Confusion free resolution

Question

I am studying the following free resolution. $R=K[x,y]$ with $K$ a field.

$$0\longrightarrow R\stackrel{\begin{pmatrix} y\\-x \end{pmatrix}}\longrightarrow R^2 \stackrel{\begin{pmatrix} x & y \end{pmatrix}}\longrightarrow R\longrightarrow R/(x,y)$$

I don't understand the first map. We construct a map $K[x,y]\to K[x,y]\oplus K[x,y]$ by sending $1\to (y,-x)$. But when we construct a map from a polynomial ring, don't we need to specify the image of $1$, $x$ and $y$ ?

Answer 1

This sequence of linear maps is very general and has nothing to do with indeterminates: it also makes sense in any ring $R$, for arbitrary elements $x,y\in R$. It simply happens that $R$ is here a polynomial ring in two indeterminates.

So, explicitly the linear map is here \begin{align} K[x,y]&\longrightarrow K[x,y]\times K[x,y], \\ F(x,y)&\longmapsto\bigl(yF(x,y),-xF(x,y)\bigr). \end{align}

Answer 2

We need to specify the image of $1,x,y$ when we consider a map from a polynomial ring as a ring. Here it is a module homomorphism and as module it is generated by $1$.