Find kernel of induced map after tensoring 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.
 A: 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 ^_^
