8
$\begingroup$

I'm very sorry to have made this obvious mistake, I should have asked the ideals also be prime. I have fixed this.

Let $R$ be the ring of complex polynomials in $n$ variables and let $I$ and $J$ be prime ideals of $R$. Consider $V(I)$ and $V(J)$, the zero set of the ideals, that is, the set of points sent to zero by all polynomials in the ideal. Gives each of these sets, $V(I)$ and $V(J)$, the subspace topology induced by the usual topology on $\mathbb{C}^n$ and then assume $V(I)$ and $V(J)$ are homeomorphic. Now consider the rings $R/I$ and $R/J$. Must they be isomorphic as rings? If so does this result have a name and can you please provide a proof or point to where I can find a proof? If not I would like a counter example.

To help explain the question further, here is a concrete example:
Say $R$ is the ring of complex polynomials in two variables and say we have the ideals generated by the polynomials $x^2+y^2-1$ and $x^2+y^2-2$. The two topologies are homeomorphic in this case and also the quotient rings are ring-isomorphic. Must it always be the case that homeomorphism implies ring-isomorphism? If not, is it the case that a stronger condition, like diffeomorphism is instead required?

$\endgroup$
  • $\begingroup$ If you allow non-algebraically closed fields (or more general rings) and work with the Zariski topology on $\operatorname{Spec}R$ (not just $k^n$) this is false: consider $\Bbb{R}[x]$ Then $I = (x)$ and $J = (x^2 + 1)$ are both prime, and $V(I)$ and $V(J)$ are homeomorphic: both are just points! But $\Bbb{R}[x]/I\cong\Bbb{R}$ and $\Bbb{R}[x]/J\cong\Bbb{C}.$ $\endgroup$ – Stahl Aug 19 at 0:19
  • $\begingroup$ While true, you don't even need non-algebraically closed fields if you're working with Spec. Just take $\mathbb{C}[x,y]$ and consider $I = (x-a,y-b)$ and $J = (x)$. These both do correspond to points in $\text{Spec } \mathbb{C}[x,y]$ but that's sort of by definition/construction. Of course the latter point is a non-closed point. $\endgroup$ – Osama Ghani Aug 19 at 0:39
  • $\begingroup$ @OsamaGhani While $J = (x)$ is itself a point in $\operatorname{Spec}\Bbb{C}[x,y],$ taking $V$ gives a line, not a point (precisely because it is not closed): $V(x) \cong\operatorname{Spec}\Bbb{C}[x,y]/(x)\cong\Bbb{A}^1_{\Bbb{C}}.$ $\endgroup$ – Stahl Aug 19 at 0:43
  • $\begingroup$ Ah once we started talking about Spec I sort of forgot that we were talking about $V$ of ideals specifically. What I meant (and was thinking about) was that $(x)$ is a point in $\text{Spec } \mathbb{C}[x,y]$ and obviously its closure is the $y$-axis, but it is a non-closed point so in that sense $(x)$ and $(x^2+1)$ are both points in $\text{Spec } \mathbb{C}[x]$. Wrong correspondence, my bad! $\endgroup$ – Osama Ghani Aug 19 at 0:48
  • $\begingroup$ Why would you consider the Euclidean topology on $\Bbb{C}^n$? $\endgroup$ – Servaes Aug 19 at 7:42
7
$\begingroup$

The answer is no! Let $k = \Bbb{C},$ and let $I = (x^2 - y^3)$ and $J = (x)$ inside $\Bbb{C}[x,y].$ First, note that $$\Bbb{C}[x,y]/I\cong\Bbb{C}[t^2,t^3]\not\cong\Bbb{C}[t]\cong\Bbb{C}[x,y]/J$$ (the former is not integrally closed, whereas the latter is). However, I claim that $V(I)$ and $V(J)$ are homeomorphic as subsets of $\Bbb{C}^2$ with its standard topology.

We have maps \begin{align*} \phi : V(x)&\to V(x^2 - y^3)\\ (0,t)&\mapsto (t^3, t^2) \end{align*} and \begin{align*} \psi : V(x^2 - y^3)&\to V(x)\\ (a,b)&\mapsto\begin{cases}(0,\frac{a}{b}),\quad b\neq 0,\\ (0,0),\quad a = b = 0.\end{cases} \end{align*}

First, note that these maps are inverses. It is clear that $\psi\circ\phi = \operatorname{id},$ and if $b\neq 0$ we compute \begin{align*} \phi\circ\psi(a,b) &= \phi(0,\frac ab)\\ &= \left(\left(\frac{a}{b}\right)^3,\left(\frac{a}{b}\right)^2\right). \end{align*} But \begin{align*} a^2 = b^3&\implies\frac{a^2}{b^2} = b\\ &\implies\left(\frac{a}{b}\right)^3 = \frac{a}{b}\cdot b = a. \end{align*} We also observe that $\psi\circ\phi(0,0) = (0,0).$

Now, all we must verify is that these maps are continuous. One sees that $\phi$ is continuous, as it is given by polynomials. The challenge is to check that $\psi$ is continuous. This is clear away from $b = 0,$ so we need only check continuity at $(a,b) = (0,0).$

Claim: The function $\psi$ is continuous at $(0,0).$

Proof: It is enough to show that each component of $\psi$ is continuous. Clearly $(a,b)\mapsto 0$ is continuous, so we need only concern ourselves with the continuity of the map $(a,b)\mapsto a/b$ at $b = 0.$

Explicitly, we need to show that for all $\epsilon > 0,$ there exists $\delta > 0$ such that if

  1. $(\alpha,\beta)\in V(x^2 - y^3),$ and
  2. $0 < \left|(\alpha,\beta)\right| < \delta,$

then $\left|\frac{\alpha}{\beta}\right| < \epsilon.$

First, observe that because $(\alpha,\beta)\in V(x^2 - y^3),$ we have $\alpha^2 = \beta^3,$ which implies $\left|\alpha\right|^2 = \left|\beta\right|^3.$ Now, set $\delta = \epsilon^2.$ We have \begin{align*} 0 < \left|(\alpha,\beta)\right| < \delta &\iff 0^2 < \left|(\alpha,\beta)\right|^2 < \delta^2\\ &\iff 0 < \left|\alpha\right|^2 + \left|\beta\right|^2 = \left|\beta\right|^3 + \left|\beta\right|^2 < \delta^2. \end{align*} This implies that $$0 < \left|\beta\right|^2(\left|\beta\right| + 1) < \delta^2,$$ and we certainly have $$\left|\beta\right|^2 \leq \left|\beta\right|^2(\left|\beta\right| + 1).$$ Putting all this together we find that if $0 < \left|(\alpha,\beta)\right| < \epsilon^2,$ then we have $$ \left|\beta\right|^2 < \epsilon^4. $$ Since both $\left|\beta\right|$ and $\epsilon$ are positive, we conclude that $$\left|\beta\right| < \epsilon^2.$$

Thus, \begin{align*} \left|\frac\alpha\beta\right|^2 &=\frac{\left|\alpha\right|^2}{\left|\beta\right|^2} \\ &= \frac{\left|\beta\right|^3}{\left|\beta\right|^2}\\ &=\left|\beta\right|\\ &<\epsilon^2. \end{align*} Taking square roots, we obtain the desired result. Phew! Q.E.D.

Remark 1: You can get easier examples over non-algebraically closed fields: for example, let $k = \Bbb{Q}.$ Then $V(x^2 + 1) = V(x^2 - 2) = \emptyset$ as subsets of $\Bbb{Q}^2,$ but $$\Bbb{Q}[x]/(x^2 + 1)\cong\Bbb{Q}[i]\not\cong\Bbb{Q}[\sqrt{2}]\cong\Bbb{Q}[x]/(x^2 - 2).$$

Remark 2: The answer is also no over general fields $k$ when $k^n$ is given the Zariski topology, but this is even easier to see: both $V(x)$ and $V(x^2 - y^3)$ are irreducible affine curves, and thus have the cofinite topology. Of course, homeomorphisms are really not what we want to consider when doing algebraic geometry (see here for a discussion).

Remark 3: Finally, the answer is also no when we work with $\operatorname{Spec}R[x_1,\dots, x_n]$ instead of $R^n.$ More generally, it is not true that if $Z_1$ and $Z_2$ are homeomorphic closed subspaces of $\operatorname{Spec}R,$ and we consider them as reduced subschemes, $\mathcal{O}_{Z_1}(Z_1)\cong\mathcal{O}_{Z_2}(Z_2).$ Indeed, let $R = k\times k'$ be the product of two non-isomorphic fields. Then $\operatorname{Spec}R = \{0\times k',k\times 0\},$ and if $Z_1 = \{0\times k'\}$ and $Z_2 = \{k\times 0\},$ then both are simply points, but by assumption, $\mathcal{O}_{Z_1}(Z_1) = k\not\cong k' = \mathcal{O}_{Z_2}(Z_2).$

Another example would be $R = \Bbb{R}[x],$ with $I = (x)$ and $J = (x^2 + 1)$. $V(x)$ and $V(x^2 + 1)$ are both points inside $\operatorname{Spec}R,$ but $\Bbb{R}[x]/(x)\cong\Bbb{R}\not\cong\Bbb{C}\cong\Bbb{R}[x]/(x^2 + 1).$

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

Not necessarily. Consider the ideals $(x)$ and $(x^2)$ in $\mathbb{C}[x]$. As rings clearly $\mathbb{C}[x]/(x) \ncong \mathbb{C}[x]/(x^2)$. If you've seen the Nullstellensatz, note that $(x^2)$ is not a radical ideal, and that $\sqrt{(x^2)} = (x)$ so that $V(x^2) = V(x) = 0$. This example should work for any other field I think.

Edit: Again, the answer is no for a stupid reason (and again you may object that you don't usually think of $(1)$, but it is sort of essential to defining the Zariski topology in the first place). I'm going to answer this in $\mathbb{R}$. If you think of the ideal $(x^2+1)$ and the ideal $(1)$, then $V(x^2+1) = V(1) = \phi$. But $\mathbb{R}[x]/(x^2+1) \cong \mathbb{C}$ whereas $\mathbb{R}[x]/(1) \cong 0$.

Oops right $(1)$ is not prime.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Its not stated in the question, but its natural to assume the ideal is radical (or even that it is prime) $\endgroup$ – vujazzman Aug 18 at 23:55
  • $\begingroup$ I personally didn't think that was the setup here. In fact my original answer was in terms of more general rings with nilpotents and how nilpotents don't affect the underlying set nor topology of schemes, but I felt that the level of this question was at a beginning classical AG level so I wasn't sure if we were assuming radicality or primeness. $\endgroup$ – Osama Ghani Aug 18 at 23:56
  • 1
    $\begingroup$ Fair enough. Lets just say I am curious about the answer under the additional restriction :) $\endgroup$ – vujazzman Aug 18 at 23:58
  • $\begingroup$ It seems OP meant for the ideals to be prime! I'll think about it and edit my answer. $\endgroup$ – Osama Ghani Aug 19 at 0:01
  • $\begingroup$ @OsamaGhani, thanks for pointing out this counter example. I should have caught my mistake sooner. $\endgroup$ – Mathew Aug 19 at 0:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.