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(see bottom for apology)

Let $Y$ be the plane curve $y = x^2$ (i.e., $Y$ is the zero set of the polynomial $f = y - x^2$). Show that $A(Y)$ is isomorphic to a polynomial ring in one variable over $K$.

In case the notation is non-standard, let $A = k[x_1, ... , x_n]$. Then $A(Y) = A / I(Y)$ is the affine coordinate ring of an algebraic subset $Y \subseteq \mathbb A^n$, where $I(Y) = \{ f \in A | f(P) = 0 \ \forall P \in Y\}$.

My initial thoughts:

I can use the information in the question to help me; I know in advance that $A(Y)$ is going to be generated by one of its elements (as a $k$-algebra). I'd like to see what the elements of $A(Y)$ look like, as this will help me to find a generator.

My attempt:

I'm quotienting out by $I(Y) = I(Z(\mathfrak{a})) = \sqrt{\mathfrak{a}}$ (Hilbert's Nullstellensatz), where $\mathfrak{a}$ is the ideal generated by $f$. It'd be nice if $\mathfrak{a}$ was a radical ideal. Suppose $g \in \sqrt{\mathfrak{a}}$. Then $g^r \in \mathfrak{a}$ for some positive integer $r$, i.e. $g(x,y)^r = h(x,y)(y-x^2)$ for some $h(x,y) \in A$. Since $(y-x^2)$ is irreducible in $A$ and it divides $g(x,y)^r$, it must divide $g(x,y)$, and so $g(x,y) \in \mathfrak{a}$. So I have that $I(Y) = \mathfrak{a}$. [if this method is correct, it can be generalised to show that an ideal of $A$ generated by an irreducible is radical]

So elements of $A(Y)$ are of the form $p(x,y) + \langle y-x^2 \rangle$. Notice that $(x + \langle y-x^2\rangle)^2 = x^2 + \langle y-x^2 \rangle = x^2 + y - x^2 + \langle y - x^2 \rangle = y + \langle y - x^2 \rangle$. Thus $x + \langle y - x^2 \rangle$ is a generator of $A(Y)$.

Let $\phi : A(Y) \to k[z]$ be the $k$-linear map specified by sending $x + \langle y - x^2 \rangle $ to $z$. This is an isomorphism of $k$-algebras.

I promise I won't ask about every exercise in this book; I just want to make sure I'm approaching the topic in the right way. Obviously what appears above isn't how I'd structure a proper solution - I included my thought processes where possible in the hope that someone might tell me I'm thinking about this in the right/wrong way.

Thanks a lot.

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You are right. I think it's easier to think in the canonical inclusion, that is $k[x]\to k[x,y]\to A(Y)$. Also, the Lin's comments are very good. Lucky with item (c)! – emiliocba Jan 21 '12 at 1:47
How do you know that $\phi$ is an isomorphism? (Even true things can be proved.) – Dylan Moreland Jan 21 '12 at 3:05
@DylanMoreland Isn't it? I can't see why it isn't, when expressing elements of $A(Y)$ as polynomials in $x + \langle y - x^2 \rangle $ . – Matt Jan 21 '12 at 3:09
«I can't see why it isn't» is not the same thing as being able to prove it is! – Mariano Suárez-Alvarez Jan 21 '12 at 4:07
up vote 7 down vote accepted

Why don't we approach this geometrically? Let us exhibit an isomorphism between $\mathbb{A}^1$ and the parabola $Y$. This is easy enough: the map $t \mapsto (t, t^2)$ is one direction and the projection $(x, y) \mapsto x$ is the other. This implies that $A(Y) \cong A(\mathbb{A}^1) \cong k[t]$, where the isomorphism takes $x \mapsto t$ and $y \mapsto t^2$.

More generally, if we have a morphism $\varphi : X \to Y$, there is a homomorphism $\varphi^* : A(Y) \to A(X)$ defined by $\varphi^*(f) = f \circ \varphi$. Explicitly in terms of generators and coordinates, if $Y \subseteq \mathbb{A}^n$ and $X \subseteq \mathbb{A}^m$ and we write $\varphi_1, \ldots, \varphi_n$ for the components of $\varphi$, $x_1, \ldots, x_m$ for the coordinate functions of $X$, and $y_1, \ldots, y_n$ for the coordinates functions of $Y$, then $\varphi^*$ has $$\varphi^*(y_j) = \varphi_j (x_1, \ldots, x_m)$$ and this data suffices to determine $\varphi^*$ because $y_1, \ldots, y_n$ generate $A(Y)$.

Of course, you probably aren't supposed to use morphisms of varieties yet. In that case the only real difficulty in the whole exercise is establishing that $A(Y) \cong k[x, y] / (y - x^2)$. But this follows from these easy facts:

  1. $y - x^2$ is an irreducible polynomial in $k[x, y]$.

  2. The principal ideal generated by an irreducible element of a UFD is prime.

  3. $k[x, y]$ is a UFD.

  4. A prime ideal is radical.

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Thanks, I'll come back to the first half of your answer once I've read the section on morphisms. My answer uses (though they're not very explicitly stated, I admit) those four facts, which is comforting I suppose. – Matt Jan 21 '12 at 1:45

Let $K$ be a commutative ring, and let $X,Y,T$ be indeterminates. By the universal property of polynomial algebras, there are unique $K$-algebra morphisms $$ c:K[T]\to K[X,Y],\quad p:K[X,Y]\to K[T] $$ satisfying $$ c(T)=X,\quad p(X)=T,\quad p(Y)=T^2. $$ Let $$ \pi:K[X,Y]\to\frac{K[X,Y]}{(Y-X^2)} $$ be the canonical projection. We claim

(A) $\ p$ induces a $K$-algebra morphism $$ \overline p:\frac{K[X,Y]}{(Y-X^2)}\to K[T], $$ (B) $\ \overline p$ and $\pi\circ c:K[T]\to K[X,Y]/(Y-X^2)$ are inverse $K$-algebra isomorphisms.

The proof is straightforward.

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