# How do modules,vector spaces, algebras,fields,rings, groups, relate to one another?

Modules, vector spaces, algebras, fields, rings, groups...

How do these basic algebraic objects relate to each other via tensor products?

Is there a way to go from one object to its generalization via a tensor product construction?

I think this is an interesting question...but I can't find many resources on the subject.

For e.g. I've seen people use the phrase 'tensoring up'.

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Well, there are a lot of ways to answer this question. What kind of answer are you looking for? "A vector space is a module over a field," that kind of thing? In what context have you seen people use the phrase "tensoring up" and what are the words they put next to it? –  Qiaochu Yuan Jun 14 '12 at 18:32
Yuan: I was more looking for a description of what happens if you tensor these objects together ...like what if you tnesor 2 fields together or a ring and an algebra , things of this nature. –  hello Jun 14 '12 at 18:39

First let's talk about the tensor product of modules. Let $R$ be a ring, let $M$ be a right $R$-module, and let $N$ be a left $R$-module. Then their tensor product $M \otimes_R N$ is an abelian group which is universal with respect to bilinear maps, that is, maps $\alpha : M \times N \to Z$ where $Z$ is an abelian group satisfying $$\alpha(m + n, -) = \alpha(m, -) + \alpha(n, -)$$ $$\alpha(-, m + n) = \alpha(-, m) + \alpha(-, n)$$ $$\alpha(mr, n) = \alpha(m, rn), r \in R.$$

It can be concretely described as the quotient of the free abelian group on formal symbols $m \otimes n, m \in M, n \in N$ by relations coming from the above requirements. The tensor product is a functor (in fact an additive functor) in both of its arguments, and consequently it can be made to inherit extra structure. For example, if $M$ is also a left $S$-module and $N$ is also a right $T$-module for rings $S, T$, then $M \otimes_R N$ canonically inherits the structure of an $(S, T)$-bimodule.

If $R$ is commutative, then there is no distinction between left and right modules, so $M$ is also a left $R$-module and $N$ is also a right $R$-module. Consequently the tensor product is an $(R, R)$-bimodule, but the left and right actions are the same, so the tensor product of two $R$-modules is an $R$-module. Moreover $M \otimes_R N \cong N \otimes_R M$ canonically; the canonical map is the unique one which sends $m \otimes n$ to $n \otimes m$.

A special case of this construction is frequently used without comment. If $\phi : R \to S$ is an arbitrary ring homomorphism, then $\phi$ equips $S$ with the structure of an $(R, S)$-bimodule (left multiplication by elements of the form $\phi(r)$ determines the left structure and right multiplication by elements of $S$ determines the right structure). Consequently, the tensor product $$M \otimes_R S$$

inherits the structure of a right $S$-module. This defines a functor $- \otimes_R S : \text{Mod-}R \to \text{Mod-}S$ called extension of scalars. Extension of scalars generalizes both induction of representations in group representation theory and complexification of real vector spaces. It is the universal way to equip $M$ with the structure of an $S$-module in a way compatible with its existing $R$-module structure (where $\phi$ determines what this compatibility means; it is usually suppressed in the notation because it is understood from context, e.g. it is usually an obvious inclusion).

Now let $R$ be commutative. An $R$-algebra, in the general sense, is an $R$-module $A$ together with an $R$-bilinear multiplication $A \times A \to A$, or equivalently by the universal property a map $A \otimes_R A \to A$ of $R$-modules. We make no additional assumptions about this map: it may be associative or it may be a Lie bracket or whatever. If $A$ and $B$ are two $R$-algebras, then on the tensor product $A \otimes_R B$ we may define an $R$-bilinear multiplication by defining $$(a_1 \otimes b_1)(a_2 \otimes b_2) = a_1 a_2 \otimes b_1 b_2$$

and extending by linearity. A more invariant way to say this is that we apply the commutativity, associativity, and functoriality of the tensor product, first to define a map $$B \otimes_R (A \otimes_R A) \otimes_R B \to B \otimes_R A \otimes_R B$$

using the multiplication on $A$, and second to define a map $$A \otimes_R (B \otimes_R B) \to A \otimes_R B$$

using the multiplication on $B$, then composing the two to get a map $$(A \otimes_R B) \otimes_R (A \otimes_R B) \to A \otimes_R B.$$

This is a general definition of the tensor product of algebras. If the multiplications on $A$ and $B$ satisfy additional properties, then so does the multiplication on $A \otimes_R B$:

• If $A$ and $B$ are both associative, then so is $A \otimes_R B$.
• If $A$ and $B$ are both unital with units $1_A, 1_B$, then so is $A \otimes_R B$ with unit $1_A \otimes 1_B$.
• If $B$ is both associative and commutative, then I believe $A \otimes_R B$ inherits any properties (in the sense of nice identities satisfied by the multiplication) that $A$ does, and it is also a $B$-algebra. This is another version of extension of scalars. For example, if $A$ is a Lie algebra over $R$, then $A \otimes_R B$ is a Lie algebra over $B$; this can be used to define the complexification of a real Lie algebra.

I think that takes care of the tensor relationships between modules, vector spaces (modules over a field), algebras, fields (just special commutative rings), and rings (algebras over $\mathbb{Z}$). Groups don't really enter into the picture except insofar as abelian groups are $\mathbb{Z}$-modules.

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To take a different tack from Qiaochu's great answer, let's take a look at these objects from the point of view of General Algebra.

All of these objects are "algebras" in the sense of General Algebra: they are a set $S$, together with a family of finitary operations on $S$ (a finitary operation on $S$ is a map $S^n\to S$, where $n$ is a nonnegative integer), together with a set of identities that are satisfied.

For example, a semigroup is an algebra of type $(2)$ (meaning it has a unique binary operation), $(S,\cdot)$, subject to the identity $(a\cdot b)\cdot c = a\cdot(b\cdot c)$ (using the infix notation). A monoid is an algebra of type $(2,0)$, $(M,\cdot,e)$, (a "nullary" operation on $M$ is a map $\{\varnothing\}\to M$, so it corresponds to a distinguished element of $M$), subject to the identities $(a\cdot b)\cdot c = a\cdot(b\cdot c)$, $a\cdot e = a$, and $e\cdot a = a$. A group is an algebra of type $(2,1,0)$ (the unary operation is the operation that maps every element to its inverse), subject to certain identities. Add the identity $a\cdot b = b\cdot a$ and you get abelian groups. A ring is an algebra of type $(2,2,1,0)$ (addition, multiplication, additive inverse, additive zero); a ring with unity would be $(2,2,1,0,0)$.

A vector space over $F$ is an algebra which, in addition to the binary, unary, and nullary operations that determine its additive structure as an abelian group, has a unary operation for every element of $F$, corresponding to scalar multiplication. The axioms of a vector space are translated into identities. $R$-modules are defined similarly, with a unary operation for each element of $R$. $K$-algebras are defined like vector spaces, but they have an extra binary operation (the product in the algebra).

Caveat. Not every standard structure can be described as a general algebra. For example, fields cannot be described as a general algebra in this way, because the multiplicative inverse is not a function on the entire field: it is undefined at $0$. There is a generalization called "Partial Algebras", which allow "partially-defined operations", but their theory is much more complicated.

Now, consider the relationship between semigroups and groups. If we look at a group, $(G,\cdot, ^{-1}, e)$, then by "forgetting" about the operations $^{-1}$ and $e$ and all the identities that are imposed on those operations, we get a structure which satisfies all the requirements for being a semigroup: we have a set, $G$, a binary operation $\cdot$, and the operation is associative. That is: every group can be considered to be a semigroup by "forgetting" part of its structure. Likewise, every abelian group can be considered to be a group by "forgetting" that we are requiring the identity $ab=ba$ to be satisfied. Every ring can be considered to be an abelian group by "forgetting" about multiplication. Every vector space over $F$ can be considered to be an abelian group by "forgetting" about scalar multiplication. Every $F$-algebra can be considered an $F$-vector space by forgetting about the multiplication of elements of the algebra. Etc.

All of this can be made precise with the notion of a forgetful functor. The notion is not a formal one, so I cannot define it directly, but to paraphrase Justice Potter Stewart, "you know it when you see it". The most common forgetful functor is the "underlying set functor", which maps an algebra to its underlying set. (The category of sets can be viewed as a category of algebras, in which the collection of operations is empty).

One nice thing about forgetful functors among categories of algebras is that they always have left adjoints. If $\mathbf{U}\colon \mathcal{A}\to\mathcal{B}$ is a forgetful functor between categories of algebras, then there exists a functor $\mathbf{F}\colon \mathcal{B}\to\mathcal{A}$ such that for all objects $A\in\mathcal{A}$ and $B\in\mathcal{B}$, $$\mathrm{Hom}_{\mathcal{B}}(B,\mathbf{U}(A)) \cong \mathrm{Hom}_{\mathcal{A}}(\mathbf{F}(B),A).$$ In this situation, $\mathbf{F}$ is said to be a left adjoint of $\mathbf{U}$ (note it occurs on the left entry of the $\mathrm{Hom}$ set).

For example, the forgetful function from $\mathsf{Group}$ to $\mathsf{Set}$ has the "free group" functor from $\mathsf{Set}$ to $\mathsf{Group}$ as an adjoint: there is a natural one-to-one correspondence between set-theoretic maps between a set $X$ and the underlying set of a group $A$, and group homomorphisms between the free group on $X$ and the group $A$. Another example: the forgetful functor $\mathbf{U}\colon\mathsf{AbGroup}\to \mathsf{Group}$ has the adjoint $\mathbf{F}\colon \mathsf{Group}\to\mathsf{AbGroup}$ that sends every group $G$ to its abelianization.

Now, suppose that $F$ and $K$ are fields, $F\subseteq K$. If you consider the categories $F$-$\mathsf{VecSpace}$ and $K$-$\mathsf{VecSpaces}$, there is a natural forgetful functor from the latter to the former: just "forget" how to multiply by scalars that are not in $F$. What is the adjoint of this forgetful functor? It's the tensor product! Given an $F$-vector space $V$, and a $K$-vector space $W$, we have (using $\mathrm{Lin}_F$ to denote the collection of $F$-linear maps), $$\mathrm{Lin}_F(V,U(W)) \cong \mathrm{Lin}_K(V\otimes_FK,W).$$ So the adjoint of the underlying set functor is the tensor product with $K$. This is the same construction for modules, where a homomorphism $f\colon A\to B$ allows us to define a forgetful functor from $B$-$\mathsf{Mod}$ to $A$-$\mathsf{Mod}$ by giving a $B$-module $M$ the $A$-module structure $a\cdot m = f(a)m$.

So one can view tensor products as a particular instance of the phenomenon of the adjoint to a forgetful functor, and so generalize them in that direction. (While there are other roles the tensor product plays; for example, among commutative rings, viewed as $\mathbb{Z}$-algebras, the tensor product is the coproduct of two objects). Personally, I would say that this is the corresponding concept to "tensoring up" in the general setting.

As ever: if you want a great introduction to General Algebra, particularly from the categorical point of view, I recommend George Bergman's An invitation to General Algebra and Universal Constructions, available from his website as PDF files. There is a discussion of forgetful functors in Chapter 6, and of adjoint functor pairs in Chapter 7.

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