RS.1 Preliminaries

Exercise RS.1.2A

Proposition

Any two initial/final objects are uniquely isomorphic.

Proof

Let $A$ and $B$ be initial. Hence, there exist unique arrows $f : A \to B$ and $g : B \to A.$ Note that $f \circ g \in \Hom(B,B)$ and $\id_B \in \Hom(B,B).$ Since $\Hom(B,B)$ is a singleton, $f \circ g = \id_B.$ An analogous argument applies to $g \circ f.$ In case $A$ and $B$ are final the same argument applies in the opposite category.

$\blacksquare$

Exercise RS.1.2B

Proposition

1. The singleton set is final in $\mathbf{Set}.$
2. The empty set is initial in $\mathbf{Set}.$
3. The zero ring is the final object in $\mathbf{Ring}$
4. The ring of integers $\mathbb{Z}$ is initial in $\mathbf{Ring}$
5. The single point space is final in $\mathbf{Top}$
6. The empty space is initial in $\mathbf{Top}.$

Proof

1.

The map $s : A \to \{\star\}$ sending $a \in A$ to $\star$ is unique and defined for any $A.$

2.

There is a unique empty function $z : \emptyset \to A$ defined by the empty graph.

3.

The zero ring $0$ has the same underlying set as $\{\star\}$ so as long as a ring homomorphism from any ring $A \in \mathbf{Ring}$ to $0$ exists it is unique due to $\mathbf{Ring}$ being concrete. Consider the quotient map $A \to A/A \cong 0.$

4.

Any homomorphism $\mathbb{Z} \to A$ must map the generator of $\mathbb{Z}$ to $1_A$ and this fully determines the homomorphism, hence it is unique. We refer to elementary ring theory as a justification for existence of such homomorphism for any ring $A$.

5.

Consider a map $f$ from a topological space $T$ to a one-point space $\mathbb{1}$ (with topology $\{\emptyset, \{\star\}\}$) defined as $f(a) = \star.$ The preimage of $\emptyset$ is necessarily $\emptyset,$ while the preimage of $\{\star\}$ is $T.$ Thus $f$ is continuous. Since the underlying set of one-point space is a singleton, there exists at most one function $T \to \mathbf{1},$ which we have constructed.

6.

The empty function is vacuously continuous. Since it is the only function from an empty set to any other set, it suffices for an initial object in $\mathbf{Top}.$

$\blacksquare$

Exercise RS.1.2C

Let $A$ be a commutative ring and $S$ a multiplicative subset of $A.$

Lemma RS.1.2C.1

Let $a_1,a_2 \in A$ and $s_1,s_2 \in S$. The relation

$$a_1/s_1 \sim a_2/s_2 \iff (\exists s \in S) : s(s_2a_1 - s_1a_2) = 0$$

is an equivalence relation.

Proof

Let $s = 1.$ Observe that $s_1a_1 - s_1a_1 = 0.$ Hence $a_1/s_1 \sim a_1/s_1,$ so $\sim$ is reflexive.

Suppose, for some $s \in S,$ $s(s_2a_1 - s_1a_2) = 0.$ In this case $ss_2a_1 = ss_1a_2$ which implies $s(s_1a_2 - s_2a_1) = 0.$ So $\sim$ is symmetric.

Let $a_3 \in A$ and $s_3 \in S,$ and suppose $a_1/s_1 \sim a_2/s_2$ and $a_2/s_2 \sim a_3/s_3.$ In this case, there exist $u, v \in S$ such that

$$u(s_2a_1 - s_1a_2) = v(s_3a_2 - s_2a_3) = 0.$$

Let $w = uvs_2 \in S.$ We can directly verify that $ws_3a_1 = ws_1a_3$ holds via the following manipulations:

$$\begin{align*} ws_3a_1 & = (uvs_2)s_3a_1 \\ & = (vs_3)us_2a_1 \\ & = (vs_3)us_1a_2 \\ & = (us_1)vs_3a_2 \\ & = (us_1)vs_2a_3 \\ & = (uvs_2)s_1a_3 \\ & = ws_1a_3. \\ \end{align*}$$

Thus $\sim$ is transitive. This completes the proof that $\sim$ is an equivalence relation.

$\blacksquare$

Lemma RS.1.2C.2

Multiplication and addition are well defined on $\sim$-equivalence classes.

Proof

Let $s_1,s_2,s_3,s_4 \in S$ and $a_1,a_2,a_3,a_4 \in A$ such that $a_1/s_1 \sim a_2/s_2$ and $a_3/s_3 \sim a_4/s_4$ (see Lemma RS.1.2C.1).

By our hypothesis, there exist $u,v \in S$ such that

$$u(s_2a_1 - s_1a_2) = v(s_3a_4 - s_4a_3) = 0.$$

Addition

Note that $a_1 / s_1 + a_3/s_3$ is definitionally equal to $(s_3a_1 + s_1a_3)/(s_1s_3).$ We would like to demonstrate that

$$(s_3a_1 + s_1a_3)/(s_1s_3) \sim (s_4a_2 + s_2a_4)/(s_2s_4).$$

That is, we want to find $w \in S$ such that

$$w(s_3a_1 + s_1a_3) \cdot (s_2s_4) = w(s_4a_2 + s_2a_4) \cdot (s_1s_3).$$

Let $w = uv.$ Then the following derivation shows that the desired equality holds.

$$\begin{align*} w(s_3 a_1 + s_1 a_3)\cdot(s_2 s_4) &= uv(s_3 a_1 + s_1 a_3)\cdot(s_2 s_4) \\ &= (uv)(a_1)s_2 s_3 s_4 + (uv)(a_3)s_1 s_2 s_4 \\ &= (v)(u s_2 a_1)s_3 s_4 + (u)(v s_4 a_3)s_1 s_2 \\ &= (v)(u s_1 a_2)s_3 s_4 + (u)(v s_3 a_4)s_1 s_2 \\ &= (uv)(a_2)s_1 s_3 s_4 + (uv)(a_4)s_1 s_2 s_3 \\ &= (uv)\bigl((a_2)s_1 s_3 s_4 + (a_4)s_1 s_2 s_3\bigr) \\ &= (uv)(a_2 s_4 + a_4 s_2)\cdot(s_1 s_3) \\ &= w(s_4 a_2 + s_2 a_4)\cdot(s_1 s_3). \end{align*}$$

Multiplication

Similarly, $(a_1 / s_1) \cdot (a_3 / s_3)$ is definitionally equal to $(a_1a_3)/(s_1s_3).$ We would like to show that

$$(a_1a_3)/(s_1s_3) \sim (a_2a_4)/(s_2s_4).$$

In other words, we want to find $w \in S$ such that

$$wa_1a_3s_2s_4 = wa_2a_4s_1s_3.$$

Let $w = uv.$ Then we derive:

$$\begin{align*} w a_1 a_3 s_2 s_4 &= uv a_1 a_3 s_2 s_4 \\ &= (u s_2 a_1)(v s_4 a_3) \\ &= (u s_1 a_2)(v s_3 a_4) \\ &= uv a_2 a_4 s_1 s_3 \\ &= w a_2 a_4 s_1 s_3. \end{align*}$$

$\blacksquare$

Lemma RS.1.2C.3

$S^{-1}A$ is a ring.

Proof

Let $a_1/s_1, a_2/s_2, a_3/s_3 \in S^{-1}A.$ We will use the relation $\sim$ to denote equivalence of elements of $S^{-1}A,$ while $=$ is reserved for definitional equality.

Associativity of multiplication

$$\begin{align*} a_1/s_1 \cdot (a_2/s_2 \cdot a_3/s_3) & = a_1/s_1 \cdot ((a_2 \cdot a_3)/(s_2 \cdot s_3)) \\ & = (a_1 \cdot (a_2 \cdot a_3))/(s_1 \cdot (s_2 \cdot s_3)) \\ & = ((a_1 \cdot a_2) \cdot a_3)/((s_1 \cdot s_2) \cdot s_3) \\ & = (a_1 \cdot a_2)/(s_1 \cdot s_2) \cdot a_3/s_3 \\ & = \boxed{(a_1/s_1 \cdot a_2/s_2) \cdot a_3/s_3} \\ \end{align*}$$

Multiplicative identity

$$\begin{align*} 1/1 \cdot a_1/s_1 &= (1 \cdot a_1)/(1 \cdot s_1) \\ &= \boxed{a_1/s_1} \\ &= (a_1 \cdot 1 )/(s_1 \cdot 1)\\ &= a_1/s_1 \cdot 1/1 \end{align*}$$

Associativity of addition

$$\begin{align*} a_1/s_1 + (a_2/s_2 + a_3/s_3) & = a_1/s_1 + (a_2s_3 + a_3s_2)/(s_2s_3)\\ & = (a_1s_2s_3 + s_1a_2s_3 + s_1a_3s_2)/(s_1s_2s_3)\\ & = (a_1s_2 + a_2s_1)/(s_1s_2) + a_3/s_3 \\ & = \boxed{(a_1/s_1 + a_2/s_2) + a_3/s_3} \end{align*}$$

Additive identity

$$\begin{align*} 0/1 + a_1/s_1 &= (0 \cdot s_1 + 1 \cdot a_1)/(1 \cdot s_1) \\ &= \boxed{a_1/s_1} \\ &= (1 \cdot a_1 + 0 \cdot s_1)/(s_1 \cdot 1)\\ &= a_1/s_1 + 0/1 \end{align*}$$

Additive inverses

Let $-a_1 \in A$ such that $a_1 + (-a_1) = 0.$

$$\begin{align*} (-a_1)/s_1 + a_1/s_1 &= ((-a_1) \cdot s_1 + s_1 \cdot a_1)/(s_1 \cdot s_1) \\ &= \boxed{0/s_1^2} \\ &= (s_1 \cdot a_1 + (-a_1) \cdot s_1)/(s_1 \cdot s_1)\\ &= a_1/s_1 + (-a_1)/s_1 \end{align*}$$

Observe that $0/s_1^2 \sim 0/1,$ since $1\cdot(0\cdot 1 - 0 \cdot s_1^2) = 1\cdot 0 = 0.$

Distributivity

$$\begin{align*} (a_3/s_3) \cdot (a_1/s_1 + a_2/s_2) &= (a_3/s_3) \cdot (a_1s_2 + s_1a_2)/(s_1s_2) \\ &= \boxed{(a_3a_1s_2 + a_3s_1a_2)/(s_1s_2s_3)} \\ &= (a_1s_2 + s_1a_2)/(s_1s_2) \cdot (a_3/s_3) \\ &= (a_1/s_1 + a_2/s_2) \cdot (a_3/s_3) \end{align*}$$

Note that $1/1 \sim x/x,$ since $1\cdot(1\cdot x - x\cdot 1) = 1 \cdot 0 = 0.$

$$\begin{align*} (a_3/s_3) \cdot (a_1/s_1) + (a_3/s_3) \cdot (a_2/s_2) & = (a_3a_1)/(s_3s_1) + (a_3a_2)/(s_2s_3) \\ & = (s_2s_3a_3a_1 + s_3s_1a_3a_2)/(s_3s_1s_2s_3) \\ & = (s_3 \cdot (s_2a_3a_1 + s_1a_3a_2))/(s_3 \cdot (s_1s_2s_3)) \\ & = (s_3/s_3) \cdot (s_2a_3a_1 + s_1a_3a_2)/(s_1s_2s_3) \\ & \sim (1/1) \cdot (s_2a_3a_1 + s_1a_3a_2)/(s_1s_2s_3) \\ & \sim \boxed{(s_2a_3a_1 + s_1a_3a_2)/(s_1s_2s_3)} \\ \end{align*}$$

$\blacksquare$

Proposition

Let $A$ be a commutative ring and $S$ a multiplicative subset of $A.$

The localization map $i : A \to S^{-1} A$ defined by $i : a \mapsto a/1$ is injective if and only if $S$ contains no zero divisors.

Proof

Suppose there is an element $z \in S$ such that for some non-zero element $u \in A$ we have $zu = 0.$ Then

$$z(1 \cdot u - 1 \cdot 0) = zu = 0$$

and hence $i(u) = i(0),$ contradicting injectivity.

Conversely, suppose $i$ fails to be injective. Then there exist distinct elements $a, b \in A$ such that $i(a) = i(b),$ hence $i(a - b) = i(0).$ For this to happen, we need $(a - b)/1 = 0/1,$ in other words there must exist $s \in S$ such that

$$s\bigl(1 \cdot (a-b) - 1 \cdot 0\bigr) = 0.$$

Hence, $s(a-b) = 0$ with $s \in S.$ Since $a$ and $b$ are distinct, $s$ is a zero divisor.

$\blacksquare$

Exercise RS.1.2D

Proposition

Let $A$ be a commutative ring and $S$ a multiplicative subset of $A.$

The localization $S^{-1} A$ is initial among $A$ algebras where each element of $S$ is invertible.

Proof

Let $U$ be a commutative ring with $u : A \to U$ a ring homomorphism such that $u(S)$ contains only units. Let $i : A \to S^{-1}A$ be the localization map.

We define $u' : S^{-1}A \to U$ as $a/s \mapsto u(s)^{-1} u(a).$

Let $a_1/s_1, a_2/s_2 \in S^{-1}A.$

Suppose $a_1/s_1 = a_2/s_2,$ that is, there exists $s \in S$ such that

$$s(s_2a_1 - s_1a_2) = 0.$$

In that case,

$$\begin{align*} u'(a_1/s_1) - u'(a_2/s_2) &= u(a_1)u(s_1)^{-1} - u(a_2)u(s_2)^{-1}\\ &= u(a_1)u(s_2)u(s_2)^{-1}u(s_1)^{-1} - u(a_2)u(s_1)u(s_1)^{-1}u(s_2)^{-1}\\ &= u(s_1s_2)^{-1} u(s_2a_1 - s_1a_2)\\ &= u(s_1s_2)^{-1} \cdot u(s)^{-1}u(s) \cdot u(s_2a_1 - s_1a_2) \\ &= u(s_1s_2)^{-1} \cdot u(s)^{-1}u(s(s_2a_1 - s_1a_2)) \\ &= u(s_1s_2)^{-1} u(s)^{-1}u(0) \\ &= 0, \\ \end{align*}$$

and hence $u'(a_1/s_1) = u'(a_2/s_2).$

Since $u$ is a ring homomorphism, $u'$ preserves $1$.

We verify that $u'$ preserves products:

$$\begin{align*} u'(a_1/s_1 \cdot a_2/s_2) & = u'((a_1a_2)/(s_1s_2)) \\ & = u(s_1s_2)^{-1}u(a_1a_2) \\ & = u(s_1)^{-1}u(s_2)^{-1}u(a_1)u(a_2) \\ & = u(s_1)^{-1}u(a_1) \cdot u(s_2)^{-1}u(a_2) \\ & = u'(a_1/s_1) \cdot u'(a_2/s_2). \\ \end{align*}$$

Similarly, we verify that $u'$ preserves sums:

$$\begin{align*} u'((a_1/s_1) + (a_2/s_2)) &= u'((s_2a_1 + s_1a_2)/(s_1s_2))\\ &= u(s_1s_2)^{-1}u(s_2a_1 + s_1a_2)\\ &= u(s_1)^{-1}u(s_2)^{-1}(u(s_2)u(a_1) + u(s_1)u(a_2))\\ &= u(s_1)^{-1}u(a_1) + u(s_2)^{-1}u(a_2)\\ &= u'(a_1/s_1) + u'(a_2/s_2). \end{align*}$$

We have verified that $u'$ is a well-defined ring homomorphism. It is also clear that $u' \circ i = u,$ since

$$(u' \circ i)(a) = u'(i(a)) = u'(a/1) = u(a).$$

We will now show uniqueness of $u'.$

Let $v : S^{-1}A \to U$ be a ring homomorphism such that $v \circ i = u.$ Note that

$$1 = v(s/s) = v(s/1 \cdot 1/s) = u(s)v(1/s),$$

hence $v(1/s) = u(s)^{-1}.$ Thus

$$v(a/s)= u(a)v(1/s) = u(a)u(s)^{-1},$$

which is identical to $u'.$

$\blacksquare$

Exercise RS.1.2E

Let $A$ be a commutative ring and let $M$ be an $A$-module. Let $S$ be a multiplicative subset of $A.$

Lemma RS.1.2E.1

There exists an $S^{-1}A$-module $S^{-1}M$ defined as follows:

1. The elements of $S^{-1}M$ are of the form $m/s$ where $m \in M$ and $s \in S$
2. $m_1/s_1 + m_2/s_2 = (s_2m_1 + s_1m_2)/(s_1s_2)$ where $m_1/s_1,m_2/s_2 \in S^{-1}M$
3. $a_1/s_1 \cdot m_2/s_2 = (a_1m_2)/(s_1s_2)$ where $m_2/s_2 \in S^{-1}M$ and $a_1/s_1 \in S^{-1}A$
4. $m_1/s_1 \sim m_2/s_2 \iff (\exists s \in S) : s(s_2m_1 - s_1m_2) =0$ where $m_1/s_1,m_2/s_2 \in S^{-1}M$

Like before, $\sim$ denotes equivalence of elements within $S^{-1}M,$ while $=$ is reserved for definitional equality.

Proof

The checks from RS.1.2C apply here almost verbatim, interpreting multiplication by $s \in S$ via $A$-action on $M.$ The relevant proofs don’t use multiplication by numerators, which makes them suitable for $S^{-1}M.$ The well-definiteness of ring multiplication can be straightforwardly reinterpreted as well-definedness of scalar multiplication.

$\blacksquare$

Proposition

There exists map $\varphi : M \to S^{-1}M$ satisfying the universal property of localization of $A$-modules with respect to $S$.

Proof

In RS.1.2E.1 we proved that $S^{-1}M$ exists. We define $\varphi : M \to S^{-1}M$ as $m \mapsto m/1.$

We define category $\mathcal{U}$ whose objects are $A$-module homomorphisms $M \to V$ for every $S^{-1}A$-module $V.$ The morphisms of $\mathcal{U}$ are commutative triangles: the morphism between $f : M \to V$ and $g : M \to U$ carries the same data as an $A$-module homomorphism $h : V \to U$ such that $h \circ f = g.$

Let $q : M \to V \in \mathcal{U}.$ We want to show that there exists unique morphism $q' : \varphi \to q,$ which can be understood as a morphism $q' : S^{-1}M \to V$ such that $q' \circ \varphi = q.$

Define $q'$ as $m_1/s_1 \mapsto s_1^{-1} \cdot q(m_1).$ We will skip the routine verification that it is indeed a homomorphism and that it is well-defined. We can see that $q' \circ \varphi = q$ since $q'(m/1) = 1^{-1} \cdot q(m) = q(m).$

Is $q'$ unique? Suppose we have $v : S^{-1}M \to V$ such that $v \circ \varphi = q.$

Observe that $v(m_1/s_1) = s^{-1}_1 \cdot v(m_1/1)$ since $s_1 \cdot v(m_1/s_1) = v(m_1/1)$ and multiplication by $s_1 \in S$ (denoted $s_1 \cdot \square : V \to V$) is an isomorphism, whose inverse we write as $s_1^{-1} \cdot \square : V \to V.$ A very similar argument applies to show that an $A$-module homomorphism between two $S^{-1}A$-modules lifts to an $S^{-1}A$-module homomorphism uniquely.

We can easily establish $q' = v$ as follows:

$$\begin{align*} v(m_1/s_1) &= s_1^{-1} \cdot v(m_1/1) \\ &= s_1^{-1} \cdot v(\varphi(m_1)) \\ &= s_1^{-1} \cdot q(m_1) \\ &= s_1^{-1} \cdot q'(\varphi(m_1)) \\ &= s_1^{-1} \cdot q'(m_1/1) \\ &= q'(m_1/s_1) \\ \end{align*}$$

$\blacksquare$

Note

The book states the universal property in terms of an $A$-homomorphism of $S^{-1}A$-modules. Since this homomorphism carries the same data as an $S^{-1}A$-homomorphism between the same modules, the same universal property applies regardless.

Exercise RS.1.2F

Proposition a/b

Localization commutes with arbitrary direct sums.

Proof

Let $A$ be a commutative ring and let $S$ be a multiplicative subset of $A.$

Let $I$ be an index set and let $M_{\square} : I \to \mathbf{A\text{-}Mod}$ be a family of $A$-modules.

We want to show that

$$S^{-1} \bigoplus_{i \in I}M_i \cong \bigoplus_{i \in I}S^{-1}M_i,$$

where the isomorphism is understood as an isomorphism of $S^{-1}A$-modules.

Let $u_i : M_i \to S^{-1}M_i$ be an $I$-indexed family of localization maps. Let

$$u : \bigoplus_{i\in I}M_i \to \bigoplus_{i \in I}S^{-1}M_i$$

be the direct sum $\bigoplus_{i \in I}u_i.$

Let $X$ be an $S^{-1}A$-module and let $v : \bigoplus_{i \in I}M_i \to X$ be an $S^{-1}A$-module homomorphism. We will use $v_i : M_i \to X$ to denote the $i$-th component of $v.$

Let $z_i : S^{-1}M_i \to X$ be the unique $S^{-1}A$-module homomorphism such that $z_i \circ u_i = v_i$ (via the universal property of $S^{-1}M_i$).

We define $z : \bigoplus_{i \in I}S^{-1}M_i \to X$ as the copairing $[z_i \mid i \in I].$ By construction, $z \circ u = v.$ Is $z$ unique? Suppose $z' : \bigoplus_{i \in I}S^{-1}M_i \to X$ is an $S^{-1}A$-module homomorphism such that $z' \circ u = v.$ Then for each component $i \in I$ we see that $z_i \circ u_i = v_i$ and $z'_i \circ u_i = v_i.$ Since the universal property of $S^{-1}M_i$ guarantees that $z_i$ is the unique such map, $z_i = z_i'.$ Hence $z = z'.$

Both $S^{-1}\bigoplus_{i\in I}M_i$ and $\bigoplus_{i\in I}S^{-1}M_i$ satisfy the universal property of localization of $\bigoplus_{i\in I}M_i,$ and thus are uniquely isomorphic, that is, $S^{-1} \bigoplus_{i \in I}M_i \cong \bigoplus_{i \in I}S^{-1}M_i.$

$\blacksquare$

Note

We use the universal property with $S^{-1}A$-module homomorphism. As in RS.1.2E, this is identical to the universal property with $A$-module homomorphism as given in the book.

Proposition c

Localization does not commute with infinite products.

Proof

Let $M_i = \mathbb{Z}$ for all $i \in \mathbb{N}.$ Let $W = \prod_{i \in \mathbb{N}} M_i.$ Note that $\mathbb{Q}$ is the localization of $\mathbb{Z}$ with respect to $\mathbb{Z}_{>0}$ (the subset of the ring containing positive elements). Let $Z = \prod_{i \in \mathbb{N}} \mathbb{Q}.$

We will consider two candidate localizations of $W,$ specifically $Z$ and $\mathbb{Z}_{>0}^{-1}W.$ Suppose, for the sake of contradiction, that the canonical localization map $\tau : \mathbb{Z}_{>0}^{-1}W \to Z$ given by $w/d \mapsto (w_1/d,w_2/d,\ldots)$ is an isomorphism with inverse $\tau^{-1} : Z \to \mathbb{Z}_{>0}^{-1}W$ .

Observe that $\xi = (1,1/2,1/3,1/4,\ldots) \in Z.$ What would $\tau^{-1}(\xi)$ look like? It has to have the form $(i_1,i_2,i_3,\ldots)/d$ for $i_n \in \mathbb{Z}, d \in \mathbb{Z}_{>0}$ such that $i_n/d =1/n$ (as a fraction) which is impossible: the denominator of left hand side is fixed, while denominators of right hand side grow unbounded. Alternatively, observe that the existence of such value would imply that there exists a non-zero $d$ such that for all $n \in \mathbb{N}$ the equality $i_n n =d$ holds which is nonsensical.

$\blacksquare$

Exercise RS.1.2G

Proposition

$\mathbb{Z}/(10) \otimes_{\mathbb{Z}} \mathbb{Z}/(12) \cong \mathbb{Z}/(2)$

Proof

Let $[a]_{10} \in \mathbb{Z}/(10)$ and $[b]_{12} \in \mathbb{Z}/(12).$ Then

$$[a]_{10} \otimes [b]_{12} = a \cdot ([1]_{10} \otimes [b]_{12}) = a \cdot (b \cdot ([1]_{10} \otimes [1]_{12})).$$

Let $S = [1]_{10} \otimes [1]_{12}.$ Observe that $S$ generates $\mathbb{Z}/(10) \otimes_{\mathbb{Z}} \mathbb{Z}/(12)$ as a $\mathbb{Z}$-module.

We find order of $S$ by a direct computation combined with a homomorphism argument:

$$\begin{align*} S + S &= [2]_{10} \otimes [1]_{12} \\ &= [12]_{10} \otimes [1]_{12} \\ &= 12 \cdot ([1]_{10} \otimes [1]_{12}) \\ &= [1]_{10} \otimes [12]_{12} \\ &= [1]_{10} \otimes [0]_{12} \\ &= 0 \cdot ([1]_{10} \otimes [1]_{12}) \\ &= 0. \end{align*}$$

Define $\tau : \mathbb{Z}^{\mathbb{Z}/(10) \times \mathbb{Z}/(12)} \to \mathbb{Z}/(2)$ as $\tau([a]_{10},[b]_{12}) = [ab]_2.$ Clearly this is an epimorphism. Since multiplication is bilinear, $\tau$ factors uniquely through $\mathbb{Z}/(10) \otimes_{\mathbb{Z}} \mathbb{Z}/(12)$. Since $\tau = \tau' \circ \pi,$ $\tau'$ must itself be an epimorphism, hence $\mathbb{Z}/(10) \otimes_{\mathbb{Z}} \mathbb{Z}/(12)$ is non-trivial.

There’s exactly one $\mathbb{Z}$-module of order $2,$ so $\mathbb{Z}/(10) \otimes_{\mathbb{Z}} \mathbb{Z}/(12) \cong \mathbb{Z}/(2).$

$\blacksquare$

Exercise RS.1.2H

Proposition 1

$- \otimes_A N$ is a covariant endofunctor on the category of $A$-modules.

Proof

Let $M_1, M_2$ be $A$-modules and $f: M_1 \to M_2$ an $A$-module homomorphism. Let $g : M_1 \otimes_A N \to M_2 \otimes_A N$ be a map defined by $m \otimes n \mapsto f(m) \otimes n.$

We verify that $g$ is well-defined:

A technicality

Formally speaking, I define $g'$ via the universal property of free object $F = A^{\oplus (M_1 \times N)}$ by specifying it on the basis (pure tensors) and extending by $A$-linearity. Then $g$ arises via the universal property of quotient, but only if $g'$ kills the the relation submodule of $F.$

This is the “well-definedness” check: I write things like $g(a_1 \otimes b + a_2 \otimes b - (a_1 + a_2) \otimes b)$ but treat tensors in $g$‘s domain as inert formal symbols, making this derivation translateable into a derivation talking about things like $g'([a_1, b] + [a_2, b] - [(a_1 + a_2), b]).$

This avoids some of the bureaucracy, such as having to introduce explicit $g'$ and a separate syntax for “formal tensors,” at the cost of requiring one to pay more attention to the context.

$$\begin{align*} g(a_1 \otimes b + a_2 \otimes b - (a_1 + a_2) \otimes b) & = g(a_1 \otimes b) + g(a_2 \otimes b) - g((a_1 + a_2) \otimes b) \\ & = f(a_1) \otimes b + f(a_2) \otimes b - f(a_1 + a_2) \otimes b \\ & = f(a_1) \otimes b + f(a_2) \otimes b - (f(a_1) + f(a_2)) \otimes b \\ & = f(a_1) \otimes b + f(a_2) \otimes b - (f(a_1)\otimes b + f(a_2) \otimes b) \\ & = 0 \end{align*}$$ $$\begin{align*} g(a \otimes b_1 + a \otimes b_2 - a \otimes (b_1 + b_2)) & = g(a \otimes b_1) + g(a \otimes b_2) - g(a \otimes (b_1 + b_2)) \\ & = f(a) \otimes b_1 + f(a) \otimes b_2 - f(a) \otimes (b_1 + b_2) \\ & = f(a) \otimes b_1 + f(a) \otimes b_2 - f(a) \otimes b_1 - f(a) \otimes b_2 \\ & = 0 \end{align*}$$ $$\begin{align*} g(v \cdot (a \otimes b) - va \otimes b) & = g(v \cdot (a \otimes b)) - g(va \otimes b) \\ & = v \cdot g(a \otimes b) - f(va) \otimes b \\ & = v \cdot (f(a) \otimes b) - f(va) \otimes b \\ & = (v \cdot f(a)) \otimes b - f(va) \otimes b \\ & = f(va) \otimes b - f(va) \otimes b \\ & = 0 \end{align*}$$ $$\begin{align*} g(v \cdot (a \otimes b) - a \otimes vb) & = g(v \cdot (a \otimes b)) - g(a \otimes vb) \\ & = v \cdot g(a \otimes b) - f(a) \otimes vb \\ & = v \cdot (f(a) \otimes b) - f(a) \otimes vb \\ & = f(a) \otimes vb - f(a) \otimes vb \\ & = 0 \end{align*}$$

Thus, given any map of $A$-modules $f: M_1 \to M_2$ we can always define a map $g : M_1 \otimes_A N \to M_2 \otimes_A N.$ We will assign this action on morphisms to $- \otimes_A N$ and then verify functorality.

Let $\id_{M_1} : M_1 \to M_1$ be the identity map. Then the map $\id_{M_1} \otimes_A N$ is given by

$$m \otimes n \mapsto \id_{M_1}(m) \otimes n = m \otimes n,$$

so $- \otimes_A N$ preserves identities.

Let $i : M_1 \to M_2,$ $j : M_2 \to M_3.$ Then $(j \circ i) \otimes_A N$ is given by $m \otimes n \mapsto j(i(m)) \otimes n,$ while $j \otimes_A N$ acts as $m \otimes n \mapsto j(m) \otimes n$ and $i \otimes_A N$ acts as $m \otimes n \mapsto i(m) \otimes n.$ It is clear that

$$(j \circ i) \otimes_A N = (j \otimes_A N) \circ i \otimes_A N.$$

Thus $- \otimes_A N$ is indeed an endofunctor on $\mathbf{A\text{-}Mod}.$

$\blacksquare$

Lemma RS.1.2H.1

Let $M_1, M_2$ be $A$-modules and $f : M_1 \to M_2.$ Then there exists a surjection $u : (\im f) \otimes_A N \to \im (f \otimes_A N).$

Proof

Let $F = A^{\oplus ((\im f) \times N)}$ be a free module generated by formal symbols $[f(m),n].$ Let $R$ be a submodule of $F$ generated by differences corresponding to bilinearity relations. Let $\pi : F \to F/R$ be the quotient map. We will identify $F/R$ with $(\im f) \otimes_A N.$ We define $u' : F \to \im (f \otimes_A N)$ on generators (and extend by universal property of free object) as $[f(m),n] \mapsto (f \otimes_A N)(\hat{m} \otimes n),$ where $\hat{m}$ is a preimage of $f(m).$ Of course $(f \otimes_A N)(\hat{m} \otimes n) = f(m) \otimes n.$

We will now check that $u'$ kills $R.$

$$\begin{align*} {} & u'([f(m_1),n] + [f(m_2),n] - [f(m_1) + f(m_2),n]) \\ ={} & u'([f(m_1),n]) + u'([f(m_2),n]) - u'([f(m_1) + f(m_2),n]) \\ ={} & f(m_1) \otimes n + f(m_2) \otimes n - (f(m_1) + f(m_2)) \otimes n \\ ={} & 0 \end{align*}$$ $$\begin{align*} {} & u'([f(m),n_1] + [f(m),n_2] - [f(m),n_1 + n_2]) \\ ={} & u'([f(m),n_1]) + u'([f(m),n_2]) - u'([f(m),n_1 + n_2]) \\ ={} & f(m) \otimes n_1 + f(m) \otimes n_2 - f(m) \otimes (n_1 + n_2) \\ ={} & 0 \end{align*}$$ $$\begin{align*} {} & u'([vf(m),n] - v[f(m),n]) \\ ={} & u'([vf(m),n]) - u'(v[f(m),n]) \\ ={} & (vf(m)) \otimes n - v \cdot (f(m) \otimes n)\\ ={} & 0 \end{align*}$$ $$\begin{align*} {} & u'([f(m),vn] - v[f(m),n]) \\ ={} & u'([f(m),vn]) - u'(v[f(m),n]) \\ ={} & f(m) \otimes (vn) - v \cdot (f(m) \otimes n)\\ ={} & 0 \end{align*}$$

Thus, $u'$ can be factored as $u : F/R \to \im (f \otimes_A N)$ satisfying $u \circ \pi = u'.$

Observe that $u'$ is a surjection, since, given $z = f(m)\otimes n \in \im(f \otimes_A N),$ $u'([f(m), n]) = z.$ Since pure tensors generate $\im (f \otimes_A N),$ this is sufficient.

$\blacksquare$

Proposition 2

Let $M' \to M \to M'' \to 0$ be an exact sequence of $A$-modules. Then $M' \otimes_A N \to M \otimes_A N \to M'' \otimes_A N \to 0$ is also exact.

Proof

Let $f : M' \to M$ and $g : M \to M''$ such that $\im f = \ker g$ and $\im g = M''.$ Let $\hat{f} : M' \to \im f$ and $\hat{g} : M \to \im g$ be surjections onto the respective images.

Let $\widehat{g \otimes_A N} : M \otimes_{A}N \to \im (g \otimes_A N)$ be surjection onto image of $g \otimes_A N.$ By RS.1.2H.1 we have a surjection $v : (\im g) \otimes_A N \to \im (g \otimes_A N).$

TODO: finish

$\blacksquare$