$SU(2)$ has a lot of interesting mathematical and physical properties. In this post we study its irreducible representations in a mathematician's way.

In this episode we focus on the rational field. What can we know about the Galois group of an irreducible polynomial with prime degree? There is a method by counting the number of nonreal roots. From this, we obtain an algorithm to compute the Galois group.

We study the Galois group of a cubic polynomial over a field with characteristic not equal to 2 and 3.

We try to prove the fundamental theorem of algebra, that the complex field is algebraically closed, using as little analysis as possible. In other words, the following proof will be *almost* algebraic.

In this post we give several forms of Masher's theorem by studying group algebra, which eventually becomes a study of semisimple rings. One can consider this post a chaotic evil introduction to representation theory or something.

This post is a continuation of a previous post about the ring of trigonometric polynomials over the real field. Now we have jumped into the complex field, and the extension is not a trivial matter.

The ring

Throughout we consider the polynomial ring

$$R=\mathbb{R}[\cos{x},\sin{x}].$$

This ring has a lot of non-trivial properties which give us a good chance to study commutative ring theory.

In this post, we study the concept of character, what it is about in abstract harmonic analysis and how to use it Galois theory.

You can find contents about Dedekind domain (or Dedekind ring) in almost all algebraic number theory books. But many properties can be proved inside ring theory. I hope you can find the solution you need in this post, and this post will not go further than elementary ring theory. With that being said, you are assumed to have enough knowledge of ring and ring of fractions (this post serves well), but not too much mathematics maturity is assumed (at the very least you are assumed to be familiar with terminologies in the linked post).$\def\mb{\mathbb}$ $\def\mfk{\mathfrak}$

Definition

There are several ways to define Dedekind domain since there are several equivalent statements of it. We will start from the one based on ring of fractions. As a friendly reminder, $\mb{Z}$ or any principal integral domain is already a Dedekind domain. In fact Dedekind domain may be viewed as a generalization of principal integral domain.

Let $\mfk{o}$ be an integral domain (a.k.a. entire ring), and $K$ be its quotient field. A Dedekind domain is an integral domain $\mfk{o}$ such that the fractional ideals form a group under multiplication. Let’s have a breakdown. By a fractional ideal $\mfk{a}$ we mean a nontrivial additive subgroup of $K$ such that

• $\mfk{o}\mfk{a}=\mfk{a}$,
• there exists some nonzero element $c \in \mfk{o}$ such that $c\mfk{a} \subset \mfk{o}$.

What does the group look like? As you may guess, the unit element is $\mfk{o}$. For a fractional ideal $\mfk{a}$, we have the inverse to be another fractional ideal $\mfk{b}$ such that $\mfk{ab}=\mfk{ba}=\mfk{o}$. Note we regard $\mfk{o}$ as a subring of $K$. For $a \in \mfk{o}$, we treat it as $a/1 \in K$. This makes sense because the map $i:a \mapsto a/1$ is injective. For the existence of $c$, you may consider it as a restriction that the ‘denominator’ is bounded. Alternatively, we say that fractional ideal of $K$ is a finitely generated $\mfk{o}$-submodule of $K$. But in this post it is not assumed that you have learned module theory.

Let’s take $\mb{Z}$ as an example. The quotient field of $\mb{Z}$ is $\mb{Q}$. We have a fractional ideal $P$ where all elements are of the type $\frac{np}{2}$ with $p$ prime and $n \in \mb{Z}$. Then indeed we have $\mb{Z}P=P$. On the other hand, take $2 \in \mb{Z}$, we have $2P \subset \mb{Z}$. For its inverse we can take a fractional ideal $Q$ where all elements are of the type $\frac{2n}{p}$. As proved in algebraic number theory, the ring of algebraic integers in a number field is a Dedekind domain.

Before we go on we need to clarify the definition of ideal multiplication. Let $\mfk{a}$ and $\mfk{b}$ be two ideals, we define $\mfk{ab}$ to be the set of all sums

$$x_1y_1+\cdots+x_ny_n$$

where $x_i \in \mfk{a}$ and $y_i \in \mfk{b}$. Here the number $n$ means finite but is not fixed. Alternatively we cay say $\mfk{ab}$ contains all finite sum of products of $\mfk{a}$ and $\mfk{b}$.

Propositions

(Proposition 1) A Dedekind domain $\mfk{o}$ is Noetherian.

By Noetherian ring we mean that every ideal in a ring is finitely generated. Precisely, we will prove that for every ideal $\mfk{a} \subset \mfk{o}$ there are $a_1,a_2,\cdots,a_n \in \mfk{a}$ such that, for every $r \in \mfk{a}$, we have an expression

$$r = c_1a_1 + c_2a_2 + \cdots + c_na_n \qquad c_1,c_2,\cdots,c_n \in \mfk{o}.$$

Also note that any ideal $\mfk{a} \subset \mfk{o}$ can be viewed as a fractional ideal.

Proof. Since $\mfk{a}$ is an ideal of $\mfk{o}$, let $K$ be the quotient field of $\mfk{o}$, we see since $\mfk{oa}=\mfk{a}$, we may also view $\mfk{a}$ as a fractional ideal. Since $\mfk{o}$ is a Dedekind domain, and fractional ideals of $\mfk{a}$ is a group, there is an fractional ideal $\mfk{b}$ such that $\mfk{ab}=\mfk{ba}=\mfk{o}$. Since $1 \in \mfk{o}$, we may say that there exists some $a_1,a_2,\cdots, a_n \in \mfk{a}$ and $b_1,b_2,\cdots,b_n \in \mfk{o}$ such that $\sum_{i = 1 }^{n}a_ib_i=1$. For any $r \in \mfk{a}$, we have an expression

$$r = rb_1a_1+rb_2a_2+\cdots+rb_na_n.$$

On the other hand, any element of the form $c_1a_1+c_2a_2+\cdots+c_na_n$, by definition, is an element of $\mfk{a}$. $\blacksquare$

From now on, the inverse of an fractional ideal $\mfk{a}$ will be written like $\mfk{a}^{-1}$.

(Proposition 2) For ideals $\mfk{a},\mfk{b} \subset \mfk{o}$, $\mfk{b}\subset\mfk{a}$ if and only if there exists some $\mfk{c}$ such that $\mfk{ac}=\mfk{b}$ (or we simply say $\mfk{a}|\mfk{b}$)

Proof. If $\mfk{b}=\mfk{ac}$, simply note that $\mfk{ac} \subset \mfk{a} \cap \mfk{c} \subset \mfk{a}$. For the converse, suppose that $a \supset \mfk{b}$, then $\mfk{c}=\mfk{a}^{-1}\mfk{b}$ is an ideal of $\mfk{o}$ since $\mfk{c}=\mfk{a}^{-1}\mfk{b} \subset \mfk{a}^{-1}\mfk{a}=\mfk{o}$, hence we may write $\mfk{b}=\mfk{a}\mfk{c}$. $\blacksquare$

(Proposition 3) If $\mfk{a}$ is an ideal of $\mfk{o}$, then there are prime ideals $\mfk{p}_1,\mfk{p}_2,\cdots,\mfk{p}_n$ such that

$$\mfk{a}=\mfk{p}_1\mfk{p}_2\cdots\mfk{p}_n.$$

Proof. For this problem we use a classical technique: contradiction on maximality. Suppose this is not true, let $\mfk{A}$ be the set of ideals of $\mfk{o}$ that cannot be written as the product of prime ideals. By assumption $\mfk{U}$ is non-empty. Since as we have proved, $\mfk{o}$ is Noetherian, we can pick a maximal element $\mfk{a}$ of $\mfk{A}$ with respect to inclusion. If $\mfk{a}$ is maximal, then since all maximal ideals are prime, $\mfk{a}$ itself is prime as well. If $\mfk{a}$ is properly contained in an ideal $\mfk{m}$, then we write $\mfk{a}=\mfk{m}\mfk{m}^{-1}\mfk{a}$. We have $\mfk{m}^{-1}\mfk{a} \supsetneq \mfk{a}$ since if not, we have $\mfk{a}=\mfk{ma}$, which implies that $\mfk{m}=\mfk{o}$. But by maximality, $\mfk{m}^{-1}\mfk{a}\not\in\mfk{U}$, hence it can be written as a product of prime ideals. But $\mfk{m}$ is prime as well, we have a prime factorization for $\mfk{a}$, contradicting the definition of $\mfk{U}$.

Next we show unicity up to a permutation. If

$$\mfk{p}_1\mfk{p}_2\cdots\mfk{p}_k=\mfk{q}_1\mfk{q}_2\cdots\mfk{q}_j,$$

since $\mfk{p}_1\mfk{p}_2\cdots\mfk{p}_k\subset\mfk{p}_1$ and $\mfk{p}_1$ is prime, we may assume that $\mfk{q}_1 \subset \mfk{p}_1$. By the property of fractional ideal we have $\mfk{q}_1=\mfk{p}_1\mfk{r}_1$ for some fractional ideal $\mfk{r}_1$. However we also have $\mfk{q}_1 \subset \mfk{r}_1$. Since $\mfk{q}_1$ is prime, we either have $\mfk{q}_1 \supset \mfk{p}_1$ or $\mfk{q}_1 \supset \mfk{r}_1$. In the former case we get $\mfk{p}_1=\mfk{q}_1$, and we finish the proof by continuing inductively. In the latter case we have $\mfk{r}_1=\mfk{q}_1=\mfk{p}_1\mfk{q}_1$, which shows that $\mfk{p}_1=\mfk{o}$, which is impossible. $\blacksquare$

(Proposition 4) Every nontrivial prime ideal $\mfk{p}$ is maximal.

Proof. Let $\mfk{m}$ be an maximal ideal containing $\mfk{p}$. By proposition 2 we have some $\mfk{c}$ such that $\mfk{p}=\mfk{mc}$. If $\mfk{m} \neq \mfk{p}$, then $\mfk{c} \neq \mfk{o}$, and we may write $\mfk{c}=\mfk{p}_1\cdots\mfk{p}_n$, hence $\mfk{p}=\mfk{m}\mfk{p}_1\cdots\mfk{p}_n$, which is a prime factorisation, contradicting the fact that $\mfk{p}$ has a unique prime factorisation, which is $\mfk{p}$ itself. Hence any maximal ideal containing $\mfk{p}$ is $\mfk{p}$ itself. $\blacksquare$

(Proposition 5) Suppose the Dedekind domain $\mfk{o}$ only contains one prime (and maximal) ideal $\mfk{p}$, let $t \in \mfk{p}$ and $t \not\in \mfk{p}^2$, then $\mfk{p}$ is generated by $t$.

Proof. Let $\mfk{t}$ be the ideal generated by $t$. By proposition 3 we have a factorisation

$$\mfk{t}=\mfk{p}^n$$

for some $n$ since $\mfk{o}$ contains only one prime ideal. According to proposition 2, if $n \geq 3$, we write $\mfk{p}^n=\mfk{p}^2\mfk{p}^{n-2}$, we see $\mfk{p}^2 \supset \mfk{p}^n$. But this is impossible since if so we have $t \in \mfk{p}^n \subset \mfk{p}^2$ contradicting our assumption. Hence $0<n<3$. But If $n=2$ we have $t \in \mfk{p}^2$ which is also not possible. So $\mfk{t}=\mfk{p}$ provided that such $t$ exists.

For the existence of $t$, note if not, then for all $t \in \mfk{p}$ we have $t \in \mfk{p}^2$, hence $\mfk{p} \subset \mfk{p}^2$. On the other hand we already have $\mfk{p}^2 = \mfk{p}\mfk{p}$, which implies that $\mfk{p}^2 \subset \mfk{p}$ (proposition 2), hence $\mfk{p}^2=\mfk{p}$, contradicting proposition 3. Hence such $t$ exists and our proof is finished. $\blacksquare$

Characterisation of Dedekind domain

In fact there is another equivalent definition of Dedekind domain:

A domain $\mfk{o}$ is Dedekind if and only if

• $\mfk{o}$ is Noetherian.
• $\mfk{o}$ is integrally closed.
• $\mfk{o}$​ has Krull dimension $1$​ (i.e. every non-zero prime ideals are maximal).

This is equivalent to say that faction ideals form a group and is frequently used by mathematicians as well. But we need some more advanced techniques to establish the equivalence. Presumably there will be a post about this in the future.

Introduction

It is quite often to see direct sum or direct product of groups, modules, vector spaces. Indeed, for modules over a ring $R$, direct products are also direct products of $R$-modules as well. On the other hand, the direct sum is a coproduct in the category of $R$-modules.

But what about tensor products? It is some different kind of product but how? Is it related to direct product? How do we write a tensor product down? We need to solve this question but it is not a good idea to dig into numeric works.

The category of bilinear or even $n$-multilinear maps

From now on, let $R$ be a commutative ring, and $M_1,\cdots,M_n$ are $R$-modules. Mainly we work on $M_1$ and $M_2$, i.e. $M_1 \times M_2$ and $M_1 \otimes M_2$. For $n$-multilinear one, simply replace $M_1\times M_2$ with $M_1 \times M_2 \times \cdots \times M_n$ and $M_1 \otimes M_2$ with $M_1 \otimes \cdots \otimes M_n$. The only difference is the change of symbols.

The bilinear maps of $M_1 \times M_2$ determines a category, say $BL(M_1 \times M_2)$ or we simply write $BL$. For an object $(f,E)$ in this category we have $f: M_1 \times M_2 \to E$ as a bilinear map and $E$ as a $R$-module of course. For two objects $(f,E)$ and $(g,F)$, we define the morphism between them as a linear function making the following diagram commutative: $\def\mor{\operatorname{Mor}}$

This indeed makes $BL$ a category. If we define the morphisms from $(f,E)$ to $(g,F)$ by $\mor(f,g)$ (for simplicity we omit $E$ and $F$ since they are already determined by $f$ and $g$) we see the composition

$$\mor(f,g) \times \mor(h,g) \to \mor(h,f)$$

satisfy all axioms for a category:

CAT 1 Two sets $\mor(f,g)$ and $\mor(f’,g’)$ are disjoint unless $f=f’$ and $g=g’$, in which case they are equal. If $g \neq g’$ but $f = f’$ for example, for any $h \in \mor(f,g)$, we have $g = h \circ f = h \circ f’ \neq g’$, hence $h \notin \mor(f,g)$. Other cases can be verified in the same fashion.

CAT 2 The existence of identity morphism. For any $(f,E) \in BL$, we simply take the identity map $i:E \to E$. For $h \in \mor(f,g)$, we see $g = h \circ f = h \circ i \circ f$. For $h’ \in \mor(g,f)$, we see $f = h’ \circ g = i \circ h’ \circ g$.

CAT 3 The law of composition is associative when defined.

---

There we have a category. But what about the tensor product? It is defined to be initial (or universally repelling) object in this category. Let’s denote this object by $(\varphi,M_1 \otimes M_2)$.

For any $(f,E) \in BL$, we have a unique morphism (which is a module homomorphism as well) $h:(\varphi,M_1 \otimes M_2) \to (f,E)$. For $x \in M_1$ and $y \in M_2$, we write $\varphi(x,y)=x \otimes y$. We call the existence of $h$ the universal property of $(\varphi,M_1 \otimes M_2)$.

The tensor product is unique up to isomorphism. That is, if both $(f,E)$ and $(g,F)$ are tensor products, then $E \simeq F$ in the sense of module isomorphism. Indeed, let $h \in \mor(f,g)$ and $h’ \in \mor(g,h)$ be the unique morphisms respectively, we see $g = h \circ f$, $f = h’ \circ g$, and therefore

$$g = h \circ h' \circ g \\ f = h' \circ h \circ f$$

Hence $h \circ h’$ is the identity of $(g,F)$ and $h’ \circ h$ is the identity of $(f,E)$. This gives $E \simeq F$.

What do we get so far? For any modules that is connected to $M_1 \times M_2$ with a bilinear map, the tensor product $M_1 \oplus M_2$ of $M_1$ and $M_2$, is always able to be connected to that module with a unique module homomorphism. What if there are more than one tensor products? Never mind. All tensor products are isomorphic.

But wait, does this definition make sense? Does this product even exist? How can we study the tensor product of two modules if we cannot even write it down? So far we are only working on arrows, and we don’t know what is happening inside an module. It is not a good idea to waste our time on ‘nonsenses’. We can look into it in an natural way. Indeed, if we can find a module satisfying the property we want, then we are done, since this can represent the tensor product under any circumstances. Again, all tensor products of $M_1$ and $M_2$ are isomorphic.

A natural way to define the tensor product

Let $M$ be the free module generated by the set of all tuples $(x_1,x_2)$ where $x_1 \in M_1$ and $x_2 \in M_2$, and $N$ be the submodule generated by tuples of the following types:

$$(x_1+x_1',x_2)-(x_1,x_2)-(x_1',x_2) \\ (x_1,x_2+x_2')-(x_1,x_2)-(x_1,x_2') \\ (ax_1,x_2)-a(x_1,x_2) \\ (x_1,ax_2) - a(x_1,x_2)$$

First we have a inclusion map $\alpha=M_1 \times M_2 \to M$ and the canonical map $\pi:M \to M/N$. We claim that $(\pi \circ \alpha, M/N)$ is exactly what we want. But before that, we need to explain why we define such a $N$.

The reason is quite simple: We want to make sure that $\varphi=\pi \circ \alpha$ is bilinear. For example, we have $\varphi(x_1+x_1’,x_2)=\varphi(x_1,x_2)+\varphi(x_1’,x_2)$ due to our construction of $N$ (other relations follow in the same manner). This can be verified group-theoretically. Note

$$\varphi(x_1+x_1',x_2)=(x_1+x_1',x_2)+N \\ \varphi(x_1,x_2)+\varphi(x_1',x_2)=(x_1,x_2)+(x_1',x_2)+N$$

but

$$\varphi(x_1+x_1',x_2)-\varphi(x_1,x_2)-\varphi(x_1',x_2)=(x_1+x_1',x_2)-(x_1,x_2)-(x_1',x_2) +N = 0+N.$$

Hence we get the identity we want. For this reason we can write

$$\begin{aligned} (x_1+x_1')\otimes x_2 &= x_1 \otimes x_2 + x_1' \otimes x_2, \\ x_1 \otimes (x_2 + x_2') &= x_1 \otimes x_2 + x_1 \otimes x_2', \\ (ax_1) \otimes x_2 &= a(x_1 \otimes x_2), \\ x_1 \otimes (ax_2) &= a(x_1 \otimes x_2). \end{aligned}$$

Sometimes to avoid confusion people may also write $x_1 \otimes_R x_2$ if both $M_1$ and $M_2$ are $R$-modules. But before that we have to verify that this is indeed the tensor product. To verify this, all we need is the universal property of free modules.

By the universal property of $M$, for any $(f,E) \in BL$, we have a induced map $f_\ast$ making the diagram inside commutative. However, for elements in $N$, we see $f_\ast$ takes value $0$, since $f_\ast$ is a bilinear map already. We finish our work by taking $h[(x,y)+N] = f_\ast(x,y)$. This is the map induced by $f_\ast$, following the property of factor module.

Trivial tensor product

For coprime integers $m,n>1$, we have $\def\mb{\mathbb}$

$$\mb{Z}/m\mb{Z} \otimes \mb{Z}/n\mb{Z} = O$$

where $O$ means that the module only contains $0$ and $\mb{Z}/m\mb{Z}$ is considered as a module over $\mb{Z}$ for $m>1$. This suggests that, the tensor product of two modules is not necessarily ‘bigger’ than its components. Let’s see why this is trivial.

Note that for $x \in \mb{Z}/m\mb{Z}$ and $y \in \mb{Z}/n\mb{Z}$, we have

$$m(x \otimes y) = (mx) \otimes y = 0 \\ n(x \otimes y) = x \otimes(ny) = 0$$

since, for example, $mx = 0$ for $x \in \mb{Z}/m\mb{Z}$ and $\varphi(0,y)=0$. If you have trouble understanding why $\varphi(0,y)=0$, just note that the submodule $N$ in our construction contains elements generated by $(0x,y)-0(x,y)$ already.

By Bézout’s identity, for any $x \otimes y$, we see there are $a$ and $b$ such that $am+bn=1$, and therefore

$$\begin{aligned} x \otimes y &= (am+bn)(x \otimes y) \\ &=am(x \otimes y)+bn (x \otimes y) \\ &= 0. \end{aligned}$$

Hence the tensor product is trivial. This example gives us a lot of inspiration. For example, what if $m$ and $n$ are not necessarily coprime, say $\gcd(m,n)=d$? By Bézout’s identity still we have

$$d(x \otimes y) = (am+bn)(x \otimes y) = 0.$$

This inspires us to study the connection between $\mb{Z}/m\mb{Z} \otimes \mb{Z}/n\mb{Z}$ and $\mb{Z}/d\mb{Z}$. By the universal property, for the bilinear map $f:\mb{Z}/m\mb{Z} \times \mb{Z}/n\mb{Z} \to \mb{Z}/d\mb{Z}$ defined by

$$(a+m\mb{Z},b+n\mb{Z})\mapsto ab+d\mb{Z}$$

(there should be no difficulty to verify that $f$ is well-defined), there exists a unique morphism $h:\mb{Z}/m\mb{Z} \otimes \mb{Z}/n\mb{Z} \to \mb{Z}/d\mb{Z}$ such that

$$h \circ \varphi(a+m\mb{Z},b+n\mb{Z}) = h((a+m\mb{Z}) \otimes(b+n\mb{Z})) = ab+d\mb{Z}.$$

Next we show that it has a natural inverse defined by

$$\begin{aligned} g:\mb{Z}/d\mb{Z} &\to \mb{Z}/m\mb{Z} \otimes \mb{Z}/n\mb{Z} \\ a+d\mb{Z} &\mapsto (a+m\mb{Z}) \otimes (1+n\mb{Z}). \end{aligned}$$

Taking $a’ = a+kd$, we show that $g(a+d\mb{Z})=g(a’+\mb{Z})$, that is, we need to show that

$$(a+m\mb{Z})\otimes(1+n\mb{Z}) = (a'+m\mb{Z}) \otimes (1+n\mb{Z}).$$

By Bézout’s identity, there exists some $r,s$ such that $rm+sn=d$. Hence $a’ = a + ksn+krm$, which gives

$$\begin{aligned} (a'+m\mb{Z}) \otimes (1+n\mb{Z}) &= (a+ksn+krm+m\mb{Z}) \otimes(1+n\mb{Z}) \\ &= (a+ksn+m\mb{Z}) \otimes (1+n\mb{Z}) \\ &=(a+m\mb{Z}) \otimes(1+n\mb{Z}) + (ksn+m\mb{Z})\otimes(1+n\mb{Z}) \\ &=(a+m\mb{Z}) \otimes (1+n\mb{Z}) \end{aligned}$$

since

$$(ksn+m\mb{Z}) \otimes (1+n\mb{Z}) =n(ks+m\mb{Z}) \otimes (1+n\mb{Z}) = (ks+m\mb{Z}) \otimes(n+n\mb{Z}) = 0.$$

So $g$ is well-defined. Next we show that this is the inverse. Firstly

$$\begin{aligned} g \circ h((a+m\mb{Z}) \otimes(b+n\mb{Z})) &= g(ab+d\mb{Z})\\ &= (ab+m\mb{Z}) \otimes (1+n\mb{Z}) \\ &=b(a+m\mb{Z}) \otimes(1+n\mb{Z}) \\ &= (a+m\mb{Z}) \otimes (b+n\mb{Z}). \end{aligned}$$

Secondly,

$$\begin{aligned} h \circ g(a+d\mb{Z}) &= h((a+m\mb{Z}) \otimes(1+n\mb{Z})) \\ &= a+d\mb{Z}. \end{aligned}$$

Hence $g = h^{-1}$ and we can say

$$\mb{Z}/m\mb{Z} \otimes \mb{Z} /n\mb{Z} \simeq \mb{Z} /\gcd(m,n)\mb{Z}.$$

If $m,n$ are coprime, then $\gcd(m,n)=1$, hence $\mb{Z}/m\mb{Z} \otimes \mb{Z}/n\mb{Z} \simeq \mb{Z}/\mb{Z}$ is trivial. More interestingly, $\mb{Z}/m\mb{Z}\otimes \mb{Z}/m\mb{Z}=\mb{Z}/m\mb{Z}$. But this elegant identity raised other questions. First of all, $\gcd(m,n)=\gcd(n,m)$, which implies

$$\mb{Z}/m\mb{Z} \otimes \mb{Z}/n\mb{Z} \simeq \mb{Z}/\gcd(m,n)\mb{Z} \simeq \mb{Z}/\gcd(n,m)\mb{Z} \simeq\mb{Z}/n\mb{Z}\otimes\mb{Z}/m\mb{Z}.$$

Further, for $m,n,r >1$, we have $\gcd(\gcd(m,n),r)=\gcd(m,\gcd(n,r))=\gcd(m,n,r)$, which gives

$$(\mb{Z}/m\mb{Z}\otimes\mb{Z}/n\mb{Z})\otimes\mb{Z}/r\mb{Z} \simeq \mb{Z}/\gcd(m,n)\mb{Z}\otimes\mb{Z}/r\mb{Z} \simeq \mb{Z}/\gcd(m,n,r)\mb{Z} \\ \mb{Z}/m\mb{Z}\otimes(\mb{Z}/n\mb{Z} \otimes\mb{Z}/r\mb{Z}) \simeq \mb{Z}/m\mb{Z} \otimes\mb{Z}/\gcd(n,r)\mb{Z} \simeq \mb{Z}/\gcd(m,n,r)\mb{Z}$$

hence

$$(\mb{Z}/m\mb{Z}\otimes\mb{Z}/n\mb{Z})\otimes\mb{Z}/r\mb{Z} \simeq \mb{Z}/m\mb{Z}\otimes(\mb{Z}/n\mb{Z}\otimes\mb{Z}/r\mb{Z}).$$

Hence for modules of the form $\mb{Z}/m\mb{Z}$, we see the tensor product operation is associative and commutative up to isomorphism. Does this hold for all modules? The universal property answers this question affirmatively. From now on we will be keep using the universal property. Make sure that you have got the point already.

Tensor product as a binary operation

Let $M_1,M_2,M_3$ be $R$-modules, then there exists a unique isomorphism

$$\begin{aligned} (M_1 \otimes M_2) \otimes M_3 &\xrightarrow{\simeq} M_1 \otimes (M_2 \otimes M_3) \\ (x \otimes y) \otimes z &\mapsto x \otimes(y \otimes z) \end{aligned}$$

for $x \in M_1$, $y \in M_2$, $z \in M_3$.

Proof. Consider the map

$$\begin{aligned} \lambda_x:M_2 \times M_3 &\to (M_1 \otimes M_2)\otimes M_3 \\ (y,z) &\mapsto (x \otimes y ) \otimes z \end{aligned}$$

where $x \in M_1$. Since $(\cdot\otimes\cdot)$ is bilinear, we see $\lambda_x$ is bilinear for all $x \in M_1$. Hence by the universal property there exists a unique map of the tensor product:

$$\overline{\lambda}_x:M_2 \otimes M_3 \to (M_1 \otimes M_2) \otimes M_3.$$

Next we have the map

$$\begin{aligned} \mu_x: M_1 \times (M_2 \otimes M_3) &\to (M_1 \otimes M_2) \otimes M_3 \\ (x,y \otimes z) &\mapsto \overline{\lambda}_x(y \otimes z) \end{aligned}$$

which is bilinear as well. Again by the universal property we have a unique map

$$\overline{\mu}_x: M_1 \otimes (M_2 \otimes M_3) \to (M_1 \otimes M_2) \otimes M_3.$$

This is indeed the isomorphism we want. The reverse is obtained by reversing the process. For the bilinear map

$$\lambda_x':M_1 \times M_2 \to M_1 \otimes (M_2 \otimes M_3)$$

we get a unique map

$$\overline{\lambda'}_x: M_1 \otimes M_2 \to M_1 \otimes (M_2 \otimes M_3).$$

Then from the bilinear map

$$\mu'_x:(M_1 \otimes M_2) \times M_3 \to M_1 \otimes (M_2 \otimes M_3)$$

we get the unique map, which is actually the reverse of $\overline{\mu}_x$:

$$\overline{\mu'}_x:(M_1 \otimes M_2) \otimes M_3 \to M_1 \otimes (M_2 \otimes M_3).$$

Hence the two tensor products are isomorphic. $\square$

Let $M_1$ and $M_2$ be $R$-modules, then there exists a unique isomorphism

$$\begin{aligned} M_1 \otimes M_2 &\xrightarrow{\simeq} M_2 \otimes M_1 \\ x_1 \otimes x_2 &\mapsto x_2 \otimes x_1 \end{aligned}$$

where $x_1 \in M_1$ and $x_2 \in M_2$.

Proof. The map

$$\begin{aligned} \lambda:M_1 \times M_2 &\to M_2 \otimes M_1 \\ (x,y) &\mapsto y \otimes x \end{aligned}$$

is bilinear and gives us a unique map

$$\overline{\lambda}:M_1 \otimes M_2 \to M_2 \otimes M_1$$

given by $x \otimes y \mapsto y \otimes x$. Symmetrically, the map $\lambda’:M_2 \times M_1 \to M_1 \otimes M_2$ gives us a unique map

$$\overline{\lambda'}:M_2 \otimes M_1 \to M_1 \otimes M_2$$

which is the inverse of $\overline{\lambda}$. $\square$

Therefore, we may view the set of all $R$-modules as a commutative semigroup with the binary operation $\otimes$.

Maps between tensor products

Consider commutative diagram:

Where $f_i:M_i \to M_i’$ are some module-homomorphism. What do we want here? On the left hand, we see $f_1 \times f_2$ sends $(x_1,x_2)$ to $(f_1(x_1),f_2(x_2))$, which is quite natural. The question is, is there a natural map sending $x_1 \otimes x_2$ to $f_1(x_1) \otimes f_2(x_2)$? This is what we want from the right hand. We know $T(f_1 \times f_2)$ exists, since we have a bilinear map by $\mu = \varphi’ \circ (f_1\times f_2)$. So for $(x_1,x_2) \in M_1 \times M_2$, we have $T(f_1 \times f_2)(x_1 \otimes x_2) = \varphi’ \circ (f_1 \times f_2)(x_1,x_2) = f_1(x_1) \otimes f_2(x_2)$ as what we want.

But $T$ in this graph has more interesting properties. First of all, if $M_1 = M_1’$ an $M_2 = M_2’$, both $f_1$ and $f_2$ are identity maps, then we see $T(f_1 \times f_2)$ is the identity as well. Next, consider the following chain

$$\cdots \to M_1 \times M_2 \xrightarrow{(f_1 \times f_2)}M_1' \times M_2' \xrightarrow{(g_1 \times g_2)}M_1'' \times M_2''\to \cdots.$$

We can make it a double chain:

It is obvious that $(g_1 \circ f_1 \times g_2 \circ f_2)=(g_1 \times g_2) \circ (f_1 \times f_2)$, which also gives

$$T(g_1 \times g_2) \circ T(f_1 \times f_2) = T(g_1 \circ f_1 \times g_2 \circ f_2).$$

Hence we can say $T$ is functorial. Sometimes for simplicity we also write $T(f_1,f_2)$ or simply $f_1 \otimes f_2$, as it sends $x_1 \otimes x_2$ to $f_1(x_1) \otimes f_2(x_2)$. Indeed it can be viewed as a map

$$\begin{aligned} T:L(M_1, M_1') \times L(M_2,M_2') &\to L(M_1 \otimes M_2, M_1' \otimes M_2') \\ (f_1 \times f_2) &\mapsto f_1 \otimes f_2. \end{aligned}$$