2008 Nov

2008 November

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

[it:1] \(A = A_1\times A_2,\) for some \(A_1\subseteq R\) and \(A_2\subseteq R',\) and
[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 have[^2]

\[\begin{aligned} \tau \,\sigma (\{a_i\} ) &= \tau ( \{a_i \otimes 1_R\}) = \tau ((a_1\otimes 1_R, 0, \dots)+(0, a_2\otimes 1_R, 0, \dots)+\cdots)\\ &= a_1\otimes (1_R, 0, \dots)+ a_2\otimes (0, 1_R, 0, \dots)+\cdots.\\[4pt] \therefore \; (\varphi\otimes 1_{\sum R})\,\tau \, \sigma (\{a_i\}) &= (\varphi\otimes 1_{\sum R})(a_1\otimes (1_R, 0, \dots)+ a_2\otimes (0, 1_R, 0, \dots)+\cdots)\\ &= (\varphi\otimes 1_{\sum R})(a_1\otimes (1_R, 0, \dots))+ (\varphi\otimes 1_{\sum R})(a_2\otimes (0, 1_R, 0, \dots)) +\cdots\\ &=\varphi(a_1)\otimes (1_R, 0, \dots)+ \varphi(a_2)\otimes (0, 1_R, 0, \dots)+\cdots. \end{aligned}\]

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 \(f(0)=0\).

[^1]: Although this may seem trivial, and to some extent it is trivial, there is something to verify here, since we could have \((a,a')\in A\), \((b,b')\in A\), so that \((a,b')\in A_1\times A_2\), and yet \((a,b')\notin A\). That is, in general, \(A\) need not be equal to the product of all first and second coordinates of elements of \(A\). (Consider, for example, the diagonal \(\{(0,0), (1,1), (2,2)\},\) which cannot be written as \(A_1 \times A_2\).)

[^2]: Again, the sums have finitely many non-zero terms, since \(\{a_i\}\) is non-zero for finitely many indices \(i\in I\). The same comment applies to the sums in the following two sets of equations.