Is there any way to show that for $a,b \in F$ , the ideal $\langle x-a , y-b\rangle$ is maximal in $ F[x,y]$ , by showing that the quotient $F[x,y]/\langle x-a , y-b\rangle$ is a field ? Is the quotient $F[x,y]/\langle x-a , y-b\rangle$ isomorphic with $F$ ? ( Here $F$ is a field )
I can show that the map $g : F[x,y] \to F$ given by $g( p(x,y))=p(a,b)$ is a surjective ring homomorphism with $\langle x-a , y-b\rangle \subseteq \ker g$ but I cannot show that the kernel is exactly that ; PLease help, Thanks in advance
There is a standard fix for this sort of thing. Let $f(x, y) \in \ker(g)$. Via the isomorphism $F[x, y] \cong (F[x])[y])$, view $f$ as a polynomial in $y$ with coefficients in $f$. Then while $F[x, y]$ is not a Euclidean domain, we can perform Euclidean division by the monic polynomial $y-b$ (since the leading coefficient of $y-b$ is a unit) to write $f$ as $f(x, y) = q(x, y)(y-b) + r(x, y)$ for some $q(x, y), r(x, y) \in F[x, y]$, where $r(x, y)$ has degree $< 1$ in $y$. In other words, $r(x, y)$ is a polynomial in $x$, which we denote simply by $r(x)$. Now, evaluating at $a, b$, we see that $f(a, b) = q(a, b)(b - b) + r(a) = 0$, i.e. $r(a) = 0$. Since $r(x) \in F[x]$, this implies $x-a \mid r(x)$, whence $f \in \langle x-a, y-b \rangle$.