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.


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 ^_^

  • 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
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    $\begingroup$ 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

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