Is $(2,x,y)$ maximal ideal in $\mathbb Z[x,y]$?

Question

I’m new to ring theory and have difficulty in showing whether an ideal $I$ is a maximal ideal of commutative ring $R$ with unity mostly when $I$ is generated by more than one element.

For example: The ideal $(x)$ is prime in the ring $\mathbb Z[x]$.

Proof: The map $\phi: \mathbb Z[x]\to \mathbb Z $ given by $\phi (f(x))=f(0)$ is an onto ring homomorphism and has kernel $(x)$. So by first isomorphism theorem, $\mathbb Z[x]/(x)\simeq \mathbb Z$. It follows that $(x)$ is prime ideal of $\mathbb Z[x]$ ($\mathbb Z$ is an integral domain and not a field).

I get stuck in the following:

Is $(2,x,y)$ a maximal ideal of $\mathbb Z[x,y]$?

I don’t know.

I plan to use the theorem that $I$ is a maximal ideal in $R$ iff $R/I$ is a field.

By third isomorphism theorem, $\frac{\mathbb Z[x,y]/(2)}{\color{red}{(2,x,y)/(2)}}\simeq \mathbb Z[x,y]/(2,x,y).\tag 1$

I observe that the numerator in LHS is isomorphic to $\mathbb Z_2[x,y]$. But I have no idea how to simplify the red part in $(1)$. So how should I simplify the red part?

I would also request to let me know some source where this type of problems (where I is generated by more than 1 element or is even more complicated than that) are explained. Thanks.

Answer 1

$\Bbb Z[x,y]/(2,x,y)\cong \Bbb Z/2\Bbb Z$ (explanation in a comment below) is a field, so the ideal is maximal. Atiyah and Macdonald is still the standard book on commutative algebra.

Since two people asked me to include my comment in the answer and since both of them can delete comments (and answers), I include it here. The comment was posted a few minutes after the answer.

Comment. Consider the homomorphism from $\Bbb Z[x,y]$ to $\Bbb Z/(2)$ which takes $x,y$ to $0$, $1$ to $1$. Such a homomorphism exists because $\Bbb Z[x,y]$ is a free commutative unital ring. The kernel $K$ of that homomorphism contains $2,x,y$, so it contains $(2,x,y)$. A polynomial not in $(2,x,y)$ must have odd free term, so it maps to $1$ (not to $0$). Hence $K=(2,x,y)$ and by the isomorphism theorem $\Bbb Z[x,y]/(2,x,y)≅\Bbb Z/(2)$.

Answer 2

The idea is that the quotient ring $\mathbb{Z}[x,y]/(2,x,y)$ is isomorphic to $\mathbb{Z}/2\mathbb{Z}$.

How can you do it? First find a candidate homomorphism $\varphi\colon \mathbb{Z}[x,y]\to\mathbb{Z}/2\mathbb{Z}$ and the choice is almost obvious: send the integers to their residue class modulo $2$, $x\mapsto0$ and $y\mapsto0$. In other words $$ \varphi(f(x,y))=[f(0,0)]_2 $$ The kernel certainly contains $2$, $x$ and $y$. Suppose $f(x,y)\in\ker\varphi$ and write it as $$ f(x,y)=g_0(x)+g_1(x)y+\dots+g_n(x)y^n $$ Then $\varphi(f(x,y))=\varphi(g_0(x))=[g_0(0]_2$ and therefore the constant term of $g_0(x)$ is even and $g_0(x)=g_0(0)+xh(x)$ for some $h(x)\in\mathbb{Z}[x]$. Thus $$ f(x,y)=g_0(0)+xh(x)+g_1(x)y+\dots+g_n(x)y^n\in(2,x,y) $$ Hence $\ker\varphi=(2,x,y)$ and we're done.

Answer 3

It is clear that $\rm (2,x,y)=(2)+(x,y)$. Let $\phi:Z[x,y]\to Z_2[x,y]$ be a natural homomorphism. It is clear that $\ker \phi=(2).$

By the third isomorphism theorem:

$$\frac{Z[x,y]}{(x,y,2)}\cong\frac{Z[x,y]/(\ker\phi)}{((x,y)+\ker\phi)/(\ker \phi)}\cong \frac{Z_2[x,y]}{(x,y)}\cong Z_2.$$

It follows that $(x,y,2)$ is maximal.