I was reading up on algebraic geometry and it was stated that if $C$ is a curve in the affine $n$-space $\mathbb{A}^n$? then all prime ideals of $k[C]$, the set of all regular functions on $C$, are maximal ideals corresponding to points in $C$. I know that if $C$ is a curve, then all prime ideals in $k[C]$ are maximal ideals since $k[C]$ is a UFD. I also know that all maximal ideals in $k[\mathbb{A}^n]=k[x_1,\cdots x_n]$ corresponds to points in $k$. However I am unable to see how this is the case for a curve $C$. I can see roughly how this passes on to the case of curves. But how does one show this in a rigorous manner? Any help/hints given would be greatly appreciated! Here we assume $k$ is an algebraically closed field.
Maximal ideals corresponding to points on curve
algebraic-geometry
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If $X=\mathrm{Spec}(A)$ is the spectrum of a discrete valuation ring, then the generic point $\eta$ is an affine open, namely $\mathrm{Spec}(k(\eta))$. Of course $\{\eta\}$ is closed in itself, but not in $\mathrm{Spec}(A)$. So this shows that a closed point in an affine open need not be closed in the ambient scheme.
This does, however, hold for schemes locally of finite type over a field $k$. Namely, if $X$ is such a $k$-scheme and $x\in X$ is closed in some affine open $\mathrm{Spec}(A)$, then $k(x)=A/\mathfrak{p}_x$ is an extension field of $k$ which is of finite type over $k$. So it is finite over $k$ by Zariski's lemma (also known sometimes as the Nullstellensatz). Now if $U=\mathrm{Spec}(B)$ is any affine open containing $x$, then $x$ corresponds to a prime $\mathfrak{p}_x^\prime$ of $B$, and $B/\mathfrak{p}_x^\prime$ is contained in $k(x)$ ($k(x)$ is the fraction field of this domain in fact). Since $k(x)/k$ is finite, we see that $B/\mathfrak{p}_x^\prime$ is a domain finite over a field, and so is itself a field, i.e., $\mathfrak{p}_x^\prime$ is a closed point of $\mathrm{Spec}(B)$. Since the affine opens cover $X$, it follows that $x$ is a closed point of $X$.
Perhaps there are other situations where closed points of affine opens are closed. I learned this particular fact from Qing Liu's book on algebraic geometry. If we're lucky maybe he will see this problem and add some further insight :)
You're thinking about this the wrong way. There is a map from $\mathbb{A}^{n+1} - \{0\} \to \mathbb{P}^n$ for any $n$, but this is not the way we realize the co-ordinate ring, and the construction one would get if one tried to blow up points into lines in this quotient would not be the affine variety corresponding to some subset of $\mathbb{P}^n$ but the pullback of the tautological bundle onto this variety.
The real maps to consider are the co-ordinate maps: $\mathbb{A}^n \to \mathbb{P}^n$ that make $\mathbb{P}^n$ a variety. These induce isomorphisms between $k[x_1, \dots, x_n]$ and $\mathcal{O}(\mathbb{P}^n - \{(x_1:\dots : x_{n+1}) | x_i = 0\})$, and this is how you prove the correspondence between maximal ideals that do not contain the trivial homogeneous ideal $R(X)_+$ (which obviously cannot be included) and points in $\mathbb{P}^n$, you use these homeomorphisms.
Best Answer
This is true only for an algebraically closed field. The reason for the assertion is that $k[C] $ is a quotient of some $k[\mathbf A^n]$, hence its spectrum corresponds to the Zariski-closed set $V(I)$ in $k[X_1,\dots,X_n]$, where $I$ is the ideal of polynomials which vanish on $C$ .
In this correspondence, maximal ideals correspond to maximal ideals (by the $3^{\textit{rd}}$ isomorphism theorem).