Is there an inclusion of ideals $(xy)\subset(x-a,y-b)$ in $K[x,y]$? Let $a,b\in K$ where $K$ is algebraically closed field. I'm trying to determine for which $a$ and $b$ there is an inclusion
$$(xy)\subset(x-a,y-b)$$
of ideals in $K[x,y]$.
Suppose that $a=0$. Given any $p(x,y)xy\in (xy)$ we see that $(p(x,y)y)x+0(y-b)\in(x,y-b)$ and by symmetry we have an inclusion if $a\neq0$ or $b\neq0$. If both are zero we have an inclusion as well.
The remaining case is when $a\neq0$ and $b\neq0$. I have an idea that there is not going to be an inclusion and here is why: If we assume by contradiction that $xy=f(x,y)(x-a)+g(x,y)(y-b)$ then if we consider the root $(a,b)$ for the right-hand side, we get an equation $ab=0$. If this is not a contradiction then it certainly seems like a hint that the inclusion does not hold, but I'm unsure how to proceed from here.
I would very much appreciate comments on this strategy/observation and if it does not yield anything about the inclusion I would also appreciate any hints that could lead one in the right direction. Thanks in advance.
 A: Your approach is fine, although it is a bit roundabout. First, it is worth noting that
$$(xy)\subset(x-a,y-b)\qquad\Leftrightarrow\qquad xy\in(x-a,y-b).$$
Now for your first case, with either $a=0$ or $b=0$, it suffices to see that then either $x\in(x-a,y-b)$ or $y\in(x-a,y-b)$, respectively, and hence $xy\in(x-a,y-b)$.
For the remaining case with $ab\neq0$, you correctly deduce that if $xy\in(x-a,y-b)$ then
$$xy=f(x,y)(x-a)+g(x,y)(y-b),$$
for some polynomials $f(x,y),g(x,y)\in K[x,y]$. Indeed plugging in $(a,b)$ yields
$$ab=f(a,b)(a-a)+g(a,b)(b-b)=0.$$
Now you are done; this directly contradicts the assumption that $ab\neq0$, so $xy\notin(x-a,y-b)$.

You could reverse the order of the two cases to get a slightly shorter variant:

Suppose $(xy)\subset(x-a,y-b)$. Then $xy\in(x-a,y-b)$ hence also
  $$xy-x(y-b)-b(x-a)=ab\in(x-a,y-b).$$
  Because $(x-a,y-b)\cap K=0$ this implies $ab=0$, so either $a=0$ or $b=0$.
Of course conversely, if either $a=0$ or $b=0$ then either 
  $$x\in(x-a,y-b)\qquad\text{ or }\qquad y\in(x-a,y-b),$$
  and hence $xy\in(x-a,y-b)$.

