What does it mean to tensor with $\mathbb{Q}$?

At our algebraic geometry seminar I often hear that something is 'tensored with $\mathbb{Q}$', e.g. a ring of endomorphisms. This phrase seems to have some intuitive meaning that I don't know. What does 'tensoring with $\mathbb{Q}$' mean?

Solution 1:

Let $M$ be a $\mathbb{Z}$-module. As $\mathbb{Q}$ is a $\mathbb{Z}$-module (i.e. it is an abelian group), we can form the tensor product $M\otimes_{\mathbb{Z}}\mathbb{Q}$ which is a $\mathbb{Z}$-module (and also a $\mathbb{Q}$-module). It consists of finite sums of expressions of the form $m\otimes q$ where $m \in M$ and $q \in \mathbb{Q}$ and we impose the rules

• $(m_1 + m_2)\otimes q = m_1\otimes q + m_2\otimes q$
• $m\otimes(q_1 + q_2) = m\otimes q_1 + m\otimes q_2$
• $(km)\otimes q = m\otimes (kq) = k(m\otimes q)$ where $k \in \mathbb{Z}$.

If $t \in M$ is a torsion element, that is $nt = 0$ for some $n \in \mathbb{N}\setminus\{0\}$, consider $t\otimes q$. Note that

$$t\otimes q = t\otimes\frac{nq}{n} = t\otimes \left(n\frac{q}{n}\right) = (nt)\otimes \frac{q}{n} = 0\otimes\frac{q}{n} = 0.$$

So any torsion elements in $M$ vanish when $M$ is tensored by $\mathbb{Q}$. That is, they contribute nothing to the module $M\otimes_{\mathbb{Z}}\mathbb{Q}$.

Solution 2:

More generally, if you have a (commutative) ring $A$, an $A$-module $M$ and an $A$-algebra $B$, you can ‘tensor $M$ with $B$ over $A$’, obtaining a $B$-module. That is the standard way to extend the scalars. A well-known example is the complexification of a real vector space, which amounts to tensoring the vector space with $\mathbf C$ over $\mathbf R$.

If $A$ is, say, a domain and the algebra is its field of fractions $K$, tensoring $M$ with $K$ is the same as constructing a module of fractions, i.e. considering elements $\dfrac ms$, where $m\in M$ and $s\in A\setminus\{0\}$, with the rule that $\dfrac ms=\dfrac{ m'}{s'}\iff \exists\, r\ne 0\,$ such that $r(s'm-sm')=0$.

There is a canonical $A$-homomorphism $M\rightarrow M\otimes_A K$ defined by $m\mapsto\dfrac m1\,$; its kernel is made up of the torsion elements of M.

Concerning rings of endomorphisms there is a canonical homomorphism: $$\operatorname{End}_A(M)\otimes_AK\rightarrow \operatorname{End}_{K}(M\otimes_AK)$$ which is injective if $M$ is a finitely generated $A$-module, and an isomorphism if it is finitely presented.