How to determine if the ideal $I = \langle x-1, y \rangle$ is a maximal ideal of $\mathbb{Q}[x, y]$ How do I determine whether $\mathbb{Q}[x, y] / I$ is a field?
Where I is generated by the Gröbner basis
$$
I = \langle x-1, y\rangle 
= \bigl\{ a(x,y)\,(x-1) + b(x,y)\,y \mid a,b \in \mathbb{Q}[x,y] \bigr\}
$$
I have a theorem that tells that the quotient ring $R/I$ is a field if and only if $I\subseteq R$ is a maximal ideal.
So I have to check whether $I\subseteq \mathbb{Q}[x,y]$ is a maximal ideal.
My understanding of a maximal ideal is that if $I\subset R$ is a proper ideal of R, then $I$ is a maximal ideal if any other proper ideal $J$ of $R$ is contained in $I$.
But how do I approach this problem? It doesn't seem easy to prove that all other proper ideals are subsets of $I$?
 A: show that the application
$$
f:\mathbb{Q}[x,y] \longrightarrow \mathbb{Q}
$$
given by $f(p(x,y))=p(1,0)$ is a surjective homomorphism with kernel $I$.
A: Since you're quotienting out by $y$, it gets sent to $0$ as an indeterminate so you can see that $$\mathbb{Q}[x,y]/(x-1,y)\simeq\mathbb{Q}[x]/( x-1)$$
You can then notice that, since you're modding out by $x-1$, you get $x=1$ by definition because modding out by $x-1$ is the same as saying $x-1=0$. We then notice that $$f:\mathbb{Q}[x]/(x-1)\longrightarrow
\mathbb{Q}(1)\simeq\mathbb{Q}$$
$$x\longmapsto 1$$
is clearly an isomorphism.
Following with KCd's comment above, one can generalize this. Let $F$ be a field. Then $$F[x_1,\ldots,x_n]/(x_1-a_1,\ldots,x_n-a_n)\simeq F$$
To see this, consider the map $$f:F[x_1,\ldots,x_n]/(x_1-a_1,\ldots,x_n-a_n)\longrightarrow F $$
$$(x_1,\ldots,x_n)\longmapsto(a_1,\ldots,a_n)$$
It's important to note that the maximal ideals in polynomial rings over algebraically closed fields are of the form $(x_1-a_1,\ldots,x_n-a_n)$. If the field isn't algebraically closed, you can get maximal ideals that are not of this form. For example $\mathbb{R}[x]/(x^2+1)\simeq \mathbb{C}$. So maximal if and only if of the form $(x_1-a_1,\ldots,x_n-a_n)$ only holds in algebraically closed fields. Of the form $(x_1-a_1,\ldots,x_n-a_n)$ implies maximal regardless of the field.
A: Here are two ways to go about this:
The first is via the first isomorphism theorem, consider the homomorphism $\phi : \Bbb Q[x,y]\to \Bbb Q$ defined by $\phi(p(x,y))=p(1,0)$ and compute its image and kernel (this is also the answer given by Thiago)
The second is to show that there is no proper ideal other than $I$ that contains $I=\langle x-1,y\rangle$ "directly".
Suppose $I\subsetneq J\trianglelefteq \Bbb Q[x,y]$ and let $p(x,y)\in J\setminus I$. Write
$p(x,y)=q_1(x,y)\cdot y+r_1(x)$ for some polynomials $q_1(x,y)\in \Bbb Q[x,y], r_1(x)\in \Bbb Q[x]$ (this is possible since $y$ is a monic polynomial over $\Bbb Q[x]$, or simply by grouping all summands which contain $y$), then by the polynomial division algorithm write $r_1(x)=q_2(x)\cdot (x-1)+r_2$, so we get:
$$
p(x,y)=q_1(x,y)\cdot y + q_2(x)\cdot(x-1) +r_2
$$
So $r_2\in J$, and since $p(x,y)\notin I$ we get $r_2\neq 0$ so it is invertible, so $J=\Bbb Q[x,y]$ (an ideal that contains an invertible element is the whole ring). Since we showed this for an arbitrary ideal $J$, we get there is no proper ideal other than $I$ that contains $I$.
