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Let's say I have a number field $k$ and a curve $C$ over $k$, can we lift $C$ to a variety $V$ over $\mathbb{C}$? It is like "complexifying" the curve. I read somewhere that it is possible to lift an elliptic curve $E$ over $k$ to an elliptic curve $\mathbb{C}/\Lambda$ where $\Lambda$ is a suitable lattice in $\mathbb{C}$ and the endomorphism on $E$ lift uniquely as well. I also want to lift not only a curve but a variety over $k$ and an endomorphism on $V$ to some variety over $\mathbb{C}$ and an endomorphism on $V$ respectively. Can this be done?

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"Lift" is not the right word; a lift is something like starting with a variety over $\mathbb{F}_p$ and lifting it to a variety over $\mathbb{Z}_p$. Lifts attach extra data and are in particular not functorial, unlike base change. – Qiaochu Yuan Aug 27 '12 at 22:01
What are these lifts? What extra data are attached to them? – Natalie Aug 27 '12 at 22:10
You can define base change from schemes over a ring $R$ to schemes over a ring $S$ from any morphism $R \to S$. Taking the natural quotient $\mathbb{Z}_p \to \mathbb{F}_p$ gives a base change map from schemes over $\mathbb{Z}_p$ to schemes over $\mathbb{F}_p$, and a lift is an inverse (on the object level) of base change. Inverses are not unique in general, so the extra data to be attached is the choice of an inverse. The basic idea is, for example, lifting a residue $\bmod p$ to an integer; one has infinitely many choices. – Qiaochu Yuan Aug 27 '12 at 22:17
up vote 7 down vote accepted

Any variety (or scheme) $X$ over a number field $k$ gives rise to a variety (or scheme) over $\mathbf{C}$ upon choosing an embedding $i:k\rightarrow\mathbf{C}$. The associated object over $\mathbf{C}$ is then $X_\mathbf{C}:=X\times_{k,i}\mathrm{Spec}(\mathbf{C})$ (this is the fibered product of $X$ and $\mathrm{Spec}(\mathbf{C})$ over $\mathrm{Spec}(k)$ where $\mathrm{Spec}(\mathbf{C})$ is regarded as a $\mathrm{Spec}(k)$-scheme via $\mathrm{Spec}(i)$. Applying $-\times_{k,i}\mathrm{Spec}(\mathbf{C})$ is called base change along $i$, or if $i$ is understood and fixed, base change from $k$ to $\mathbf{C}$. Base change is functorial in that if you have a morphism $f:X\rightarrow Y$, you get a morphism $f_\mathbf{C}:X_\mathbf{C}\rightarrow Y_\mathbf{C}$ as $f_\mathbf{C}=f\times 1_{\mathrm{Spec}(\mathbf{C})}$.

In concrete terms, if $X$ is the variety cut out in $\mathbf{A}_k^n$ by an ideal $I=(f_1,\ldots,f_m)$ for $f_1,\ldots,f_m\in k[X_1,\ldots,X_n]$, then $X_\mathbf{C}$ is the variety cut out by the same polynomial equations $f_j=0$ viewed over $\mathbf{C}$. Similarly for a variety in $\mathbf{P}_k^n$.

Regarding an elliptic curve $E/k$, the base change $E_\mathbf{C}$ is an elliptic curve over $\mathbf{C}$. The group of $\mathbf{C}$-rational points $E_\mathbf{C}(\mathbf{C})$ has a canonical structure of compact, complex Lie group (it is the analytification of $E_\mathbf{C}$) and admits an ``analytic uniformization" $E_\mathbf{C}(\mathbf{C})\cong\mathbf{C}/\Lambda$ for a lattice $\Lambda$. This uniformization is explained, e.g., in Silverman's first book, and many other references for elliptic curves.

EDIT: To answer the question in the comments about dimension, if your definition of dimension is the transcendence degree of the function field, then you probably want to assume your variety is geometrically integral (one definition of a $k$-variety is a finite type, separated, geometrically integral $k$-scheme, although sometimes "geometrically integral" is replaced with various weaker conditions). If $X$ is a geometrically integral $k$-scheme of finite type, then for any non-empty open $U$ of $X$, $\dim(U)=\dim(X)$ (this is true for an integral $X$ but we want geometrically integral to ensure that the definition of dimension we're using makes sense after base change). We then have $\dim(U_\mathbf{C})=\dim(X_\mathbf{C})$, and we can therefore assume $X=\mathrm{Spec}(A)$ is affine, the spectrum of a finitely generated domain over $k$. Now the result follows from Noether normalization: there is a finite injection $k[X_1,\ldots,X_n]\rightarrow A$, and tensoring with $\mathbf{C}$ gives a finite injection $\mathbf{C}[X_1,\ldots,X_n]\rightarrow A\otimes_k\mathbf{C}$. Dimension is preserved in finite ring extensions, so $\dim(A)=\dim(k[X_1,\ldots,X_n])=n=\dim(\mathbf{C}[X_1,\ldots,X_n])=\dim(A\otimes_k\mathbf{C})$.

FURTHER EDIT: It occurs to me that dimension is preserved in great generality for base change along algebraic extension (which a number field to $\mathbf{C}$ definitely isn't, but just for completeness I thought I'd add this here). If $X$ is any locally of finite type $k$-scheme, $k$ any field at all, and $K/k$ is algebraic, then $\dim(X)=\dim(X_K)$. For the proof, take an affine open cover $X=\bigcup_iU_i$, so that $X_K=\bigcup_i(U_i)_K$ is an affine open cover, and thus $\dim(X)=\sup_i\dim(U_i)$ and $\dim(X_K)=\sup_i\dim((U_i)_K)$. It therefore suffices to prove $\dim(U_i)=\dim((U_i)_K)$ for all $i$, so we may assume that $X=\mathrm{Spec}(A)$ is affine, the spectrum of a finite type $k$-algebra (so necessarily $\dim(A)<\infty$). The map $A\rightarrow A_K$ is an integral extension (integral because it is the base change of integral $k\rightarrow K$ and injective because $K$ is $k$-flat). Integral ring extensions preserve dimension, so we win.

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Thanks! I'll take a look at Silverman's book. – Natalie Aug 27 '12 at 21:44
If my variety over $k$ is of a certain dimension, then after the base change to $\mathbb{C}$ does my new variety have the same dimension? – Natalie Aug 27 '12 at 22:11
@Natalie: what's your definition of the dimension of a variety over a non-algebraically closed field? – Qiaochu Yuan Aug 27 '12 at 22:18
Take any scheme $X$ of finite type, separated, integral over an algebraic number field $k$. What kind of scheme $X_{\mathbf{C}}$ is? Is $X_{\mathbf{C}}$ reduced? What are the irreducible components of $X_{\mathbf{C}}$? – Makoto Kato Aug 27 '12 at 23:35
Dear @Makoto, It will stay reduced, because over a perfect field, reduced is the same as geometrically reduced, but it need not stay irreducible. Take, for example, $X=\mathbf{Q}(i)$ as a $\mathbf{Q}$-scheme. Over $\mathbf{C}$ you get two reduced (closed) points. – Keenan Kidwell Aug 27 '12 at 23:41

Just to give a more elementary explanation: if you have a set of equations with coefficients in a field $k$, then you can also think of these equations as having coefficients in $l$, for any extension field $l$ of $k$. So if $V$ is a variety over $k$, cut out by some equations with coefficients in $k$, then thinking of these equations instead as having coefficients in $l$, we can think of $V$ as being a variety over $l$ instead.

Exactly how this works on a technical level depends on your precise definition of variety over $k$. (If you work scheme-theoretically, this corresponds to base-change of schemes. If you work with a more traditional language as in Weil's foundations, then the points of the variety don't actually change.) But the intuitive idea is the one in the paragraph above, and should hopefully be pretty clear.

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