Understanding Ideals and Quotients in Polynomial Rings of Multiple Variables

Question

I am trying to prove that $I =(x^2+1,y-1)$ is a maximal ideal in $\mathbb{Q}[x,y]$, but I am having a hard time understanding what this ideal even looks like. I know that I can prove it's a maximal ideal by proving $\mathbb{Q}[x,y]/I$ is a field, but I'm also having a hard time understanding what this quotient looks like, or what it is isomorphic to.

Basically, some intuition into dealing with ideals generated by multiple polynomials, as well as how to deal with quotient rings would be much appreciated.

Edit: My first thoughts are that $\mathbb{Q}[x,y]/(x^2+1,y-1) \simeq \mathbb{Q}[x]/(x^2+1) \simeq \mathbb{Q}[i],$ which is a field, but I am not sure how to make these isomorphisms rigorous.

Also, is there a way to show this ideal is maximal other than showing the quotient is a field? Or is this the best way?

Answer 1

Perhaps the intuition you would like comes from algebraic geometry? There is a correspondence called the Nullstellensatz which gives a connection between prime ideals and algebraic sets in affine space over the given algebraically closed field. This is one way to visually 'see' what an ideal looks like.

It is a fact that under this correspondence, points are associated with maximal ideals. Thus determining whether or not an ideal is maximal is equivalent to determining the solution set of the system of equations $x^2 + 1 = 0, y-1 = 0$, and determining if the solution set consists of just finitely many points, or an algebraic curve.

Now the Nullstellensatz requires the ground field to be algebraically closed, but I don't think this is a problem, as in your edit, you seem to be willing to talk about extending $\mathbb{Q}$ by adding $i$ formally in order to produce this isomorphism, so let's just go ahead and think about the algebraic closure of the rational numbers.

In this case, I feel it is easy to see that the system of equations has only two solutions and that you've more or less properly identified them - $(i,0)$ and $(-i, 0)$.

As for your question about how to make those isomorphisms rigorous, the first I feel is obvious, and the second uses the fact that $x^2+1 = (x+i)(x-i)$, and so we can split this into $\mathbb{Q}[x]/(x+i) \oplus\mathbb{Q}[x]/(x-i)$ which is in turn isomorphic to two copies of $\mathbb{Q}$ together, with the multiplication relation being exactly that of the multiplication with complex numbers.

Answer 2

@luthien Your idea is correct. The maps you are thinking of are indeed isomorphisms. One way to show they are isomorphisms is to start with the map $f : \mathbb{Q}[x,y] \to \mathbb{Q}[i]$ given by $x \mapsto i$, $y \mapsto 1$. Then $x^2+1$ and $y-1$ are in the kernel of $f$.

Claim: The kernel of $f$ is the ideal generated by $x^2+1$ and $y-1$, namely, the ideal $I$.

I'm sure there is a "nice" way to do this. Here is a crude way.

Certainly $I \subseteq \ker f$. Conversely suppose $p = p(x,y) \in \ker f$. Say the highest total degree of any term of $p$ is $d$, and suppose by induction that every polynomial in $\ker f$ of degree strictly less than $d$ is in $I$ (the cases of polynomials of degree $0,1,2$ being "easy"). Let $c x^a y^{d-a}$ be one of the terms of degree $d$ appearing in $p$. If $a \geq 2$, then $c x^{a-2} y^{d-a} (x^2+1) \in I \subseteq \ker f$, and subtracting it from $p$ amounts to changing this term into $-c x^{a-2} y^{d-a}$, which has smaller total degree. Or if $a < 2$, $d-a \geq 1$, then we can similarly change $y$ into $1$. (If $a < 2$ and $d-a < 1$, then $d < 3$ and we are in the "initial" cases.) After changing enough of these terms we get a polynomial $p'$ that differs from $p$ by an element of $I$, and by induction $p' \in I$, so $p \in I$ too.

Claim: The map $f$ is surjective.

Immediate.

By the First Isomorphism Theorem, the quotient ring $\mathbb{Q}[x,y]/I$ is isomorphic to the image $\mathbb{Q}[i]$, which is a field. So $I$ is maximal.

Alternative approaches? Well, you could suppose $I \subseteq J \subseteq \mathbb{Q}[x,y]$, and suppose $p \in J$, $p \notin I$, of smallest possible total degree (among all elements in $J \setminus I$). A similar "reduction" argument (modifying $p$ by replacing $x^2$ with $-1$ and $y$ with $1$) shows that $p$ is at most linear in $x$, and constant in $y$. If $0 \neq p = ax+b \in J$, then $(ax-b)p = a^2x^2-b^2 \in J$ too. And $J$ contains $x^2+1$ since $I \subseteq J$. From $a^2x^2-b^2$ and $x^2+1$ we get $-(a^2+b^2)=(a^2x^2-b^2)-a^2(x^2+1) \in J$. So $1 \in J$. That is a pretty direct proof that $I$ is maximal — it avoids quotient rings. It's beyond me to judge if one way or the other is "best".