Let $J \subset k[x_1,\dots,x_n], V(J) = Z \in \mathbb{A}_k^n$, and $A(Z) := k[x_1,\dots,x_n] /J$. Then $f \in A(Z)$ becomes a function $f: Z \to k^1$.

Question

Here $V(J)$ is the vanishing locus of $J \subset k[x_1,\dots,x_n]$. The above is a remark from my lecture that I can’t wrap my head around. If $f \in A(Z)$, shouldn’t $f$ be a coset of $J$ in $k[x_1,\dots,x_n]$, i.e. $f = p + J$, for $p \in k[x_1,\dots,x_n]$? Then how can it be a function?

Also my professor has the habit of writing $J \subset k[x_1,\dots,x_n]$. But technically, for $k[x_1,\dots,x_n]/J$ to make sense, I suppose $J$ must be an ideal of $k[x_1,\dots,x_n]$, and not just a subset?

Answer 1

1. Yes, $p+J\in k[x_1,\dots,x_n]/J$ is technically a coset. It is the subset of $k[x_1,\dots,x_n]$ consisting of all polynomials that can be written as $$p+q$$ where $q\in J$. Now, $V(J)$ is the vanishing locus of $J$, i.e. every $q\in J$ evaluates to $0$ on every point of $V(J)$. Thus, $p+q$ evaluates to whatever $p$ evaluates to, and consequently $p+J$ defines a function on $V(J)$.

2. Yes, $J$ is an ideal. It's not uncommon to just write $\subset$ since there is little risk of confusion, especially if the ideal is denoted by a common letter $(I,J,\mathfrak{p}$ etc).