Directly computing the ideal of a variety Let $k$ be an algebraically closed field, let $f \in k[x,y]$ be an irreducible polynomial, and let $C = V(f)$ be the variety it defines. Show directly (without using the Nullstellensatz) that $I(C) = (f)$.
The fact that $(f) \subset I(C)$ is clear, but I'm not sure how to show the other inclusion. I know that I need to show that for $g \in I(C)$, $g = hf$ for some $h \in k[x,y]$.
So, write $g = qf + r$ where all letters are polynomials. It will be enough to show $r = 0$. Since $g(x) = 0$ for all $x \in V(f)$, we see that $r(x) \in I(C)$. I'm not sure how to show that $r$ is identically zero though.
 A: I hope someone will post a nicer answer. But for now here is a slightly crude idea.
Suppose $g \in I(C)$, but $g \notin (f)$. In particular $g$ and $f$ have no common factor in $k[x,y]$—this is where we use irreducibility of $f$. (If $f$ and $g$ had a common factor, that common factor would have to be $f$, and we'd have $f \mid g$.) Consider $f, g \in k[x][y]$, i.e., as polynomials in $y$ with coefficients in $k[x]$. We have $k[x][y] \subset k(x)[y]$, which is a Euclidean domain since $k(x)$ is a field. And $g$ and $f$ still have no common factor in $k(x)[y]$, since if $p(x,y)/q(x)$ were a common factor of $g$ and $f$, then $p$ would be a common factor of $g$ and $f$ in $k[x,y]$.
So, in $k(x)[y]$, $(f,g)=1$. That means there are $p,q \in k(x)[y]$ such that $pf+qg=1$. Then clearing denominators, $p'f+q'g=d$, where $p',q' \in k[x,y]$ and $d \in k[x]$, and $d \neq 0$. Every point of $I(C)$ is a vanishing point of $f$ and $g$, hence also of $d$. So $I(C)$ is contained in the union of finitely many "vertical" lines, namely $V(D)$ (one line for each root of $d$).
Finally we use the closedness of $k$. First of all, for each $a \in k$, the vertical line $x=a$ must intersect $I(C)$, since the univariate polynomial $f(a,y)$ has roots in $k$. Second, $k$ is infinite; there are infinitely many "vertical" lines. This shows that $I(C)$ is not contained in any finite union of vertical lines. Contradiction.
Well, having written all that, I will say that it should be possible to simplify things if you can use resultants. The polynomial $d$ is a resultant of $f$ and $g$; you have to show that the resultant is not identically zero, which should not be too hard I think... (We got $d \neq 0$ because it is a common denominator for $p$ and $q$.)
