# 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 $$I$$?

• No, that's not a correct definition of a maximal ideal. What you say only happens if $R$ is a local ring, and $I$ is the unique maximal ideal of $R$. Commented May 20, 2022 at 23:28
• Find a homomorphism out of $\mathbf Q[x,y]$ onto a field with kernel $I$. Hint: at what point in $\mathbf Q^2$ do all elements of $I$ vanish?
– KCd
Commented May 20, 2022 at 23:40
• A more general result: for every field $K$ and every $(c_1,\ldots,c_n)$ in $K^n$, the ideal $(x_1-c_1,\ldots,x_n-c_n)$ in $K[x_1,\ldots,x_n]$ is maximal.
– KCd
Commented May 20, 2022 at 23:43

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 $$I$$.

• +1. The intuition behind this approach is that quotienting by $\langle y \rangle$ “sets $y$ equal to $0$” and quotienting by $\langle x-1 \rangle$ “sets $x$ equal to $1$”. Then it’s just a matter of turning this intuition into an actual homomorphism, and by the first isomorphism theorem, into an isomorphism. Commented May 21, 2022 at 0:32

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.

• Note that you are guaranteed a field when you mod out by irreducible polynomials because polynomial rings over fields are Euclidean domains. Commented May 21, 2022 at 18:37

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 $$I$$.