Show that $ \operatorname{Hom}(K[s]/I, A)$ is isomorphic to $Z(I;A)$ We have a field $K$, and an ideal $I \subseteq K[s],$ $s=(s_1,..., s_n)$ . The exercise requires me to prove that for any $A$ algebra over $K$,
$Z(I; A) = \operatorname{Hom}(K[s]/I, A),$ and here $Z(I; A)$ is the zero set of $I$ ideal in $A^n$.
I understand what zeros are for $I$ in a ring but what does it mean for an ideal to have zeros in an algebra?
 A: Recall the universal property of the polynomial ring $K[X_1,...,X_n]$. It says, given a $K$-algebra $A$ and a function $\phi:\{1,...,n\} \rightarrow A$ there is a unique $K$-algebra homomorphism $\Phi: K[X_1,...,X_n] \rightarrow A$ satisfying $\Phi(X_i) = \phi(i)$. Hence we can evaluate a polynomial $f\in K[X_1,...,X_n]$ at a point $(a_1,...,a_n)\in A^n$ by considering the image of $f$ under the map $\Phi$ induced by the map $\phi(i)=a_i$. So in other words every point $a\in A^n$ induces a unique homomorphism
$$\begin{array}{rcl}
\operatorname{ev}_a: K[X_1,...,X_n] & \longrightarrow & A\\
f & \longmapsto & f(a_1,...,a_n)
\end{array}$$
and every homomorphism $\Phi: K[X_1,...,X_n] \rightarrow A$ yields a point $(\Phi(X_1),...,\Phi(X_n))\in A^n$. Thus we have mutually inverse maps
$$\begin{array}{rcl}
\operatorname{Hom}(K[X_1,...,X_n],A) & \longleftrightarrow & A^n\\
\Phi & \longmapsto & (\Phi(X_1),...,\Phi(X_n))\\
\operatorname{ev}_a & \leftmapsto & a=(a_1,...,a_n)
\end{array}$$
Now if we restrict this bijection to
$$Z(I;A)=\{(a_1,...,a_n) \in A^n \mid \forall f\in I: f(a_1,...,a_n)=0\}$$ on the right side, which set of homomorphisms do we get on the left? For this we will need another universal mapping property, namely that of the quotient ring. Let $I$ be an ideal in some ring $R$. Given any $R$-module $M$ we have that any morphism $\psi: R \rightarrow M$ satisfying $\psi\vert_I = 0$ induces a unique morphism $\Psi: R/I \rightarrow M$ satisfying $\Psi \circ \pi = \psi$ (where $\pi: R \rightarrow R/I$ denotes the quotient map). In other words we have mutually inverse maps
$$\begin{array}{rcl}
\operatorname{Hom}(R/I, M) & \longleftrightarrow & \{\psi \in \operatorname{Hom}(R,M)\mid \psi\vert_I =0\}\\
\Psi & \longmapsto & \Psi\circ \pi\\
\Psi & \mapsfrom & \psi
\end{array}$$
Can you see how this helps you with the exercise?
