@@ -7,3 +7,32 @@ We define and construct tensor products for modules over a ring, and then for al
\subsection{Definition/Construction}
\input{./files/05/05a-construction}
\subsection{Functoriality of the tensor product}
If clear from the context, we will omit the ring $R$ in the notation $M \otimes_R N$ and instead write $M \otimes N$ and just write linear instead of $R$-linear.
@@ -85,6 +85,10 @@ Before we give the definition, we make a small detour to the category $\Set$\foo
\end{rem}
\textbf{Notation\footnote{This is neither standard nor used by the lecturer, but we will sometimes use the Lambda-like notation for function application for higher-order functions.
So for $f \in\Hom(X,\Hom(Y,Z))$, we write $f\ x$ for $f(x)\in\Hom(Y,Z)$ and $f\ x \ y$ for $(f(x))(y)\in Z$}}:
\begin{thm}[Whitney]
Let $M,N$ be $R$-modules.
Then there exists an $R$-module $M \otimes_R N$ ($M$ tensor $N$ over $R$) and a bilinear map
...
...
@@ -144,4 +148,67 @@ In other words, there is a natural equivalence
We construct $M \otimes N$ as a quotient of a ``huge'' free module with a submodule that imposes the bilinearity by applying the $\Hom$ properties of free modules aswell as those of quotient modules.
Let $\tilde{T}$ be the free $R$-module with basis induced by the elements of $M \times N$ and write $[(m,n)]$ for the basis vector corresponding to $(m,n)$.
Let $Z \subseteq\tilde{T}$ be the submodule generaated by the elements
The proof is elementary and follows from uniqueness.
In particular, if $u,v$ are isomorphisms, then $u \otimes v$ is an isomorphism with inverse $u^{-1}\otimes v^{-1}$.
We can thus say that the tensor product is a functor from the category $R_{\text{Mod}}\times R_{\text{Mod}}\to R_{\text{Mod}}$ that sends pairs of $R$-modules $(M,N)$ to $M \otimes N$ and a morphism $(u,v)$ to $u \otimes v$.
((m_i)_{i \in I},n) \mapsto\sum_{i}' m_i \otimes n
\end{align*}
which is well-defined because all but finitely many $m_i$ are zero.
By the universal propety, this defines a linear map
\begin{align*}
g: \left(
\bigoplus_{i \in I}M_i
\right)
\otimes N
\to\bigoplus_{i \in I}M_i \otimes N
\end{align*}
One easily checks that $f$ and $g$ are inverses of eachother:
\begin{align*}
(f \circ g)((m_i)_{i \in I}\otimes n)
&= f\left(
\sum_{i \in I}'m_i \otimes n
\right)
\\
&= \sum_{i\in I}' f_i(m_i \otimes n)\\
&= \sum_{i\in I}' \phi_i(m_i) \otimes n\\
&= \left(
\sum_{i \in I}' \phi_i(m_i) \otimes n
\right)\\
&= (m_i)_{i\in I}\otimes n
\end{align*}
and for the other way, it suffices to show this for generators of the module:
\begin{align*}
(g \circ f)(m_j \otimes n) = g(\phi_j(m_j) \otimes n) = m_j \otimes n, \quad\text{for}\quad j \in I
\end{align*}
\end{proof}
\begin{cor}[]
If $M$ is free with basis $(m_i)_{i \in I}$ and $N$ is free with basis $(n_j)_{j \in J}$, then $M \times N$ is free with basis $(m_i \times n_j)_{(i,j)\in I \times J}$.
In particular, if $K$ is a field and $V,W$ finite-dimensional $K$-vector spaces. Then so is $V \times_K W$ and
\begin{align*}
\dim(V \otimes_K W) = (\dim V) (\dim W)
\end{align*}
\end{cor}
\begin{proof}[]
Let $M_i = R e_i, N_j = R f_j$ so that
$M =\bigoplus_{i\in I}M_i$, $N =\bigoplus_{j \ni J}N_j$. By the proposition, we get an isomorphism
\begin{align*}
M \otimes N = \left(
\bigoplus_{i\in I}M_i
\right) \otimes\left(
\bigoplus_{j \in J}N_j
\right)
\iso\bigoplus_{(i,j) \in I \times J} M_i \otimes N_j
\end{align*}
From the second part of the proposition, we see that
\begin{align*}
M_i \otimes N_j = Re_i \otimes N_j \iso N_j
\end{align*}
so it is free of rank $1$.
Since it is generated by $e_i \otimes f_j$, this element is a basis, the above means that $M \otimes N$ is free with basis $(e_i \otimes f_j)_{(i,j)\in I \times J}$.
\end{proof}
\begin{xmp}[]
Let $K$ be a field and take $M = N = K^{2}$ with basis $(e_1,e_2)$.
Then $M \otimes_K N$ is a $K$-vector space of dimension $4$ with basis
v = ae_1 \otimes e_1 + b e_1 \otimes e_2 + c e_2 \otimes e_1 + d e_2 \otimes e_2
\end{align*}
is a pure tensor if there exist $x_1,x_2,y_1,y_2$ such that $a,b,c,d$ satisfy
\begin{align*}
a = x_1y_1, \quad b = x_1y_2, \quad c = x_2y_1, \quad d = x_2y_2
\end{align*}
Which requires that $ad - bc =0$.
\end{xmp}
Note for pure tensors, the representation $v = v_1\otimes v_2$ is not unique, as we could also write $v = xv_1\otimes x^{-1}v_2$ for a unit $x \in K^{\times}$.
