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I have some basic familiarity and intuition with algebraic varieties and am wanting to start on schemes. I am looking at section 2.2 of Hartshorne, in particular the definition of the sheaf of rings of $\text{spec}A$. For a ring $A$, this consists of functions, \begin{equation} \label{regulartionfunction} s: U \longrightarrow \bigsqcup_{\mathfrak{p} \in U}A_{p} \end{equation} satisfying certain local conditions. I am trying to connect this back to the case of affine varieties where the maps are from an open set $U$ into the ground field, say $k$. Suppose $A$ is some polynomial ring, for simplicity start with $\mathbb{C}[x]$. The points of an affine variety correspond to the maximal ideals $\mathfrak{m}$, which look like $\left< x-a \right>$. My reasoning (which I think has some kind of problem), was as follows:

The prime ideals of $\mathbb{C}[x]_{\mathfrak{m}}$ are in one-one correspondence with the prime ideals of $\mathbb{C}[x]$ contained in $\mathfrak{m}$. Since the prime ideals of $\mathbb{C}[x]$ are precisely the maximal ones, then $\mathbb{C}[x]_{\mathfrak{m}}$ has precisely one maximal ideal, which is $(0)$, hence making it a field. If that is correct, then bringing this in line with above definition for schemes in the case that $A = \mathbb{C}[x]$ would result in $s$ being a map from $U$ onto a field.

However, this reasoning seems to break down for polynomial rings in more than one variable. Also, I am not even convinced my above reasoning is correct, since I only seemed to use the fact that $\mathbb{C}[x]$ was a PID, and am not sure all localizations of PIDs are fields.

So if anyone could shed some light on how the map $s$ reduces to the regular functions we are familiar with for affine varieties, and also sure up my reasoning about localizations of polynomial rings by maximal ideals I would be very appreciative.

Thanks

Luke

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Your description of the ideal structure of $\mathbb{C}[x]_\mathfrak{m}$ is incorrect. $\mathbb{C}[x]_\mathfrak{m}$ is a local ring which is not a field- it contains the ideal $\mathfrak{m}$ which is different from the zero ideal and not the whole ring. (Exercise: verify this with $\mathbb{C}[x]_{(x)}$, where every element not divisible by $x$ is a unit- show that $1\notin(x)$ and $x\notin(0)$.)

The way to recover the notion of functions with which you're familiar (target being a field) is to evaluate our function in to local rings by tensoring with the residue field at the point you're interested in. This gives more information because the local rings are larger and you can track more stuff using them.

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