Theorem 11, Section 4.5 of Hungerford's Algebra

Question

Let $R$ be a ring with identity. If $A$ is a unitary right $R$-module and $F$ is a free left $R$-module with basis $Y$, thell every element $u$ of $A\otimes_R F$ may be written uniquely in the form $u =\sum_{i=1}^n a_i\otimes y_i$, where $a_i\in A$ and the $y_i$ are distinct elements of $Y$.

Proof: For each $y\in Y$, let $A_y$ be a copy of $A$ and consider the direct sum $\sum_{y\in Y} A_y$. We first construct an isomorphism $\theta : A\otimes_R F \cong \sum_{y\in Y}A_y$ as follows. Since $Y$ is a basis, $\{y\}$ is a linearly independent set for each $y\in Y$. Consequently, the $R$-module epimorphism $\varphi : R\to Ry$ given by $r\mapsto ry$ (Theorem 1.5) is actually an isomorphism. Therefore, by Theorem 5.7 there is for each $y\in Y$ an isomorphism $$A\otimes_R Ry \xrightarrow{1_A\otimes \varphi^{-1}} A\otimes_R R\cong A_y.$$ Thus by Theorems 5.9 and I.8.10 there is an isomorphism $\theta$: $$A\otimes_R F\cong A\otimes_R ( \sum_{y\in Y}Ry) \cong \sum_{y\in Y} A\otimes_R Ry\cong \sum_{y\in Y}A_y.$$ Verify that for every $a\in A$, $z\in Y$, $\theta (a\otimes z) = \{u_y\} \in \sum A_y$ , where $u_z = a$ and $u_y = 0$ for $y\neq z$; in other words, $\theta (a\otimes z) = \iota_z(a)$, with $\iota_z :A_z\to \sum A_y$ the canonical injection. Now every nonzero $v\in \sum A_y$ is a finite sum $v = \iota_{y_1} (a_1) +…+ \iota_{y_n} (a_n)=\theta (a\otimes y_1)+…+\theta (a\otimes y_n)$ with $y_1,…,y_n$ distinct elements of $Y$ and $a_i$ uniquely determined nonzero elements of $A$. It follows that every element of $A\otimes_R F$ (which is necessarily $\theta^{-1}(v)$ for some $v$) may be written uniquely as $\sum_{i=1}^n a_i\otimes y_i$.

Author proof of $\theta: A\otimes_R F\cong \sum_{y\in Y}A_y$ is messy, I mean, it make use of multiple non trivial results. Below I have written “standard” proof of $A\otimes_R F\cong \sum_{y\in Y}A_y$. Author has used standard approach in all of previous theorems which involved constructing isomorphism on tensor product.

Answer 1

We can simplify the proof by showing that $Y^{(A)}=\sum_{y\in Y}A_y$ satisfies the defining property of the tensor product.

We have the obvious $R$-balanced bilinear map $$ i:A\times F\to Y^{(A)}, \quad (a,f) \mapsto (ar_y) \quad \textrm{for } f=\sum_yr_yy. $$ Given any $R$-balanced bilinear map $\theta:A\times F\to M$, we have a linear map $$ \hat\theta:Y^{(A)}\to M, \quad (a_y)\mapsto\sum_y\theta(a_y,y), $$ satisfying $\hat\theta i=\theta$. Moreover, $\hat\theta$ is unique. Thus, by the uniqueness of the tensor product, there exists a unique isomorphism $\phi:A\otimes_RF\to Y^{(A)}$ such that $a\otimes f\mapsto(ar_y)$ for $f=\sum_yr_yy$.

Answer 2

Let $u=\sum_{i=1}^na_i\otimes f_i \in A\otimes_R F$. Since $Y$ is basis of $F$, we have $f_i=\sum_{j=1}^{n_i}r_{ij}y_{ij}$ where $r_{ij}\in R$, $y_{ij}\in Y$. So $$\begin{align} u &= \sum_{i=1}^na_i\otimes f_i \\ &= \sum_{i=1}^na_i\otimes \left( \sum_{j=1}^{n_i}r_{ij}y_{ij} \right) \\ &= \sum_{i=1}^n\sum_{j=1}^{n_i}a_i\otimes r_{ij} y_{ij} \\ &= \sum_{(i,j)} a_i\otimes r_{ij} y_{ij} \\ &= \sum_{(i,j)}a_ir_{ij}\otimes y_{ij} \\ &= \sum_{(i,j)}a_{ij} \otimes y_{ij}. \end{align}$$ Thus $A\otimes_R F$ is generated by $\{a\otimes y\mid a\in A \text{ and } y\in Y\}$.

Define $\psi :A\times F\to \sum_{y\in Y}A_y$ given by $\psi (a,f)=\psi (a,\sum_{i=1}^nr_iy_i) = \{ar_y\}_{y\in Y}$, where $r_{y_i}=r_i$ and $r_y=0$ for all $y\in Y-\{y_1,...,y_n\}$. Note $\psi$ is well defined function, since $Y$ is basis of $F$. We show $\psi$ is middle linear map. Let $a_1,a_2,a\in A$, $f,g\in F$ and $r\in R$. Then $$\psi (a_1+a_2,f)=\{(a_1+a_2)r_y\}=\{a_1r_y+a_2r_y\} =\{a_1r_y\}+\{a_2r_y\} = \psi(a_1,f)+\psi (a_2,f);$$ $$\psi(a,f+g)=\{a(r_y+s_y)\}=\{ar_y +as_y\}=\{ar_y\} + \{as_y\}=\psi (a,f) +\psi (a,g);$$ $$\psi (ar,f)=\{(ar)r_y\}=\{a(rr_y)\}=\psi(a,rf).$$ Thus $\psi$ is middle linear map. By universal property of tensor product, $\exists !$ group homomorphism $\alpha: A\otimes_RF\to \sum_{y\in Y}A_y$ such that $\alpha (a\otimes y)=\psi (a,y)=\iota_y(a)$.

Define $\beta: \sum_{y\in Y}A_y\to A\otimes_R F$ given by $\beta (\{a_y\}_{y\in Y})=\sum_{y\in Y}a_y\otimes y$. let $\{a_y\},\{b_y\}\in \sum A_y$. Then $$\beta (\{a_y\}+\{b_y\})=\beta (\{a_y + b_y\})=\sum_{y\in Y}(a_y+b_y)\otimes y=\sum_{y\in Y} a_y\otimes y +\sum_{y\in Y} b_y\otimes y=\beta(\{a_y\})+\beta (\{b_y\}).$$ Thus $\beta$ is well defined group homomorphism.

We calim $\beta \alpha$ and $\alpha \beta$ are respective identities. let $a\otimes y\in A\otimes_R F$. Then $\beta \alpha (a\otimes y)=\beta (\iota_y (a))=a\otimes y$. Thus $\beta \alpha =1_{A\otimes_R F}$. Let $\{a_y\}$. Then $\alpha \beta (\{a_y\})=\alpha (\sum_y a_y\otimes y)=\sum_y \alpha (a_y\otimes y)=\sum_y \iota_y (a_y)=\{a_y\}$. Thus $\alpha \beta =1_{\sum A_y}$. Hence $A\otimes_R F\cong \sum_{y\in Y}A_y$.