WJD Open Notebook

2008 November {#sec:Nov2008}

All rings are unitary, i.e. have an identity.

List the ideals of the rings:

$\Z/12\Z$ $M_2(\R)$ (the ring of $2\times 2$ real matrices) $\Z/12\Z \times M_2(\R)$.

An ideal is, in particular, a subgroup of the additive group. Since $\langle \Z_{12}; +, 0\rangle$ is cyclic, all subgroups are cyclic. Thus, the subgroups of $\langle \Z_{12}; +, 0\rangle$ are as follows:

• $(0) = {0}$

• $(1) = {0, 1, \dots, 11}$

• $(2) = {0, 2, 4, 6, 8, 10}$

• $(3) = {0, 3, 6, 9}$

• $(4) = {0, 4, 8}$

• $(6) = {0, 6}$

Of course, 5, 7, and 11 are relatively prime to 12, so $(1) = (5) = (7) = (11) = \Z_{12}$. The remaining generators, 8, 4, and 9, are also redundant: $(8) = (4)$, $(9) = (3)$, and $(10)=(2)$. It is easy to verify that each of the 6 distinct subgroups listed above—call them ${G_i : 1\leq i\leq 6}$—have the property that, if $r\in \Z_{12}$ and $a\in G_i,$ then $r a \in G_i$. Thus, each $G_i,$ $1\leq i \leq 6,$ is an ideal of $\Z_{12}$.

(Remark: This also shows $\Z_{12}$ is a PID, hence the proper prime ideals are the maximal ideals, $(2)$ and $(3)$.)

By the following lemma, since $\R$ is a field (and thus has no non-trivial proper ideals), the only ideals of $M_2(\R)$ are $M_2(0)$ and $M_2(\R)$.

Let $R$ be a commutative ring with $1_R \neq 0$. Then the ideals of $M_2(R)$ are precisely the subsets $M_2(J)\subseteq M_2(R)$, where $J$ is an ideal of $R$.

Suppose $J$ is an ideal and $A_1, A_2 \in M_2(J)$. Then $A_1-A_2$ has all elements in $J$, since $J$ is, in particular, a subgroup. Therefore $M_2(J)$ is a subgroup of $\langle M_2(R); +, \mathbf{0}\rangle$. If $M = (m_{ij}) \in M_2(R)$ and $A = (a_{ij}) \in M_2(J)$, then $MA = (\sum_{k=1}^2 m_{ik}a_{kj})$. Clearly, all the elements of $MA$ are in $J,$ since $J$ is an ideal and $a_{ij}\in J$. Similarly, all the elements of $AM$ are in $J$. This proves that $M_2(J)$ is an ideal of $M_2(R)$ whenever $J$ is an ideal of $R$.

Now, let $\mathscr{I} \subseteq M_2(R)$ be an ideal of $M_2(R)$, and let $J$ be the set of all $a\in R$ such that $a$ is an element of some $A\in \mathscr{I}$. Clearly $0\in J,$ since $A\in \mathscr{I}$ implies $\mathbf{0} = A - A \in \mathscr{I}$. Let $a, b \in J$. Suppose $a$ is the $ij^{th}$ entry of $A\in \mathscr{I}$ and $b$ is the $kl^{th}$ entry of $B\in \mathscr{I}$. Let $M_{ij}$ be a matrix with 1 in the $ij^{th}$ position and 0 elsewhere. Then $M_{1i}AM_{j1} = \begin{pmatrix}a & 0\0&0\end{pmatrix} \in \mathscr{I}$ and $M_{1k}BM_{l1} = \begin{pmatrix}b & 0\0&0\end{pmatrix} \in \mathscr{I}$. Therefore, $\begin{pmatrix}a-b & 0\0&0\end{pmatrix}\in \mathscr{I}$, so $a-b\in J$, whence, $J$ is a subgroup. Fix $r\in R$. Then $\begin{pmatrix}r&0\0&0\end{pmatrix}\begin{pmatrix}a&0\0&0\end{pmatrix} = \begin{pmatrix}ra&0\0&0\end{pmatrix} \in \mathscr{I}$, so $ra \in J$, and $J$ is an ideal.

By the following lemma, and parts [a.]{} and [b.]{}, the ideals of $\Z/12\Z \times M_2(\R)$ are all sets of the form $I \times J$, where $I\in {(0), (1), (2), (3), (4), (6)}$ and $J \in {M_2(0), M_2(\R)}$.

Let $R$ and $R'$ rings. Then the ideals of $R\times R'$ have the form $I\times J$, where $I$ is an ideal of $R$ and $J$ is an ideal of $R'$.

Let $A \subseteq R\times R'$ be an ideal. we will show

1. [it:1] $A = A_1\times A_2,$ for some $A_1\subseteq R$ and $A_2\subseteq R',$ and

2. [it:2] $A_1$ and $A_2$ are ideals of $R$ and $R'$, respectively.

([it:1]) Define $$A_1 = {a\in R: (a,a') \in A \text{ for some } a'\in R'} \quad \text{and} \quad A_2 = {a'\in R': (a,a') \in A \text{ for some } a\in R}.$$ That is, $A_1$ (resp. $A_2$) is the set of all first (resp. second) coordinates of elements in $A$. I claim that $A = A_1 \times A_2$.¹ Fix $a\in A_1$ and $b'\in A_2$. We show $(a,b')\in A$. Since $a\in A_1,$ there is some $a'\in A_2$ such that $(a,a')\in A$. Similarly, $(b,b')\in A,$ for some $b\in A_1$. Since $A$ is an ideal, $(1,0)\cdot(a,a') = (a,0)\in A$, and $(0,1)\cdot(b,b') = (0,b')\in A$. Therefore, $(a,0) + (0,b') = (a, b') \in A,$ as claimed. This proves $A_1\times A_2 \subseteq A$. The reverse inclusion is obvious.

([it:2]) Fix $a, b \in A_1$. Then, there exist $a', b' \in A_2$ such that $(a,a')\in A$ and $(b,b')\in A$. Now, since $A$ is, in particular, a subgroup of the additive group of $R\times R'$, we have $(a,a')- (b,b') = (a-b, a'-b')\in A$. Therefore, $a-b \in A_1$, so $A_1$ is a subgroup of $\langle R, +, 0\rangle$. Fix $r\in R$. Then $(ra,a')= (ra,1a') = (r,1)\cdot(a,a') \in A,$ since $A$ is an ideal of $R\times R'$. Therefore, $ra\in A_1,$ which proves that $A_1$ is an ideal of $R$. The same argument, [mutatis mutandis]{}, proves that $A_2$ is an ideal of $R'$. This completes the proof of (ii) and establishes the lemma.

Let $R$ be a commutative ring and let $A, B,$ and $C$ be $R$-modules with $A$ a submodule of $B$. Which (if any) of the following conditions guarantee that the natural map $A \otimes_R C \rightarrow B \otimes_R C$ is injective?

(a) $C$ is free,

(b) $C$ is projective.

I will prove a slightly more general result (as the increased generality comes at no additional cost). That is, if $\varphi: A \rightarrow B$ is an $R$-module monomorphism, then, under conditions (a) or (b), the natural map $\varphi \otimes 1_C: {\ensuremath{A\otimes_R}}C \rightarrow {\ensuremath{B\otimes_R}}C$ in injective. (To answer the question above, take $\varphi$ to be the inclusion map.)

We begin with the simplest case, where $C = R$.

: If $\varphi: A \rightarrow B$ is an $R$-module monomorphism, then the natural map $\varphi \otimes 1_R: {\ensuremath{A\otimes_R}}R \rightarrow {\ensuremath{B\otimes_R}}R$ is injective.

: Define $\alpha_A: {\ensuremath{A\otimes_R}}R \cong A$ by $\alpha_A(a\otimes r) = ar$, and $\alpha_B: {\ensuremath{B\otimes_R}}R \cong B$ by $\alpha_B(b\otimes r) = br$. (See lemma [lemma-A4] below for proof that these are, indeed, isomorphisms.) I claim that the following diagram of $R$-module homomorphisms is commutative: $$\begin{CD} {\ensuremath{A\otimes_R}}R @>{\alpha_A}>> A \ @V{{\ensuremath{\varphi \otimes 1_R}}}VV @VV{\varphi}V \ {\ensuremath{B\otimes_R}}R @>{\alpha_B}>> B \end{CD}$$ Fix $a\in A$ and $r\in R$. By definition, $({\ensuremath{\varphi \otimes 1_R}})(a\otimes r) = \varphi(a) \otimes 1_R(r)$. Therefore, $(\alpha_B, ({\ensuremath{\varphi \otimes 1_R}}))(a\otimes r) = \alpha_B(\varphi(a) \otimes r) = \varphi(a)r$. Following the diagram in the other direction, $(\varphi, \alpha_A)(a\otimes r) = \varphi(ar) = \varphi(a)r,$ since $\varphi$ is an $R$-module homomorphism. Therefore, $(\alpha_B, ({\ensuremath{\varphi \otimes 1_R}}))(a\otimes r) = (\varphi, \alpha_A)(a\otimes r)$. Since $a\in A$ and $r\in R$ were chosen arbitrarily, $\alpha_B, ({\ensuremath{\varphi \otimes 1_R}})$ and $\varphi, \alpha_A$ agree on generators, so the diagram is commutative.

Now $\alpha_A$ is an isomorphism, so $\alpha_A$ and $\varphi$ are both injective. Therefore, by lemma [lemma-A1] (below), $\varphi, \alpha_A$ is injective. Then, by commutativity of the diagram, $\alpha_B,({\ensuremath{\varphi \otimes 1_R}})$ is injective. Finally, lemma [lemma-A3] implies that ${\ensuremath{\varphi \otimes 1_R}}$ is injective, which proves claim 1.

Next, let $C$ be a free $R$-module. Then $C \cong \sum_{i\in I}R,$ for some index set $I$. For ease of notation, in the sequel, $\sum$ denotes $\sum_{i\in I}$.

: If $\varphi: A \rightarrow B$ is an $R$-module monomorphism, then the natural map $\varphi \otimes 1_{\sum R}: {\ensuremath{A\otimes_R}}\sum R \rightarrow {\ensuremath{B\otimes_R}}\sum R$ is injective.

: Define $\Phi: \sum A \rightarrow \sum B$ by $\Phi({a_i}) = {\varphi(a_i)}$ for each ${a_i} \in \sum A$. Then $\Phi$ is an $R$-module monomorphism. This is seen as follows: $(\forall {a_i}, {a'_i} \in \sum A);(\forall r\in R)$ $$\Phi({a_i} + {a'_i}) = \Phi({a_i + a'_i}) = {\varphi(a_i + a'_i)} = {\varphi(a_i) + \varphi(a'_i)} = {\varphi(a_i)} + {\varphi(a'_i)} = \Phi({a_i}) + \Phi({a'_i}),$$ $$\text{ and } ; \Phi(r{a_i} ) = \Phi({ra_i})= {\varphi(ra_i)} = {r\varphi(a_i)} = r{\varphi(a_i)} = r\Phi({a_i} ).$$ If $\Phi({a_i} ) = {0} \in \sum B,$ then ${\varphi(a_i)} = 0,$ which implies $\varphi(a_i) = 0$ for all $i\in I$, and therefore, ${a_i} = 0 \in \sum A$. Thus, $\Phi$ is a monomorphism.

By (the proof of) lemma [lemma-A4], the map $\sigma: \sum A \rightarrow \sum({\ensuremath{A\otimes_R}}R)$ given by $\sigma({a_i}) = {a_i \otimes 1_R}$ is an isomorphism. By (the proof of) lemma [lemma-A5], there is an isomorphism $\tau: \sum({\ensuremath{A\otimes_R}}R) \rightarrow {\ensuremath{A\otimes_R}}\sum R,$ defined on generators by $$\tau(a\otimes r, 0, \dots) = a \otimes (r,0,\dots), \quad \tau(0, a\otimes r, 0, \dots) = a \otimes (0,r,0,\dots), \dots, ; \text{ and }$$ $$\begin{aligned} \tau(a_1\otimes r_1, a_2\otimes r_2, \dots) &= \tau((a_1\otimes r_1, 0, \dots) + (0, a_2\otimes r_2, 0, \dots) + \cdots)\ &= \tau(a_1\otimes r_1, 0, \dots) + \tau(0, a_2\otimes r_2, 0, \dots) + \cdots\ &= a_1\otimes (r_1, 0, \dots) + a_2\otimes (0,r_2, 0, \dots) + \cdots. \end{aligned}$$ (The sum has finitely many non-zero terms, since ${a_i}$ is non-zero for finitely many indices $i\in I$.)

Consider the following diagram of $R$-module homomorphisms: $$\label{eq:CD2} \begin{CD} \sum A @>{\sigma}>> \sum ({\ensuremath{A\otimes_R}}R) @>{\tau}>> {\ensuremath{A\otimes_R}}\sum R \ @V{\Phi}VV @. @VV{\varphi \otimes 1_{\sum R}}V \ \sum B @>{\bar{\sigma}}>> \sum ({\ensuremath{B\otimes_R}}R) @>{\bar{\tau}}>> {\ensuremath{B\otimes_R}}\sum R \end{CD}$$ where $\bar{\sigma}$ and $\bar{\tau}$ are the same maps as their bar-less counterparts, but defined on $\sum B$ and $\sum ({\ensuremath{B\otimes_R}}R)$, respectively. I claim that diagram ([eq:CD2]) is commutative. Indeed, for any ${a_i} \in \sum A$, we have² $$\begin{aligned} \tau ,\sigma ({a_i} ) &= \tau ( {a_i \otimes 1_R}) = \tau ((a_1\otimes 1_R, 0, \dots)+(0, a_2\otimes 1_R, 0, \dots)+\cdots)\ &= a_1\otimes (1_R, 0, \dots)+ a_2\otimes (0, 1_R, 0, \dots)+\cdots.\[4pt] \therefore ; (\varphi\otimes 1_{\sum R}),\tau , \sigma ({a_i}) &= (\varphi\otimes 1_{\sum R})(a_1\otimes (1_R, 0, \dots)+ a_2\otimes (0, 1_R, 0, \dots)+\cdots)\ &= (\varphi\otimes 1_{\sum R})(a_1\otimes (1_R, 0, \dots))+ (\varphi\otimes 1_{\sum R})(a_2\otimes (0, 1_R, 0, \dots)) +\cdots\ &=\varphi(a_1)\otimes (1_R, 0, \dots)+ \varphi(a_2)\otimes (0, 1_R, 0, \dots)+\cdots. \end{aligned}$$ In the other direction, $\bar{\sigma},\Phi({a_i}) = \bar{\sigma}({\varphi(a_i)}) = {\varphi(a_i)\otimes 1_R},$ so $$\begin{aligned} \bar{\tau} ,\bar{\sigma}, \Phi({a_i} ) &= \bar{\tau} ( {\varphi(a_i) \otimes 1_R}) = \tau ((\varphi(a_1)\otimes 1_R, 0, \dots)+(0, \varphi(a_2)\otimes 1_R, 0, \dots)+\cdots)\ &=\varphi(a_1)\otimes (1_R, 0, \dots)+ \varphi(a_2)\otimes (0, 1_R, 0, \dots)+\cdots, \end{aligned}$$ which proves that $\bar{\tau} ,\bar{\sigma}, \Phi$ and $(\varphi\otimes 1_{\sum R}),\tau , \sigma$ agree on generators. Therefore, diagram ([eq:CD2]) is commutative.

Now, $\bar{\sigma}$ and $\bar{\tau}$ are $R$-module isomorphisms, and $\Phi$ is an $R$-module monomorphism. Therefore, by lemma [lemma-A1] below, $\bar{\tau} ,\bar{\sigma}, \Phi$ is an $R$-module monomorphism, so commutativity of ([eq:CD2]) implies that $\varphi \otimes 1_{\sum R} , \tau , \sigma$ is also an $R$-module monomorphism. Finally, since $\sigma$ and $\tau$ are isomorphisms, so is $\tau, \sigma$. Whence, $\varphi \otimes 1_{\sum R}$ is injective, by lemma [lemma-A2]. This which proves claim 2 and completes part (a) of the problem.

: If $C$ is a projective $R$-module and $\varphi: A \rightarrow B$ is an $R$-module monomorphism, then the natural map $\varphi \otimes 1_C: {\ensuremath{A\otimes_R}}C \rightarrow {\ensuremath{B\otimes_R}}C$ is injective.

: Since $C$ is projective, there exists a free $R$-module $F$ and a (projective) $R$-module $D$ such that $F = C\oplus D$ (lemma [lemma-A6]). Therefore, $${\ensuremath{A\otimes_R}}F = {\ensuremath{A\otimes_R}}(C\oplus D) \cong ({\ensuremath{A\otimes_R}}C)\oplus ({\ensuremath{A\otimes_R}}D) ; \text{ and } ; {\ensuremath{B\otimes_R}}F = {\ensuremath{B\otimes_R}}(C\oplus D) \cong ({\ensuremath{B\otimes_R}}C)\oplus ({\ensuremath{B\otimes_R}}D),$$ where the isomorphisms are given by lemma [lemma-A5].

By claim 2 above, the natural map $\varphi \otimes 1_F : {\ensuremath{A\otimes_R}}F \rightarrow {\ensuremath{B\otimes_R}}F$ is injective. Consider the diagram $$\label{eq:CD3} \begin{CD} {\ensuremath{A\otimes_R}}C @>{\iota_1}>> ({\ensuremath{A\otimes_R}}C)\oplus({\ensuremath{A\otimes_R}}D) @>{\tau}>> {\ensuremath{A\otimes_R}}(C\oplus D)\ @V{\varphi \otimes 1_C}VV @. @VV{\varphi \otimes 1_F}V \ {\ensuremath{B\otimes_R}}C @>{\iota_1}>> ({\ensuremath{B\otimes_R}}C)\oplus({\ensuremath{B\otimes_R}}D) @>{\bar{\tau}}>> {\ensuremath{B\otimes_R}}(C\oplus D)\ \end{CD}$$ By (the proof of) lemma [lemma-A5] the map $\tau: ({\ensuremath{A\otimes_R}}C)\oplus({\ensuremath{A\otimes_R}}D) \rightarrow {\ensuremath{A\otimes_R}}(C\oplus D)$ given by $\tau(a_1\otimes c, a_2\otimes d) = a_1\otimes (c,0) + a_2\otimes (0,d)$ is an $R$-module isomorphism. Given $a\otimes c \in {\ensuremath{A\otimes_R}}C$, then, $$(\varphi \otimes 1_F), \tau, \iota_1 (a\otimes c) = (\varphi \otimes 1_F), \tau (a\otimes c, 0) = (\varphi \otimes 1_F)(a\otimes (c, 0)) = \varphi(a)\otimes (c, 0).$$ Similarly, there exists $\bar{\tau}: ({\ensuremath{B\otimes_R}}C)\oplus({\ensuremath{B\otimes_R}}D) \cong {\ensuremath{B\otimes_R}}(C\oplus D)$, with $\bar{\tau}(b_1\otimes c, b_2\otimes d) = b_1\otimes (c,0) + b_2\otimes (0,d)$, and $$\bar{\tau}, \iota_1, (\varphi \otimes 1_C)(a\otimes c) = \bar{\tau}, \iota_1, (\varphi(a) \otimes c) = \bar{\tau}(\varphi(a) \otimes c, 0) = \varphi(a)\otimes (c, 0).$$ Therefore, $(\varphi \otimes 1_F), \tau, \iota_1$ and $\bar{\tau}, \iota_1, (\varphi \otimes 1_C)$ agree on generators of ${\ensuremath{A\otimes_R}}C$, so diagram ([eq:CD3]) is commutative.

By lemma [lemma-A1], $(\varphi \otimes 1_F), \tau, \iota_1$ is injective, so, by commutativity, $\bar{\tau}, \iota_1, (\varphi \otimes 1_C)$ is injective. It now follows from lemma [lemma-A3] that $\varphi \otimes 1_C$ is injective, which completes the proof of claim 3 and part (b) of the problem.

The following six lemmas are used in the answer to problem 3 given above. The first is a standard theorem about projective modules, the proof of which is not hard, and can be found, in Hungerford. The next two lemmas ([lemma-A4] and [lemma-A5]) are also standard, but I haven’t seen them proved in detail elsewhere and, as the solution given above makes repeated use of the maps defined in proving these lemmas, I include detailed proofs below. (I hope they are correct!) The last three lemmas are trivial verifications.

[lemma-A6] Let $R$ be a ring. The following conditions on an $R$-module $P$ are equivalent:

(i) $P$ is projective;

(ii) every short exact sequence of $R$-modules $0\rightarrow A \xrightarrow{f} B\xrightarrow{g} P \rightarrow 0$ is split exact (hence $B\cong A\oplus P$);

(iii) there is a free $R$-module $F$ and an $R$-module $N$ such that $F\cong N\oplus P$.

[lemma-A4] If $R$ is a commutative ring with $1_R$ and $A$ is a unitary $R$-module, then ${\ensuremath{A\otimes_R}}R \cong A$.

Define $\alpha: A\times R \rightarrow A$ by $\alpha(a,r) = ar$. Since $\alpha$ is clearly bilinear, there exists a unique $R$-module homomorphism $\bar{\alpha}: {\ensuremath{A\otimes_R}}R \rightarrow A$ such that $\bar{\alpha} \iota = \alpha$, where $\iota : A\times R \rightarrow {\ensuremath{A\otimes_R}}R$ is the canonical bilinear map. Define $\beta: A\rightarrow {\ensuremath{A\otimes_R}}R$ by $\beta(a) = a\otimes 1_R$. Then it is easy to verify that $\beta$ is an $R$-module homomorphism and that $$\beta \bar{\alpha}(a\otimes r) = \beta(ar) = ar{\ensuremath{\otimes 1_R}}= a\otimes r \quad (\forall a\in A) (\forall r\in R).$$ Therefore, $\beta\bar{\alpha}$ is the identity on generators of ${\ensuremath{A\otimes_R}}R$. Also, $\bar{\alpha} \beta(a) = \bar{\alpha}(a{\ensuremath{\otimes 1_R}}) = a1_R = a,$ so $\bar{\alpha} \beta = 1_A$. Therefore, $\bar{\alpha}: {\ensuremath{A\otimes_R}}R \cong A$.

[lemma-A5] Let $R$ be a commutative ring with $1_R$ and let $A, B, C$ be unitary $R$-modules. Then ${\ensuremath{A\otimes_R}}(B\oplus C) \cong ({\ensuremath{A\otimes_R}}B)\oplus ({\ensuremath{A\otimes_R}}C)$.

Recall that if $\varphi_1: A_1\rightarrow D$ and $\varphi_2: A_2\rightarrow D$ are (group) homomorphisms, then there is a unique homomorphism $\Phi :A_1 \oplus A_2 \rightarrow D$ such that $\Phi \iota_i = \varphi_i,(i=1, 2)$, where $\iota_i : A_i \rightarrow A_1 \oplus A_2$ are the canonical injections. In other words, $\exists ! , \Phi \in \Hom(A_1 \oplus A_2,D)$ such that the following diagram is commutative: $$\xymatrix{A_1 \ar[dr]{\varphi_1}\ar[r]^-{\iota_1} &A_1\oplus A_2 \ar[d]^{\Phi}& A_2 \ar[l]-{\iota_2} \ar[dl]^{\varphi_2}\ & D & }$$ We can apply this theorem to the Abelian group $({\ensuremath{A\otimes_R}}B)\oplus ({\ensuremath{A\otimes_R}}C)$. Let $\varphi_1: A\times B \rightarrow {\ensuremath{A\otimes_R}}(B\oplus C)$ be given by $\varphi_1(a,b)=a\otimes (b,0),$ and let $\varphi_2: A\times C \rightarrow {\ensuremath{A\otimes_R}}(B\oplus C)$ be given by $\varphi_2(a,c)=a\otimes (0,c)$. It is easily verified that $\varphi_i$ are bilinear and thus induce $R$-module homomorphisms $\bar{\varphi_1}: {\ensuremath{A\otimes_R}}B \rightarrow {\ensuremath{A\otimes_R}}(B\oplus C)$ and $\bar{\varphi_2}: {\ensuremath{A\otimes_R}}C \rightarrow {\ensuremath{A\otimes_R}}(B\oplus C)$ such that $\bar{\varphi_i} \kappa_i = \varphi_i,$ where $\kappa_1: A\times B \rightarrow {\ensuremath{A\otimes_R}}B$ and $\kappa_2: A\times C \rightarrow {\ensuremath{A\otimes_R}}C$ are the canonical bilinear maps. Therefore, the two universal properties combine to give a unique homomorphism $\Phi: ({\ensuremath{A\otimes_R}}B) \oplus ({\ensuremath{A\otimes_R}}C) \rightarrow {\ensuremath{A\otimes_R}}(B\oplus C)$ such that the following diagram is commutative: $$\xymatrix{{\ensuremath{A\otimes_R}}B \ar[dr]{\bar{\varphi_1}} \ar[r]^-{\iota_1} & ({\ensuremath{A\otimes_R}}B)\oplus ({\ensuremath{A\otimes_R}}C) \ar[d]^{\Phi}& {\ensuremath{A\otimes_R}}C \ar[dl]^{\bar{\varphi_2}} \ar[l]-{\iota_2} \ A\times B \ar[u]^{\kappa_1} \ar[r]-{\varphi_1} & {\ensuremath{A\otimes_R}}(B\oplus C) & A\times B \ar[u]{\kappa_2} \ar[l]^-{\varphi_2}}$$ Now, for all $a\in A$ and $b\in B$, $\iota_1(a\otimes b) = (a\otimes b, 0)$ and $\bar{\varphi_1}(a\otimes b) = \bar{\varphi_1}\kappa_1(a, b) = \varphi_1(a,b) = a\otimes(b,0)$, so, $\Phi(a\otimes b, 0) = a\otimes (b,0)$. Similarly, $\Phi(0, a\otimes c) = a\otimes (0,c),$ for all $a\in A$ and $c\in C$. Therefore, $$\Phi(a\otimes b, a'\otimes c) = \Phi(a\otimes b,0)+ \Phi(0, a'\otimes c) = a\otimes (b,0) + a'\otimes (0,c) \quad (\forall a, a' \in A, b\in B, c\in C).$$

Next, define $\psi: A\times (B\oplus C) \rightarrow ({\ensuremath{A\otimes_R}}B) \oplus ({\ensuremath{A\otimes_R}}C)$ by $\psi(a, (b,c)) = (a\otimes b, a\otimes c)$. Again, $\psi$ is bilinear, so there is a unique $\Psi: {\ensuremath{A\otimes_R}}(B\oplus C) \rightarrow (B\oplus C) \rightarrow ({\ensuremath{A\otimes_R}}B) \oplus ({\ensuremath{A\otimes_R}}C)$ such that $\Psi \kappa = \psi$ where $\kappa: A\times (B\oplus C) \rightarrow {\ensuremath{A\otimes_R}}(B\oplus C)$ is canonical. To complete the proof, we check that $\Psi \Phi$ and $\Phi \Psi$ given the appropriate identity maps: $$\Psi \Phi (a\otimes b, 0) = \Psi (a\otimes (b, 0)) = (a\otimes b, a\otimes 0) = (a\otimes b, 0), \text{ and }$$ $$\Psi \Phi (0,a\otimes c) = \Psi (a\otimes (0,c)) = (a\otimes 0, a\otimes c) = (0, a\otimes c).$$ Thus, $\Psi \Phi$ is the identity on generators of $({\ensuremath{A\otimes_R}}B) \oplus ({\ensuremath{A\otimes_R}}C)$, so $\Psi \Phi = 1_{({\ensuremath{A\otimes_R}}B) \oplus ({\ensuremath{A\otimes_R}}C)}$. Finally, $$\Phi \Psi (a\otimes (b, c)) = \Phi (a\otimes b, a\otimes c) = \Phi( (a\otimes b, 0) +(0, a\otimes c))= a\otimes (b, 0) +a\otimes (0,c)= a\otimes (b, c).$$ Thus $\Phi \Psi$ is the identity on generators of ${\ensuremath{A\otimes_R}}(B\oplus C),$ so $\Phi \Psi = 1_{{\ensuremath{A\otimes_R}}(B\oplus C)}$. This proves that $\Phi: ({\ensuremath{A\otimes_R}}B) \oplus ({\ensuremath{A\otimes_R}}C) \cong {\ensuremath{A\otimes_R}}(B\oplus C)$.

[lemma-A1] If $h\in \HomR(A,B),, g\in \HomR(B,C),$ and $f\in \HomR(C,D)$ are injective, then $f g h \in \HomR(A, D)$ is injective.

$f g h(a) = 0 \Rightarrow g h (a) = 0$ (since $f$ is injective) $\Rightarrow h (a) = 0$ (since $g$ is injective) $\Rightarrow a = 0$ (since $h$ is injective).

[lemma-A2] If $g\in \HomR(A,B),, f\in \HomR(B,C),$ if $g$ is surjective, and if $f g$ is injective, then $f$ is injective.

Suppose $f(b) = 0$. Since $g$ is surjective, there is an $a\in A$ such that $g(a) = b$. Then $fg(a) = f(b) = 0,$ which implies $a=0,$ since $f g$ is injective. Therefore, $b = g(a) = g(0) = 0$ since $g$ is a homomorphism.

[lemma-A3] If $g\in \HomR(A,B),, f\in \HomR(B,C),$ and if $f g$ is injective, then $g$ is injective.

$\ker g = {a\in A: g(a) = 0} \subseteq {a\in A: f g(a) = 0} = \ker fg = {0}$. The set containment holds since $f$ is a homomorphism, so $f(0)=0$.