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I'm trying to learn by myself some algebraic geometry, since I will probably be taking an elementary course next semester.

I have some trouble understanding the following:

Let $Y=Z(y-x^{2})$ then by definition $Y$ is algebraic set in $\mathbb{A}^{2}$. The author finds the coordinate ring $K[x,y]/I(Y)$ where $I(Y)=\{f \in K[x,y]: f=0 \ \textrm{in Y}\}$.

Claim: $K[x,y]/I(y) \cong K[t]$ where $t$ is an indeterminate.

Define a map $\phi: K[x,y] \rightarrow K[t]$ by $x \mapsto t$ and $y \mapsto t^{2}$. Then $f$ is a ring morphism by the universal property.


1) Why it suffices to say where to send $x$ and $y$?

2) How do you check using the universal property that this is indeed a ring morphism? I don't know how to apply this property

3) What is the reason of sending $x$ to $t$ and $y \mapsto t^{2}$,i.e how do you figure out the images of $x$ and $y$ under $\phi$ ?

I would really appreciate if you can please explain as much as possible, hope that is not too much to ask.

Thanks in advance

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To learn algebraic geometry by yourself you could use some of the first books I recommend in my Amazon list: Above all, you should try in this order Beltrametti et al., Hulek, Shafarevich and Perrin. You can also search the web and download the pdf courses by Dolgachev, Gathmann, Milne and Fulton (and Debarre in French or Manetti in Italian) – Javier Álvarez Feb 20 '11 at 1:32
up vote 4 down vote accepted
  1. It suffices to say where $x$ and $y$ go because they generate $K[x, y]$ (by the universal property).

  2. The universal property tells you that a $K$-algebra homomorphism out of a polynomial ring over $K$ is determined by the images of the indeterminates.

  3. The targets of $x$ and $y$ are chosen so that the ideal of the homomorphism is $I(Y)$. A different way of seeing this is to note that there is a (surjective) morphism $\mathbb{A}^1 \to Y$ sending $t \mapsto (t, t^2)$ — a morphism of affine varieties corresponds to a $K$-algebra homomorphism of the coordinate rings (but in the opposite direction!) and vice-versa.

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so the map $\phi$ is viewed as a $K-algebra$ morphism? – user6495 Feb 20 '11 at 2:38
@user6495 That's right. – Alex B. Feb 20 '11 at 3:31

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