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So in algebraic geometry, the prime spectrum of the ring of polynomials over a field $k$ is isomorphic to the set of $k$-homomorphisms from the ring of polynomials to $k$, and we do this by using the fact that the residue field (localise w.r.t. some prime ideal, then mod out the ideal generated by this in the localisation) is isomorphic to $k$, and hence any polynomial goes to an element of $k$, via the equivalence relations. This all makes sense as the ideal generated by a prime ideal of the polynomial ring in its localisation is clearly maximal, but I'm stuck on a very basic example:

Let $\mathbb{C}[x,y]$ be the polynomial ring over the complex numbers and $(x-1)$ be the prime ideal. What number does $y$ go to? A $\mathbb{C}$-algebra homomorphism is defined completely by its action on the generators of the polynomial ring (so just the $n$ variables, and taking $n=2$ we get $x\mapsto1$, but $y\mapsto?$), but I can't figure out how this works in this simple case. It should maybe be zero (otherwise how do you invert $1 + y$?), but I don't know. Can anyone explain this?

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  • $\begingroup$ The quotient C[x, y]/(x - 1) is isomorphic to C[y], not to C. $\endgroup$ Nov 17, 2010 at 1:57
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    $\begingroup$ As Akhil Mathew's answer shows, your first sentence is not correct. This is why you are stuck; you are trying to prove something that is not true (as Qiaochu's Yuan's comment shows). $\endgroup$
    – Matt E
    Nov 17, 2010 at 4:07

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The maximal ideals in a polynomial ring of an algebraically closed field $k$ is isomorphic to the set of $k$-homomorphisms from the polynomial ring into $k$. (This is the Nullstellensatz.) This is not true for arbitrary prime ideals, nor is it true for nonalgebraically closed fields. (Consider for instance the ideal $(X^2 +1) \subset \mathbb{R}[X]$; this morally wants to correspond to the points $\pm i$, but it doesn't, since those don't live in the reals. Nonetheless, it happens that the scheme $\mathrm{Spec} \mathbb{R}[X]$ includes all prime ideals, so it includes a copy of $\pm i$ glued together, if you want to call it that.) In the terminology of algebraic geometry, one would say that the point you described is not "rational over $\mathbb{C}$." Such points do exist, and make schemes of even finite type over an algebraically closed field different from varieties. (Fortunately, it turns out that closed points alone determine many topological properties---the precise statement is that the closed points (or maximal ideals) are dense in any constructible sets.)

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    $\begingroup$ would you explain the first line how that isomorphism is? $\endgroup$
    – Myshkin
    May 18, 2012 at 7:48

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