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$(A)$. $\endgroup$ Dec 20, 2015 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. $\endgroup$ Dec 20, 2015 at 17:41
  • $\begingroup$ Sorry, whats not isomorphic to what over $\mathbb{C}$? $\endgroup$ Dec 20, 2015 at 17:43
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    $\begingroup$ 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, 2015 at 17:45

1 Answer 1


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$(A)$. $\endgroup$ Dec 20, 2015 at 19:47
  • $\begingroup$ Thanks very much for your example! $\endgroup$ Dec 20, 2015 at 19:47

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