In class, we defined an affine algebraic variety to be a $k$-ringed space $(V, \mathcal{O}_V)$ where $V$ is an algebraic set in $\bar{k}^n$ defined by a system of polynomial equations over $k$, and the sheaf of regular functions $\mathcal{O}_V$ that assigns an open subset of $V$ to the set of regular functions on that open set.

We then defined a morphism of affine algebraic varieties $(V, \mathcal{O}_V)$, $(W, \mathcal{O}_W)$ to be given by a morphism of the underlying ringed spaces, i.e. a continuous map (which I'm assuming means continuous with respect to the Zariski topology) $\phi: V \to W$, together with a family of $k$-algebra homomorphisms $\phi_U : \mathcal{O}_W(U) \to \mathcal{O}_V(\phi^{-1} (U))$.

This definition is very abstract and I'm having a bit of a hard time wrapping my head around it. However, we were given some equivalent definitions of this, as well as the following correspondence:

For affine algebraic varieties $(V, \mathcal{O}_V)$, $(W, \mathcal{O}_W)$, there is a $1-1$ correspondence between the morphisms of ringed spaces between $(V, \mathcal{O}_V)$, $(W, \mathcal{O}_W)$, and the $k-$algebra homomorphisms $k[W] \to k[V]$, where $k[W], k[V]$ are the coordinate rings of $W, V$ respectively (so $k[W] = k[x_1 , \dots , x_n]/I(W)$).

So, just to make sure I am understanding this correctly. Let's say I am defining a morphism of the affine algebraic varieties of the circle $x^2 + y^2 = 1$ over $\mathbb{C}$ to the hyperbola $xy = 1$. Does the correspondence stated above mean that it is enough to just write a ring homomorphism $\phi : \mathbb{C}[x,y]/(x^2 + y^2 - 1) \to \mathbb{C}[x,y]/(xy-1)$?


1 Answer 1


More or less correct, the thing to note is however, that the functor from affine varieties to rings is contravariant. Meaning that the correspondence is arrow reversing. Intuitively this corresponds to the pullback of functions.

  • $\begingroup$ Ahh I see. So a morphism of affine algebraic varieties $(V, \mathcal{O}_V) \to (W, \mathcal{O}_W)$ corresponds to a $k$-algebra homomorphism in the opposite direction $k[W] \to k[V]$? $\endgroup$
    – Bastiza
    Mar 8, 2021 at 20:59
  • $\begingroup$ Yes! That is correct $\endgroup$
    – LBE
    Mar 10, 2021 at 12:02

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