If two zero sets are homeomorphic, is the ring of polynomials over the sets also homeomorphic?

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 $\Bbb{R}[x]/(x)\cong\Bbb{R}\not\cong\Bbb{C}\cong\Bbb{R}[x]/(x^2 + 1).$

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.