y+1$.