If we take the definition of a variety as a reduced integral scheme of finite type over an algebraically closed field $k$, then a variety is in particular a scheme over $k$, so is a scheme $X$ with a morphism $X \to Spec(k)$. As I'm understanding it, more emphasis should be put on the morphism instead of the underlying scheme since we can have non-isomorphic schemes over a base $S$ say $\pi,\eta : X \to S$. Then even if we have a finitely generated reduced $k$-algebra $A$, can we still have non-isomorphic $k$-schemes $\eta, \pi : Spec (A) \to Spec (k)$? It seems to me like if this were the case it would be unexpected, since there is just one $k$ variety called Spec$(A)$, but I don't really have anything to base this on other than the fact that the relative point of view is really never applied to varieties.

This is equivalent to putting two non-isomorphic $k$-algebra structures on $A$, where $k$ is algebraically closed, so I the question I'm really interested in is can we have a ring $A$ and two monomorphisms $i_1,i_2 :k \to A$ so that there is no automorphism $\varphi$ of $A$ with $\varphi \circ i_2 = i_1$?

  • $\begingroup$ Nah this isn't quite what I'm looking for: conjugation is an $\mathbb{R}$-algebra automorphism, but $\mathbb{R}$ isn't algebraically closed and this isn't an automorphism of $\mathbb{C}$ which is a $k$-algebra automorphism for any algebraically closed $k$. What I'm trying to get at, maybe badly, is that specifying $A$ should unambiguously define an affine $k$ variety Spec$(A)$. But if we have two non-isomorphic k-algebra structures we get two non-isomorphic k-schemes. I think this shouldn't be possible, since in the theory of varieties, without reference to schemes, $A$ does give us Spec$(A)$. $\endgroup$ – CameronJWhitehead Dec 20 '15 at 17:38
  • $\begingroup$ The conjugation automorphism $\mathbb{C} \to \mathbb{C}$ does define a $\mathbb{C}$-scheme Spec$\mathbb{C} \to$ Spec$\mathbb{C}$, but this scheme is isomorphic to the one given by the identity map. $\endgroup$ – CameronJWhitehead Dec 20 '15 at 17:41
  • $\begingroup$ Sorry, whats not isomorphic to what over $\mathbb{C}$? $\endgroup$ – CameronJWhitehead Dec 20 '15 at 17:43
  • 1
    $\begingroup$ Let me think for a moment. I think I might have said something wrong and I think I understand your question somewhat better than I did originally. My feeling is that conjugation is still the thing to look at, in the sense that the $\mathbf{C}$-structure determines whether conjugation is a morphism at all and you do have to make a choice. But I'm not sure that that answers the question. $\endgroup$ – Hoot Dec 20 '15 at 17:45

Here is a non-affine example, although I think (but haven't checked) that it produces an affine example after removing a point: take an elliptic curve $E$ over $k$ whose $j$-invariant $j(E)$ is moved by some automorphism $g : k \to k$ of $k$. Then applying $g$ to the coefficients of a polynomial defining $E$ produces a new elliptic curve $gE$ with $j$-invariant $j(gE) = g j(E) \neq j(E)$. So $E$ and $gE$ are not isomorphic as varieties over $k$, but they are isomorphic as schemes.

In the setting of varieties the point of the structure morphism to $\text{Spec } k$ is mostly to force morphisms to be $k$-linear. Without it you just get the wrong notion of morphism of varieties over $k$.

  • $\begingroup$ I think I understand my problem now, which is a bit of a silly one: In my copy of Kempf's Algebraic Varieties they call the functor from reduced finitely generated $k$-algebras to affine varieties `Spec', when they really mean maxspec. So there is no reason to think there shouldn't be multiple $k$-variety structures on Spec$(A)$. $\endgroup$ – CameronJWhitehead Dec 20 '15 at 19:47
  • $\begingroup$ Thanks very much for your example! $\endgroup$ – CameronJWhitehead Dec 20 '15 at 19:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.