チンカス
$A$ is a commutative ring with unit.
$I,J\in\textsf{Set}$.$(U,I):\textbf{Set}\rightarrow A\texttt{-}\textbf{Mod}$ be forgetfull functor and $(F,{\large\text{ᖷ}}):A\texttt{-}\textbf{Mod}\rightarrow\textbf{Set}$ be free functor with unit $u_\bullet$ s.t. corresponding basises are $u_I=(b_i)_{i\in I},\,u_J=(d_j)_{j\in J},\,u_{I\times J}=(q_{i,j})_{(i,j)\in I\times J}$. Then following holds:
- $F(I\times J)\cong_{A\texttt{-}\text{Mod}}F(I)\otimes_A F(J)$ and
- $(F(I)\otimes_A F(J),(b_i\otimes d_j)_{(i,j)\in I\times J})$ is free object of $I\times J$. In other words, $(b_i\otimes d_j)_{(i,j)\in I\times J}$ become a basis of $F(I)\otimes_A F(J)$.
Below, we are going to give you a categorical proof. One will say it's very straightforward elite proof yeah.
First, $(F,{\large\text{ᖷ}})\dashv(U,I)$, so $(F,{\large\text{ᖷ}})$ preserves arbitrally colimit, especially preserves coproduct. Assume $(\coprod_{i\in I}J,(\iota_i)_{i\in I})$ coproduct of $(J)_{i\in I}$ in $\mathbf{Set}$, then $\big(F(\coprod_{i\in I}J),({\large\text{ᖷ}}\,(\iota_i))_{i\in I}\big)$ is coproduct of $(F(J))_{i\in I}$ in $A\texttt{-}\textbf{Mod}$. Here, we are going to concretely write down coproduct in $\mathbf{Set}$ like
$$
\begin{eqnarray}
\left\{
\begin{array}{l}
\displaystyle\coprod_{i\in I} X_i\coloneqq\bigcup_{i\in I}\set{i}\times X_i \\
\iota_i(x)\coloneqq(i,x)
\end{array}
\right.
\end{eqnarray}
$$
In present case, $\coprod_{i\in I} J=I\times J$. Let $(\bigoplus_{i\in I}F(J),(\jmath_i)_{i\in I})$ is coproduct of $(F(J))_{i\in I}$, because of its universal property, following holds[1]:
$$ (\exists! h)\bigg[F(I\times J)\overset{h}{\cong}_{A\texttt{-}\text{Mod}}\bigoplus_{i\in I}F(J),\; (\jmath_i)_{i\in I}=\big(h\circ_{A\texttt{-}\text{Mod}}{\large\text{ᖷ}}(\iota_i)\big)_{i\in I} \bigg]$$
As well as this, using univesal property of copower $I\odot F(J)$[2],
$$ (\exists! v)\bigg[\bigoplus_{i\in I}F(J)\overset{v}{\cong}_{A\texttt{-}\text{Mod}} F(I)\otimes_A F(J),\; \big(I(-_1\otimes -_2)(b_i)\big)_{i\in I}=(v\circ_{A\texttt{-}\text{Mod}}\jmath_i)_{i\in I} \bigg]$$
Finally, because of construction of free functor, following diagram commutes:
\begin{xy}
\xymatrix { J \ar[d]_{\iota_i} \ar[r]^{d_\bullet\;\;\;\;\;\;\;} & U(F(J)) \ar[d]^{I({\large\text{ᖷ}}(\iota_i))} \\
I\times J \ar[r]^{q_\bullet\;\;\;\;\;\;\;} & U(F(I\times J)) }
\end{xy}
It means $$ \big(I({\large\text{ᖷ}}(\iota_i))\circ_{\text{Set}}d_\bullet\big)_{i\in I}=\big(q_\bullet\circ_{\text{Set}}\iota_i\big)_{i\in I}$$
By using above, we can conclude $F(I\times J)\overset{v\circ h}{\cong}_{A\texttt{-}\text{Mod}}F(I)\otimes_A F(J)$ and
$\forall i\in I$,
\begin{eqnarray}
I(-_1\otimes -_2)(b_i) &=& v\circ_{A\texttt{-}\text{Mod}}\jmath_i\\
&=& v\circ_{A\texttt{-}\text{Mod}} h\circ_{A\texttt{-}\text{Mod}} {\large\text{ᖷ}}(\iota_i)
\end{eqnarray}
especially $\forall i\in I,\,\forall j\in J$,
\begin{eqnarray}
b_i\otimes d_j &=& I(b_i\otimes -)(d_j)\\
&=& I\big(I(-_1\otimes -_2)(b_i)\big)(d_j)\\
&=& I(v\circ_{A\texttt{-}\text{Mod}} h\circ_{A\texttt{-}\text{Mod}} {\large\text{ᖷ}}(\iota_i))(d_j)\\
&=& \big(I(v\circ_{A\texttt{-}\text{Mod}}h)\circ_{\text{Set}} I({\large\text{ᖷ}}(\iota_i))\circ_\text{Set} d_\bullet\big)(j)\\
&=& \big(I(v\circ_{A\texttt{-}\text{Mod}}h)\circ_{\text{Set}} q_\bullet \circ_{\text{Set}}\iota_i \big)(j)\\
&=& \big(I(v\circ_{A\texttt{-}\text{Mod}}h)\circ_{\text{Set}} q_\bullet\big)(\iota_i(j))\\
&=& \big(I(v\circ_{A\texttt{-}\text{Mod}}h)\circ_{\text{Set}} q_\bullet\big)(i,j)\\
\end{eqnarray}
Now, $(F(I\times J),q_\bullet)$ satisfies universal property of free object of $I\times J$, so $(F(I)\otimes_A F(J),(b_i\otimes d_j)_{(i,j)\in I\times J})$ satisfies same one iff there exists $\phi$ such that $F(I\times J)\overset{\phi}{\cong}_{A\texttt{-}\text{Mod}}F(I)\otimes_A F(J)$ and $$ (b_i\otimes d_j)_{(i,j)\in I\times J}=I(\phi)\circ_{\text{Set}}(q_{i,j})_{(i,j)\in I\times J}$$ (
See this
, Theorem 4). It is obviously true by substituting $\phi=v\circ_{A\texttt{-}\text{Mod}} h$.
$$ \text{dim}(F(I)\otimes_A F(J))=|I\times J|$$
If $I$ and $J$ is finite, then
$$ \text{dim}(F(I)\otimes_A F(J))=\text{dim}(I)\,\text{dim}(J)$$
\begin{eqnarray} \text{Hom}_{A\texttt{-}\text{Mod}}\Big(\bigoplus_{i\in I}N,\cdot\Big)&\cong& \prod_{i\in I}\text{Hom}_{A\texttt{-}\text{Mod}}(N,\cdot)\\ &\cong& \text{Map}(I,\text{Hom}_{A\texttt{-}\text{Mod}}(N,\cdot)) \end{eqnarray} so it is precisely coproduct of $I$th $N$. Anyway, by universal property of tensor product, we obtain this:
\begin{eqnarray} \text{Hom}_{A\texttt{-}\text{Mod}}(F(I)\otimes_A N,\cdot)&\cong& \text{Hom}_{A\texttt{-}\text{Mod}}(F(I),[N,\cdot])\\ &\cong& \text{Map}(I,U([N,\cdot]))\\ &=&\text{Map}(I,\text{Hom}_{A\texttt{-}\text{Mod}}(N,\cdot)) \end{eqnarray} so $F(I)\otimes_A N$ is also $I\odot N$. According to this Theorem 4, we get this proposion.
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