Two polynomials $f,g \in K[x,y]$ ring. Prove that $K[x,y]/(f,g)$ is finite dimensional vector space 
Let $f,g \in K[x,y]$ be polynomials with no common factor. Prove that $K[x,y]/(f,g)$ is a finite dimensional vector space.

I know there are non-zero (this word is correct?) $r(x)$ and $s(x)$  in the ideal $(f,g)$. And i think that the quotient $K[x,y]/(r(x),s(y))$ might have finite dimension; but I truly don't know how to prove this.
 A: Here's a complete proof. First, regard $f$ and $g$ as elements of $K(x)[y]$. This is a PID, so there exist $a, b \in K(x)[y]$ such that $af + bg = 1$. Clearing denominators, we conclude that there is some polynomial in $(f, g)$ of the form $r(x)$. Similarly, by regarding $f$ and $g$ as elements of $K(y)[x]$, we conclude that there is some polynomial in $(f, g)$ of the form $s(y)$. Now Mohan's argument shows that $K[x, y]/(f, g)$ is finite-dimensional because it is a quotient of $K[x, y]/(r, s)$, which has dimension $\deg r \deg s$.
Geometrically, $K[x, y]/(f, g)$ is the ring of functions on the scheme-theoretic intersection of the curves cut out by $f$ and $g$ in the affine plane $\mathbb{A}^2$, which (as long as $F$ and $g$ have no common factors) we expect to be a finite set of possibly "fat" points (e.g. $K[x, y]/(f, g)$ could have nilpotents; take, for example, $f = y - x^2, g = y + x^2$). The fact that $(f, g)$ contains $r(x), s(y)$ reflects (up to some "fat") the fact that these points only take finitely many $x$ and $y$ coordinates.
A: If $r(x),s(y)$ are non-constant polynomials, then it is easy to show that $k[x,y]/(r,s)$ has finite dimension. Since $k[x]/r(x)$ has dimension $\deg r(x)$ and $k[x]/(r(x)[y]/s(y)$ has a basis with dimension $\deg s(y)$ over $k[x]/r(x)$, we have $\dim k[x,y]/(r(x),s(y))=\deg r(x)\deg s(y)$, we have what we need.
