Disclaimer: Though I have been re-reading my notes, and have scanned the relevant texts, my commutative algebra is quite rusty, so I may be overlooking something basic.

I want to show $\mathbb{Q} \simeq \mathbb{Q}[x,y]/\langle x,y \rangle$ is not projective as a $\mathbb{Q}[x,y]$ module. I've tried two methods, neither of which gets me to the conclusion.

I first tried what seems to be sort of standard when proving that something is not projective: show that the lifting of the identity yields a contradiction. So I let $\pi: \mathbb{Q}[x,y] \to \mathbb{Q}[x,y]/\langle x,y \rangle$ be my surjection given by $f \mapsto \bar{f}$ and the identity map is $id: \mathbb{Q}[x,y]/\langle x,y \rangle \to \mathbb{Q}[x,y]/\langle x,y \rangle$. So all I need to show is that a homomorphism $\phi: \mathbb{Q}[x,y]/\langle x,y \rangle \to \mathbb{Q}[x,y]$ such that $\pi \circ \phi =id$ does not exist. But if $$\pi(f) = \bar{f} = \overline{a_0+a_{10}x+a_{01}y+a_{11}xy+\cdots+a_{n0}x^n + a_{0n}y^n} = \bar{a_0}$$ then doesn't the map $\bar{a_0} \mapsto a_0$ work? After all, $$ (\pi\circ \phi)(\bar{a_0}) = \pi(a_0) = \bar{a_0} = id(\bar{a_0}).$$I was concerned at first about this not being well defined, but since every element of a particular coset has the same constant term, it does not depend on choice. So either I have already made a mistake, or this is just the wrong map from which to derive a contradiction.

The next thing I tried used a different characterization of projective modules: that $P$ is a projective $R$-module iff there is a free module $F$ and an $R$-module $K$ such that $F \simeq K\oplus P$. In our case, this means there is a free module $F$ and a $\mathbb{Q}[x,y]$-module $K$ such that $$ \mathbb{Q}[x,y] \oplus \cdots \oplus \mathbb{Q}[x,y] \simeq F \simeq K \oplus \mathbb{Q}[x,y]/\langle x,y \rangle \simeq K \oplus \mathbb{Q}.$$ From here, my concern is that I am waving my hand too much when I say: obviously this cannot be true, since every element of the LHS, which is a tuple of polynomials, cannot be broken up with one chunk in $K$ and the other in $\mathbb{Q}$. Do agree? If so, how can I make this argument more rigorous?

One more trouble: nowhere in either of these methods did I explicitly use that the polynomial ring here is only in two variables. The fact that the question did not use $\mathbb{Q}[x_1,\ldots,x_n]$ instead of $\mathbb{Q}[x,y]$ worries me.

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    $\begingroup$ You might try using the fact that projective modules are always flat. $\endgroup$ – user641 May 30 '12 at 1:19
  • $\begingroup$ Thanks again for the prompt and excellent answers: +1's all around. I accepted the one by @wxu because it was the first. $\endgroup$ – Derek Allums May 30 '12 at 22:42
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    $\begingroup$ Every submodule of $\mathbb{Q}[x,y]^I$ is infinite-dimensional over $\mathbb{Q}$ (by considering degree), and so every projective module, a summand of a free module, is also infinite-dimensional over $\mathbb{Q}$. $\endgroup$ – Joshua Mundinger Dec 14 '18 at 1:40
  • $\begingroup$ @JoshuaMundinger Thanks - been a few years since I've thought about this, but are you saying that if $M := \mathbb{Q}[x, y]/\langle x, y\rangle$ were projective, it'd be infinite dimensional over $\mathbb{Q}$, but of course this is false since $M \simeq \mathbb{Q}$? $\endgroup$ – Derek Allums Dec 14 '18 at 13:39

As Steve D said, you can use the fact that projective modules are always flat.

Consider the map $\mathbb{Q}[x,y]\to \mathbb{Q}[x,y]$ defined by multiplying $x$. This is an injective $\mathbb{Q}[x,y]$-module map, while tensoring $\mathbb{Q}$ will give an injective map, but it is NOT. So $\mathbb{Q}$ is not flat as $\mathbb{Q}[x,y]$-module.

However, following by your methods. The map defined in your first method is not a $\mathbb{Q}[x,y]$-module map, since if $1$ sends to $1$, then $0=\overline{x}$ sends to $x=0$, contradiction.

In your second method, you can always find an element $(0,1)$ in RHS, this element is a torsion element, $x(0,1)=0$. But there is no element in LHS corresponding to it, since every element in LHS is not torsion.

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  • $\begingroup$ +1 I like your proof that $\mathbb{Q}$ is not flat. Thanks for the comments on my attempts at solutions as well. As usual, the solution seems so obvious now. $\endgroup$ – Derek Allums May 30 '12 at 2:13

I'm denoting your ring with $R$ and your ideal with $I$. We'll just need that $R$ is an integral domain, and that $I$ is a nontrivial ideal.

If $R/I$ is a projective $R$ module, then the following exact sequence splits

$$ 0\rightarrow I\rightarrow R\rightarrow R/I\rightarrow 0 $$

But integral domains have no proper direct summands, so $I$ would have to be trivial (a contradiction).

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  • $\begingroup$ @rschwieb (1) So you actually showed that $R/I$ is never flat over $R$, where $R$ is an integral domain and $I$ is a non-zero ideal of $R$? (2) Can you please explain why an integral domain has no proper direct summands. $\endgroup$ – user237522 Jun 24 '15 at 15:24
  • $\begingroup$ @user237522 (1) I didn't mention anything about flatness. (2) If A and B are nonzero ideals of a domain, then $\{0\}\neq AB\subseteq A\cap B$, so you can't form a direct sum of the two ideals. $\endgroup$ – rschwieb Jun 24 '15 at 15:45
  • $\begingroup$ @rschwieb (1) Truly, it was a typo; I wanted to write "projective" not "flat", namely: $R/I$ is never projective over $R$, where $R$ is an integral domain and $I$ a non-zero ideal of $R$. If I am not wrong, I think I can prove the following: $R/I$ is never flat over $R$, where $R$ is a noetherian integral domain and $I$ a non-zero ideal of $R$. Reason: By a known criterion for flatness, flatness would imply $I=I^2$, but in a noetherian integral domain this would imply $I=0$ (since such $I$ must be generated by an idempotent of $R$). (2) Thank you very much. $\endgroup$ – user237522 Jun 24 '15 at 17:06
  • $\begingroup$ @user237522 (1) OK, yes, I have shown it for the projective case. (2) Interesting, I am not very well versed in flatness. I know in non-Noetherian domains you can have $I^2=I$, so your inclusion of Noetherianness is very necessary... $\endgroup$ – rschwieb Jun 24 '15 at 17:16

A projective module over a domain has no nonzero torsion element, since it is a submodule of a free module.
But every element of your module is a torsion element: it is killed by $x$.

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I believe the (injective) map $\bar{a_0} \mapsto a_0$ does not work, since it is not a map of $\mathbb Q[x,y]$-modules. In particular, we have that $x\bar{a_0}=0$ but $xa_0\neq0$ for every $a_0\in\mathbb Q[x,y]^{\oplus 1}$, and a $\mathbb Q[x,y]$-linear map would require that $x\bar{a_0}\mapsto xa_0$.

In any case, you should try the third method, which relies on the fact that if $0\to A\to B\to C\to 0$ is a short exact sequence of $R$-modules which splits in the sense that there is a map $C\to B$ that inverts the map $B\to C$, then in fact $B\cong A\oplus C$. This allows you to reformulate the first property of projective modules to say that any short exact sequences involving a projective module as a final term splits (the lift of the identity map gives the splitting).

Hence, all you need do is find an exact sequence of $\mathbb Q[x,y]$-modules $0\to A\to B\to\mathbb Q[x,y]/\left<x,y\right>\to0$ such that $B\not\cong A\oplus\mathbb Q[x,y]/\left<x,y\right>$.

The obvious choices are $A=\left<x,y\right>$ and $B=\mathbb Q[x,y]$, and this reduces the problem to showing that $\left<x,y\right>\oplus\mathbb Q[x,y]/\left<x,y\right>$ is not isomorphic to $\mathbb Q[x,y]$, i.e. is not free of rank $1$. This is easy as you take an arbitrary element in the direct sum and show that you cannot get both $1$ and $x$ by multiplying with elements of $\mathbb Q[x,y]$. If I am not mistaken, this should work for showing that for any domain $R$ and proper ideal $I$, $R/I$ is not projective as an $R$-module.

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  • $\begingroup$ +1 Regarding your first and last sentences: you're right, as also pointed out below by wxu and rschwieb, respectively. Thanks for the answer. $\endgroup$ – Derek Allums May 30 '12 at 2:19

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