So, given a field $K$ and a polynomial $f(x,y)$ in the ring of polynomials $K[X,Y]$, I am trying to understand why the following statement is true:
If $f(a,b) = 0$ for $(a,b) \in K \times K$, then $(f) \subset (x-a,y-b)$. I know that roots for a polynomial $g(x)$, give linear factors, but that's not true for polynomials over more than one indeterminates, is it?
A useful trick is to think about the isomorphism $K[x,y] \cong K[x][y]$, i.e.
a polynomial in $x$ and $y$ can be thought of as a polynomial in $y$ with coefficients that are polynomials in $x$.
For a polynomial $g(x)$, we can use the division algorithm to write $g(x) = q(x)(x-a) + r$, where $r$ is a constant. This shows $g(a) = 0$ iff $r=0$ iff $x-a$ divides g.
For a polynomial $f(x,y)$ we can do the same thing, using our trick. Thinking of $f(x,y)$ as a polynomial in $y$ with coefficients in $x$, the $r$ above is now a constant in $K[x][y]$, i.e. a polynomial in $x$. Thus we can write $f(x,y) = q(x,y)(y-b) + r(x)$. Now dividing $r(x)$ by $(x-a)$ we have that $f(x,y) = q(x,y)(y-b) + s(x)(x-a) + t$, with $t \in K$. We see that $f(a,b) = 0$ iff $t=0$.
So we have actually proved the more precise statement that $$(f) \subset (x-a)K[x] + (y-b)K[x,y]$$
Indeed this argument generalizes perfectly well to any amount of variables and any commutative ring with identity.
Hint: Polynomials of 2 variables are linear combinations of $1,\, x,\, y,\, xy,\, x^2,\, y^2,\, x^2y, \dots$.
But instead, we could also choose a basis as $(x-a)^n(y-b)^m$, $\, m, n\ge0$.
By a change of variables we may assume $a=0, b=0$. Then $f(0,0)=0$ is equivalent to saying that the coefficient of $x^0 y^0$ in $f$ is $0$, and it's easy to write $f$ in the form $a(x,y) x + b(x,y) y$. If $f$ is contained in the ideal generated by $x$ and $y$, so is the ideal generated by $f$.
