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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

user26857
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2 Answers2

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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 $f \in \langle x-a, y-b \rangle$.

Alex Wertheim
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  • Well , so though $F[x,y]$ is not a ED , we can use division algorithm with the quotient polynomial $y-b$ as its leading coefficient is a unit and since $F[x,y]$ is a polynomial ring over $F[x]$ which is a commutative ring with unity , we are exploiting that ... –  May 07 '16 at 08:38
  • That's right. I used ''Euclidean division'' to mean this, since it's the same division procedure, but I'll edit the post for clarity. – Alex Wertheim May 07 '16 at 18:21
  • @AlexSir,can you recommend me some books where I can learn this stuff? Every ring theory book mentions only single variable case.It does not go to several variable rings. – Koushik_halder Dec 23 '22 at 16:18
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Hint: You can prove it first for $a=b=0$ and use then the ring isomorphism

$$\phi:F[x,y] \rightarrow F[x,y], f(x,y)\mapsto f(x-a, y-b)$$

to conclude the general case.

Severin Schraven
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