3
$\begingroup$

Let $R=k[x,y]$ and $I=(x,y)$. Consider the map $$R^2 \xrightarrow{\phi:(f(x,y),g(x,y))\mapsto xf(x,y)+yg(x,y)}I$$

I am trying to show that after tensoring by $R/I$, the kernel of the induced map is isomorphic to $R/I$ which I am pretty sure it is.

$\endgroup$
5
$\begingroup$

Welcome to MSE!

This is actually false, and this problem is a great example of the power of abstract nonsense. Start with the exact sequence

$$ 0 \to R \xrightarrow{f \mapsto (yf, \ -xf)} R^2 \xrightarrow{(f,g) \mapsto xf + yg} I \to 0 $$

Now since tensoring is right exact, we get a new exact sequence

$$ R \otimes R/I \to R^2 \otimes R/I \to I \otimes R/I \to 0 $$

of course, we know how to compute tensor products with $R/I$, and we find

$$ R/I \to (R/I)^2 \to I \big / I^2 \to 0 $$

Lastly, we know $R = k[x,y]$ and $I = (x,y)$, so we can actually compute these quotients too.

$$ k \to k^2 \to (x,y) \big / (x,y)^2 \to 0 $$

Now, what are our maps?

Well we're viewing $k$ as $k[x,y] \big / (x,y)$. That is, the constant polynomials. So our old map $f \mapsto (yf, -xf)$ always outputs a pair of polynomials with $0$ constant term. This becomes the $0$ map from $k$ to $k^2$.

At this point we can stop, because we see that $k$ is not the kernel of the resulting map $k^2 \to (x,y) \big / (x,y)^2$. If we wanted to go further, though, we would see this map sends a pair of constant polynomials $(c_1, c_2) \mapsto c_1 x + c_2 y$. We have to quotient out any quadratic terms, but there aren't any! So we see this map is actually injective, and $k^2 \cong (x,y) \big / (x,y)^2$ as $R$-modules.

That is, unwinding all this, $R^2 \otimes R/I \cong I \otimes R/I$ as $R$-modules, and this isomorphism came from the induced map. So the kernel of the induced map is $0$ and not $R/I$.

There are faster ways to see this (using some algebraic geometry, for instance), but I think working things out like this is instructive. In general, you should reach for exact sequences, rather than the definition of tensor product, in basically every situation.


I hope this helps ^_^

$\endgroup$
4
  • 2
    $\begingroup$ No, in fact $\text{Tor}_1(I, R/I) = R/I$. Remember that we get an exact sequence $\text{Tor}_1(R^2, I) \to \text{Tor}_1(I, R/I) \to R \otimes R/I \to R^2 \otimes R/I \to I \otimes R/I \to 0$. We computed the right half of this sequence to be $k \to k^2 \to I/I^2 \to 0$, and $k \to k^2$ should be the $0$ map. So the map $\text{Tor}_1(R^2, I) \to k$ should be surjective to make the sequence exact. We also know that $\text{Tor}_1(R^2,I) = 0$ since $R^2$ is free, so $\text{Tor}_1(I, R/I) \to k$ should be injective. That tells us that $\text{Tor}_1(I, R/I) \cong R/I \cong k$. $\endgroup$ Nov 27 '21 at 21:31
  • 1
    $\begingroup$ Yes, I meant $\text{Tor}_1(I, R/I)$, good catch. As for your other questions, it might be worth putting those in a new question rather than in the comments here. $\endgroup$ Nov 27 '21 at 22:31
  • 2
    $\begingroup$ It is the same map as before, $f \mapsto (yf, -xf)$, but now we've quotiented out the ideal $(x,y)$, so every polynomial of degree $\geq 1$ becomes $0$. But $yf$ and $-xf$ both have $0$ constant term, so when we kill the higher degree polynomials, we get $0$ for both. Thus our map becomes $f \mapsto (0,0)$, which is the $0$ map. $\endgroup$ Nov 27 '21 at 22:34
  • 1
    $\begingroup$ Maybe it is worth adding that here $\otimes = \otimes_R$ (e.g. if $\otimes = \otimes_k$ then we get a different result since $R/I \simeq k$ is trivially $k$-flat). $\endgroup$
    – qualcuno
    Nov 27 '21 at 22:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.