Why is $(X,Y)$ not a projective $\mathbb{Z}[X,Y]$-module?

Question

This was an exam problem I had which stumped me. The question was to prove that the ideal generated by $X$ and $Y$ in $\mathbb{Z}[X,Y]$ is not a projective $\mathbb{Z}[X,Y]$-module.

I was trying to exhibit an exact sequence with fourth term $(X,Y)$ which did not split, but hit a dead end.

With some heavy machinery, you can prove that the only projective/invertible ideals in $\mathbb{Z}[x,y]$ are principal ideals. See this post http://math.stackexchange.com/questions/76957/modules-over-kx-y — Prahlad Vaidyanathan, May 10 '14 at 07:29
@PrahladVaidyanathan, that post is about $k[x,y]$ with $k$ a field. — Mariano Suárez-Álvarez, May 10 '14 at 08:09
@MarianoSuárez-Alvarez : Read the "Edit" portion of the answer. It refers to UFDs in general. — Prahlad Vaidyanathan, May 10 '14 at 08:48

score 7 · Accepted Answer · edited May 10 '14 at 07:54

7

Let $R=\mathbb Z[X,Y]$ and $I=(X,Y)$. There is a short exact sequence of left $R$-modules $$0\to R\xrightarrow{f} R\oplus R\xrightarrow{g}I\to 0$$ with $f(a)=(Ya,-Xa)$ and $g(a,b)=aX+bY$ for all $a$, $b\in R$.

Check that it doesn't split.

edited May 10 '14 at 07:54

user26857

52,094

answered May 10 '14 at 06:30

Mariano Suárez-Álvarez

135,076

2

Thanks, I tried this: Let $s\colon R\oplus R\to R$ be such that $sf=1_R$. Suppose $s(1,0)=p$ and $s(0,1)=q$ for some polynomials $p$ and $q$. Then $s(Y,0)=pY$ and $s(0,-X)=-qX$ since $s$ is a module homomorphism. Then $s(Y,-X)=pY-qX=sf(1)=1$. But this is impossible since $pY-qX$ has zero constant term. Is this correct? – YN Chew May 10 '14 at 17:36

score 4 · Answer 2 · answered May 10 '14 at 06:33

Let's write $R=\mathbf Z[X,Y]$, $I=(X,Y)$. Two strategies:

First strategy: Let $M$ be the free $R$-module of rank $2$, with generators $u$ and $v$. Consider the exact sequence

$$0 \to K \to M \to I \to 0 $$

where $u \mapsto X$, $v \mapsto Y$, and $K$ is the kernel. Prove that $K$ is free of rank $1$, generated by $Yu-Xv$. Then prove that the sequence is not split on the left. (In a splitting $M\to K$, where could $u$ and $v$ go? Look at degrees of polynomials...)

Second strategy: Prove that $I$ is not flat. Can you find an ideal $J\subseteq R$ such that the natural map $J\otimes_R I \to I$ is not injective?

Thanks Bruno, is this idea like this: The generated module by $Yu-Xv$ is in the kernel. On the other hand, suppose $pu+qv$ is in the kernel. Then $pX+qY=0$. Then $pX=-qY$, so I think I can write $p=p'Y$ and $q=q'X$ for polynomials $p'$ and $q'$. Then $p'YX=-q'XY$, so $p'=-q'$. But then $pu+qv=p'Yu+q'Xv=p'Yu-p'Xv=p'(Yu-Xv)$? Then I think the rest follow by my comment on Mariano's answer. — YN Chew, May 10 '14 at 20:41

score 1 · Answer 3 · answered Feb 06 '20 at 10:10

1

It is not flat because $I\otimes I\to I$ not injective because $X\otimes Y-Y\otimes X\neq0$ but $XY-YX=0$. For $X\otimes Y-X\otimes Y\neq0$ I hope the argument here works well for $\mathbb{Z}$ instead of $k$.

answered Feb 06 '20 at 10:10

Gabriel Soranzo

2,770
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Why is $(X,Y)$ not a projective $\mathbb{Z}[X,Y]$-module?

3 Answers3