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Consider a variety $p:X\longrightarrow\operatorname{Spec K}$ where $X$ is an integral scheme and $p$ is a separated morphism of finite type. Now chose an element $\sigma\in\operatorname{Aut}(K)\setminus\{\operatorname{id}\}$, then we can construct another variety over $K$ namely $\operatorname{Spec}(\sigma)\circ p: X^\sigma\longrightarrow\operatorname{Spec}(K)$, where, despite the two names, we have that $X^\sigma=X$ as schemes. These two varieties are different (in general not even isomorphic) because the structural morphisms are distinct, but they are defined by the same underlying scheme. I have proved that, in the framework of algebraic subsets (so working with the classical/old concept of variety), this switch from $X$ to $X^\sigma$ is equivalent to changing through $\sigma$ the coefficients of the polynomials that define our varieties. Formally this is clear, but it is hard to understand how the concept of variety depends so heavily on the structural morphism to $\operatorname{Spec}{K}$. Simply modifying the morphism, but mantaining the same scheme $X$, we obtain two different objects, and this is so strange!! Can you point out some other examples of the importance of the structural morphism in the modern definition of algebraic varieties? I'd like also some practical enlightenments that help me to figure out the above situation.

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  • $\begingroup$ Dubious: are you sure that the varieties are in general not isomorphic ? Can you give examples ? $\endgroup$ – Boccherini Dec 19 '16 at 11:27
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If you have a vector space $V$ over $K$, you can construct another vector space $V^\sigma$ by letting $V=V^\sigma$ as additive groups, but redefining the scalar multiplication as $c\cdot v = c^\sigma v$. The vector spaces $V$ and $V^\sigma$ are abstractly isomorphic because they have the same dimension. In fact, the identity map (as additive groups) $V \to V^\sigma$ takes a basis $\beta$ of $V$, to a basis $\beta^\sigma$ of $V^\sigma$. Nevertheless, the identity map $V \to V^\sigma$ is not $K$-linear, so $V$ and $V^\sigma$ are not canonically isomorphic as vector spaces over $K$. (More precisely, the endo-functor $V\mapsto V^\sigma$ of the category of $K$-vector spaces is not isomorphic to the identity functor.)

This is exactly the same phenomenon. If you can get comfortable with the above, you should have no problem swallowing that pill.

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  • $\begingroup$ Bruno Joyal: They are not canonically isomorphic, but still they are isomorphic as varieties, right ? $\endgroup$ – Boccherini Dec 19 '16 at 11:25
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Let's look at the affine case; let $X=\mbox{Spec}(A)$ and $Y=\mbox{Spec}(B)$ be two affine varieties. Variety morphisms $X\to Y$ (in the classical sense) correspond to $K$-algebra homomorphisms $B\to A$, but scheme morphisms $X\to Y$ correspond to ring homomorphisms $B\to A$; this shows that in general there are more scheme morphisms than variety morphisms.

The $K$-algebra structure on $\mbox{Spec}(A)$ is uniquely given by a homomorphism $K\to A$; that is, a scheme morphism $\mbox{Spec}(A)\to\mbox{Spec}(K)$. A ring homomorphism $B\to A$ is a $K$-algebra homomorphism if and only if it commutes with the homomorphisms $K\to A$ and $K\to B$.

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