In the Wikipedia article on schemes where we give a motivation of nilpotence for transitioning from varieties to schemes, it says that the ring of regular functions (i.e. coordinate ring) of the closed subscheme of $\mathbb{A}^1_{\mathbb{C}}$ defined by $x^2=0$ (as a variety just $V(x^2)=\{0\}$) is $\mathbb{C}[x]/(x^2)$. But if I were to use Hilbert's Nullstellensatz, I'd get
$$ I(V(x^2)) = \sqrt{(x^2)} = (x) \implies \mathcal{O}_{V(x^2)}(V(x^2)) \cong \mathbb{C}[V(x^2)] = \mathbb{C}[x]/I(V(x^2))=\mathbb{C}[x]/(x) \cong \mathbb{C} $$
where $\mathbb{C}[V(x^2)]$ is the coordinate ring of our variety.
On the other hand, if we start by defining the scheme rather than the variety, which I think would be defined by writing $R= \mathbb{C}[x]/(x^2)$ and then saying that the scheme is $\text{Spec}(R)$, then the ring of regular functions (coordinate ring from previous paragraph) will be
$$ \mathbb{C}[\text{Spec}(R)] = \mathcal{O}_{\text{Spec}(R)}(\text{Spec}(R)) \cong R \ncong \mathbb{C}. $$
My question is: is the moral of the story here that $(\text{Spec}(R), \mathcal{O}_{\text{Spec}(R)})$ as a pair remembers more information (e.g. about nilpotency) than the pair $(V(I), \mathcal{O}_{V(I)})$? I'm a bit confused because I've given my variety a structure sheaf, and a scheme is roughly a space with a structure sheaf, but on the other hand I'm describing my "space" algebraically via prime ideals (taking Spec) along with a structure sheaf on that, and this seems to remember more information. Is it the fact that before you even look at the structure sheaf of the variety, it has already forgotten all of the nilpotency information, i.e.
$$ \{0\} = V(x) = V(x^2) = V(x^3) = \cdots $$
so when you take the structure sheaves of $V(x^n)$, they'll all be the same because the nilpotency information has already been forgotten?
Thank you.
Best Answer
Exactly. Let's take a look at the examples you consider $R_n=\mathbb{C}[x]/(x^n)$. Prime ideals $\mathfrak{p}\subset R_n$ are in one-to-one correspondence with prime ideals $(x^n)\subseteq\mathfrak{P}\subset\mathbb{C}[x]$. Now, any such $\mathfrak{P}$ is prime so that $(x^n)\subset\mathfrak{P}$ implies $(x)\subseteq \mathfrak{P}$. Conversely, if $(x)\subseteq\mathfrak{P}$ we clearly have $(x^n)\subset\mathfrak{P}$.
What does this mean? We just exhibited a homeomorphism between the underlying topological spaces $Spec(R_n)$ and $Spec(R_1)$ - I just presented the bijection as sets, you can verify that everything is continuous.
On the other hand, as you already noted, the structure sheaves of both schemes are different.
To conclude let me just mention that it's always possible to endow a closed subset of a scheme with its "reduced subscheme structure". This process is basically taking the radical as you do at the beginning of your post.