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Let $R = \mathbb Z[X,Y]$. Construct an exact sequence of $R$-modules $$0 \to R\to R\oplus R \to R\xrightarrow{f} \mathbb Z \to 0,$$ where $f(g(X,Y)) = g(0,0)$ for all $g \in R$. Here $\mathbb Z$ is viewed as an $R$-module via $X\cdot 1 = Y\cdot 1 = 0$.

The only choice of $f$ is to actually define $f(g(x,y)) = g(0,0)$ for all $g \in R$, and it can easily be verified that this $f$ is surjective and is an $R$-module homomorphism. I am not sure how I can find the appropriate maps $f_1: R \to R \oplus R$ and $f_2 : R \oplus R \to R$ that makes the above sequence exact. I thought of defining the map $f_1$ to be $g \mapsto (g,0) \in R \otimes R$ and $f_2$ to be $(g,h) \mapsto xh(x,y)$, but this does not work since (for example) $y \in \ker f = \{ g(x,y) \in R \mid g(0,0) = 0 \}$ but $y \not\in Im f_2$. I get a similar result if I define $f_2$ to be $(g,h) \mapsto yh(x,y)$.

Is there anything I can do?

EDIT: Thanks to @morrowmh, here is a solution:

Let $f_1(1) := (y,x)$ and extend the definition of $f_1$ to the whole domain $R$ as to make $f_1$ an $R$-module homomorphism, i.e. $f_1(h(x,y)) = (h(x,y)y, h(x,y)x)$. Define $f_2((1,0)) := x$ and $f_2((0,1)) := y$ and extend the definition of $f_2$ to its whole domain $R \oplus R$ as to make $f_2$ an $R$-module homomorphism. Then we may check that the sequence thus defined is an exact sequence.

Squirrel-Power
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  • Can you give a more explicit description of the kernel of $f$? That may help you write down the map $R \oplus R \to R$, which is essential in coming up with the correct map $R \to R \oplus R$. – Alex Wertheim Feb 27 '24 at 03:49
  • This might help: https://math.stackexchange.com/questions/1854588/ker-phi-x-1-a-1-x-n-a-n-for-a-ring-homomorphism-phi-rx-1?noredirect=1&lq=1 – morrowmh Feb 27 '24 at 03:50
  • @AlexWertheim yes of course, $\ker f = { g(x,y) \in R \mid g(0,0) = 0 }$ – Squirrel-Power Feb 27 '24 at 07:38
  • @morrowmh thank you, I looked at the link you provided, but I still am not sure how to solve this question. That the kernel of $f$ is generated by $x$ and $y$ is evident, but how can make come up with an appropriate $f_1$ and $f_2$? – Squirrel-Power Feb 27 '24 at 07:41
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    @Squirrel-Power $R\oplus R$ is a free $R$-module with basis $e_1:=(1,0)$ and $e_2:=(0,1)$. By the universal property of free $R$-modules, the assignment $e_1\mapsto x$ and $e_2\mapsto y$ gives a well-defined $R$-module homomorphism $R\oplus R\stackrel{f_2}{\to}R$ such that $\text{im}(f_2)=\langle x,y\rangle=\ker(f)$. – morrowmh Feb 28 '24 at 00:24
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    Then compute (i.e. find a generating set for) $\ker(f_2)$ and play the same game again to define $f_1$. – morrowmh Feb 28 '24 at 00:31

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