Prove that $\mathbb{Q}[x,y]/ \langle x+y \rangle \cong \mathbb{Q}$

Question

My direction: Set mapping $\varphi:\mathbb{Q}[x,y] \to \mathbb{Q}$ that $\varphi$ is surjective homomorphic and $\ker{\varphi}=\langle x+y \rangle$. Since Noether Theorem, I have $\mathbb{Q}[x,y]/\langle x+y \rangle \cong \mathbb{Q}$.

I have try many mapping $\varphi$ but not working. In my attempt to prove this problem, I have found that $\mathbb{Q}[x,y]/\langle x,y \rangle \cong \mathbb{Q}$ by set mapping $\varphi: \mathbb{Q}[x,y]\to \mathbb{Q}, \varphi(f)=f(0,0)$.

Answer 1

The claim is false as stated. We actually have $\Bbb{Q}[x,y]/\langle x+y\rangle\simeq \Bbb{Q}[x]$. The isomorphism follows from the fact that the kernel of the surjective homomorphism $$ \phi:\Bbb{Q}[x,y]\to \Bbb{Q}[x], f(x,y)\mapsto f(x,-x) $$ is the ideal generated by $x+y$.

• $\phi$ is surjective, because $\phi(g(x))=g(x)$ for all polynomials $g(x)$.
• The kernel is generated by $x+y$, because we can view $\Bbb{Q}[x,y]$ as $\Bbb{Q}[x][y]$. That is, the ring of polynomials in $y$ with coefficients coming from the integral domain $R=\Bbb{Q}[x]$. Viewed this way, we see that $\phi$ amounts to evaluating the polynomials at the point $y=-x$. The kernel of an evaluation homomorphism $R[y]\to R$ at the point $y=a$, $a\in R$, is more or less automatically generated by $y-a$. A proof comes from the polynomial divison algorithm. A polynomial $f(y)\in R[y]$ can be written in the form $f(y)=q(y)(y-a)+f(a)$ for some polynomial $q(y)\in R[y]$. Observe that the leading coefficient of $y-a$ is equal to $1$. When $R$ is not a field (as is the case here) we need that assumption for the usual long division of polynomials to work.

Answer 2

As stated in the first answer, the claim is false.

What is true is that $\mathbb{Q}[x,y]/\langle x,y\rangle\cong\mathbb{Q}$. You can prove this by considering $\varphi:\mathbb{Q}[x,y]\to\mathbb{Q},\;\varphi(f)=f(0,0) $. To compute $\ker(\varphi)$, note that $x,y\in\ker(\varphi)$, so $\langle x,y\rangle\subset\ker(\varphi)$. On the other hand, if $f\in\mathbb{Q}[x,y]$ with $f(0,0)=0$, then $f(x,y)$ is of the form $\sum_{i,j}q_{i,j}x^iy^j$ where $i+j>0$.

Edit: As Ennar points out at the comments: a very smart way to see that $\mathbb{Q}[x,y]/\langle x+y\rangle$ cannot be isomorphic to $\mathbb{Q}$ is the following: We know that a ring modulo an ideal is isomorphic to a field if and only if the ideal is maximal. But $\langle x+y\rangle\subset \langle x,y\rangle\subset\mathbb{Q}[x,y]$ and these inclusions are all strict, hence $\langle x+y\rangle$ is not maximal, thus our quotient cannot be a field, while $\mathbb{Q}$ is a field.