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This was an extra tutorial question in my algebraic geometry course,

We have the following subsets of $\mathbb{C^2}$

$$X_1 = \{( (x,y) \ | \ y = 0 \}$$

$$X_2 = \{( (x,y) \ | \ y -x^2 = 0 \}$$

$$X_3 = \{( (x,y) \ | \ xy = 1 \}$$

so $X_1$ is the $x$-axis, $X_2$ is the parabola and $X_3$ is the hyperbola in $\mathbb{C^2}$. We're asked to calculate the function algebras $\mathcal{O}(X_1),\mathcal{O}(X_2), \mathcal{O}(X_3)$, and show that $X_1$ and $X_2$ have isomorphic function algebras, but that $\mathcal{O}(X_3)$ is not isomorphic to either.

For the function algebras I first get the ideals

$$ I_1 = \langle y \rangle, \quad I_2 = \langle y-x^2 \rangle, \quad I_3 = \langle xy-1 \rangle $$ immediate from the definition of the zero sets. Then the function algebras are,

$$\mathcal{O}(X_1) = \mathbb{C}[x,y]/\langle y \rangle \cong \mathbb{C}[x] $$

$$\mathcal{O}(X_2) = \mathbb{C}[x,y]/\langle y-x^2 \rangle \cong \mathbb{C}[x] $$

but for the third,

$$\mathcal{O}(X_3) = \mathbb{C}[x,y]/\langle xy-1 \rangle \cong \ ? $$

I think I understand $\mathcal{O}(X_3)$ is the quotient of $\mathbb[x,y]$ with no mixed monomial terms, but I don't know how to actually write that as a quotient. I'm hesitant to claim it's isomorphic to $\mathbb{C}[x,\frac{1}{x}]$? (Is this then the same as the fraction field?)

Finally for the isomorphisms, $\mathcal{O}(X_1) \cong \mathcal{O}(X_2)$ is clear assuming I got those quotients right, but how do I show $\mathcal{O}(X_3)$ is not isomorphic to the other two, precisely?

(Also one other more broad question. This course has emphasized the equivalence of algebra and geometry (the geometry should be reflected in the algebra, and vice versa) so what does it mean geometrically for the function algebras of the parabola and the $x$-axis to be isomorphic?)

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  • $\begingroup$ Ah, typo, thanks for pointing that out! $\endgroup$ Commented Aug 30, 2017 at 3:59
  • $\begingroup$ Your equation for the hyperbola is wrong. $\endgroup$ Commented Aug 30, 2017 at 4:15
  • $\begingroup$ Ah dang my bad, fixed. $\endgroup$ Commented Aug 30, 2017 at 4:32
  • $\begingroup$ Okay that should all be fixed up $\endgroup$ Commented Aug 30, 2017 at 4:39

2 Answers 2

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The map $\mathbb C\to \mathcal O(X_1)$ restricts to an inclusion of groups of units $\mathbb C^\times\to \mathcal O(X_1)^\times$ which is surjective, but the map $\mathbb C\to \mathcal O(X_3)$ restricts to a non-surjective map $\mathbb C^\times\to \mathcal O(X_3)^\times$.

To show this, compute explicitly the groups of units.

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Yes indeed, $\mathbb{C}[x,y]/\langle xy - 1 \rangle$ is the set of complex polynomials which are sums of powers of $x$ plus sums of powers of $y$. That is, $\mathbb{C}[x] + \mathbb{C}[y]$. (It's not $\mathbb{C}[x] \oplus \mathbb{C}[y]$ because the constant polynomials don't have unique representations in this direct sum. I suppose one could write $\mathbb{C} \oplus x\mathbb{C}[x] \oplus y\mathbb{C}[y]$. Meh.)

Suppose $\mathbb{C}[z] \cong \mathbb{C}[x] + \mathbb{C}[y]$. Then there is an algebra isomorphism $\varphi: \mathbb{C}[z] \rightarrow \mathbb{C}[x] + \mathbb{C}[y]$. Since $\varphi$ is a bijection, there are $f,g \in \mathbb{C}[z]$ such that $\varphi(f) = x$ and $\varphi(g) = y$. Then $\varphi(fg) = xy = 1$. But $\varphi$ is an injection, so $fg = 1$ and we see that both $f$ and $g$ are units in $\mathbb{C}[z]$, so are elements of $\mathbb{C}$. This forces $\varphi^{-1}(\mathbb{C}[x] + \mathbb{C}[y]) \subseteq \mathbb{C} \subsetneq \mathbb{C}[z]$, so $\varphi$ is not an isomorphism.

There ought to be a sneaky way to use the involution $x \leftrightarrow y \in \mathrm{Aut}(\mathcal{O}(X_3))$. I just don't see how to show there is no involution in $\mathbb{C}[x]$.

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