Homomorphisms from quotient $R$ module to the base ring Let $R=k[x,y]$. Let $M=k[x,y]/(x,y)$ where $k$ is a field as an $R$-module.
Would $\text{Hom}_R(M,R) \cong \{0\}$?
If so, how do I proceed to show this?
I know I need to find all $R$-module homomorphism $\phi:k[x,y]/(x,y) \to k[x,y]$. I guess I should consider the image of $\overline{1}$ under this map. Then would $\phi(\overline{x})=x\phi(1)=0 \implies \phi(1)=0 \implies \phi=0$?
 A: So first of all, if $I = (x,y)$ then $k[x,y]/I \cong k$. Every polynomial $f(x,y)$ can be written as its constant term, $f(0,0)$ plus an element of $I$. Therefore $f(x,y) + I = f(0, 0) + I$ and simply evaluating at $x = 0$ and $y = 0$ gives an isomorphism between $k[x,y]/I$ and $k$.
But even putting that aside, it is theorem that homomorphisms from $A/B \to C$ correspond one to one with homomorphisms $A \to C$ which are $0$ on all of $B$.
So here, a homomorphism $k[x,y]/I \to k[x,y]$ corresponds to a homomorphism $\phi : k[x,y] \to k[x,y]$ such that $\phi(x) = \phi(y) = 0$.
And ok, so $x$ and $y$ go to $0$ so what about the constants? Well, what would be a homomorphism from $k \to k[x,y]$? This is a $k[x,y]$-morphism but it is also a $k$-morphism and $k$-modules are just vector spaces. So what sort of vector space homomorphisms are there from $k \to k[x,y]$. And like you say, those morphisms are entirely determined by $\phi(1)$.
Let's say $\phi(1) = g$. Then there is certainly a $k$-linear map $k \to k[x,y]$ where $\phi(t) = tg$. We should also check when this is $k[x,y]$-linear. Meaning, we need $0 = \phi(x) = x\phi(1) = 0$ (and the same with $y$). So $x\phi(1) = 0$ and therefore $\phi(1) = 0$.
