Consider the affine plane over $k$, i.e. Spec $k[x,y]$. There are three kinds of prime ideals: $(0)$, $(x-a,y-b)$, and $(f(x,y))$, for $f$ irreducible. Let the ideal $(f(x,y))$ correspond to the point $\eta$. Standard algebraic geometry textbooks say that the closure of $\eta$ consists of $\eta$ together with all closed points $(a,b)$ s.t. $f(a,b)=0$. In particular, if $f(a,b)=0$, then $(x-a,y-b)$ contains $(f(x,y))$.
Could someone explain in detail why this is true? I have never seen an explicit explanation, and this fact does not strike me as completely obvious.
Since $\dim k[x,y]/(f)=1$ you know the primes properly lying above $(f(x,y))$ are maximal. But, since $k=\overline{k}$ all maximal ideals are of the form $(x-a,y-b)$. Since $(x-a,y-b)$ is the kernel of the map $k[x,y]\to k:g\mapsto g(a,b)$ we see that $(x-a,y-b)\supseteq (f(x,y))$ if and only if $f(a,b)=0$.
This should intuitively be what must happen since the whole intuition of (plane) curves is that the points of $V(f)$ are in correspondence to the maximal ideals of $k[x,y]/(f)$. But the points of $V(f)$ are just $(a,b)$ where $f(a,b)=0$, and the max ideals of $k[x,y]/(f)$ are just $(x-c,y-d)/(f)$ for some $c,d$. Thus, the $c,d$ should be precisely the points $(c,d)\in V(f)$ which are precisely the $(c,d)$ such that $f(c,d)=0$.