Projective Morphism

Why is this section called "Projective Morphism"?

The naming of Projective Morphisms (as elements in the image of ${\tau }_{X,Y}$) is deeply connected to the Dual Basis Lemma. This lemma reveals that if $X$ is a finitely generated projective module, then ${\mathrm{Hom}}_R(X,Y)\simeq X^{\ast }{\otimes }_{R}Y$, meaning every such homomorphism arises from tensoring linear functionals on $X$ with elements in $Y$.

Proposition

Let $R$ be a commutative ring with $1$. For $R$-modules $X,Y$ and $M$, we have an $R$-morphism

$$\begin{array}{rl}{\mathrm{Hom}}_R(X,M){\otimes }_R{\mathrm{Hom}}_R(M,Y)& \to {\mathrm{Hom}}_R(X,Y),\\ \phi \otimes \psi & \mapsto \psi \phi .\end{array}$$

The image of this morphism is comprised of those morphism from $X$ to $Y$ which factor through some finite power of $M$.

\begin{proof} An element of the image is ${\sum }_{i\in I}{\psi }_i{\phi }_i$, where ${\psi }_i:X\to M$ and ${\phi }_i:M\to Y$. Give a morphism $\phi :X\to M^I,x\mapsto ({\phi }_i(x))$ and a morphism $\psi :M^I\to Y,(m_i)\to \sum {\psi }_i(m_i)$. Now we finish the proof. \end{proof}

projective morphisms

Recall that the dual of an $R$-module $X$ is defined as $X^{\ast }={\mathrm{Hom}}_R(X,R)$. Taking $M=R$, we obtain a natural $R$-linear map

$${\tau }_{X,Y}:X^{\ast }{\otimes }_{R}Y⟶{\mathrm{Hom}}_R(X,Y),\phi \otimes y\mapsto (x\mapsto \phi (x)\cdot y).$$

The image of ${\tau }_{X,Y}$ is denoted by ${\mathrm{Hom}}_R^{\text{pr }}(X,Y)$, and its elements are called projective morphisms from $X$ to $Y$.

Lemma

For any $R$-module $M$, ${\mathrm{Hom}}_R^{\mathrm{pr}}(M,M)$ is a $2$-sided ideal of the ring ${\mathrm{End}}_R(M)={\mathrm{Hom}}_R(M,M)$.

\begin{proof} It is easy to check for any $\phi \in {\mathrm{Hom}}_R^{\mathrm{pr}}(M,M)$ and $\psi \in {\mathrm{End}}_R(M)$, one have $\phi \circ \psi ,\psi \circ \phi \in {\mathrm{Hom}}_R^{\mathrm{pr}}(M,M)$.
\end{proof}

dual basis lemma

Let $P$ be an $R$-module. TFAE:

• $P$ is a finitely generated projective module.
• ${\mathrm{Hom}}_R^{\mathrm{pr}}(P,P)={\mathrm{End}}_R(P)$.
• For any $R$-module $X$, the morphism ${\tau }_{X,P}:X^{\ast }{\otimes }_{R}P\to {\mathrm{Hom}}_R(X,P)$ is an isomorphism.
• For any $R$-module $X$, the morphism ${\tau }_{P,X}:P^{\ast }\otimes X\to {\mathrm{Hom}}_R(P,X)$ is an isomorphism.
• The morphism ${\tau }_{P,P}:P^{\ast }{\otimes }_{R}P\to {\mathrm{End}}_R(P)$ is an isomorphism.

\begin{proof} i)→ii) Since $P$ is a finitely generated projective $R$-module, there exists $n$ such that $R^n\simeq P\oplus Q$, that is, $P$ is a direct summand of $R^n$. Define $\iota :P\hookrightarrow R^n$ and $f=\iota \circ \phi$. Notice that ${\tau }_{P,P}(f\otimes {\mathrm{id}}_P)(x)=f(x)\cdot {\mathrm{id}}_P=\phi (x)$, so $\phi ={\tau }_{P,P}(f\otimes {\mathrm{id}}_P)\in {\mathrm{Hom}}_R^{\mathrm{pr}}(P,P)$. Now we finish the proof.

ii)→iii) Assume ${\mathrm{id}}_P={\tau }_{P,P}({\sum }_{i\in I}{\phi }_i\otimes m_i)$, then $v={\phi }_i(v)m_i$ for any $v\in P$. Consider the morphism

$$\begin{array}{rl}\sigma :{\mathrm{Hom}}_R(X,P)& \to X^{\ast }\otimes P,\\ \alpha & \mapsto \sum _{i\in I}{\phi }_i\alpha \otimes m_i.\end{array}$$

Now we check that $\sigma$ is the inverse of ${\tau }_{X,P}$. Notice that

$$\begin{array}{rl}{\mathrm{Hom}}_R(X,P)& \stackrel{\sigma }{\to }{X}^{\ast }{\otimes }_{R}P\stackrel{{\tau }_{X,P}}{\to }{\mathrm{Hom}}_R(X,P),\\ \alpha & \mapsto \sum _i{\phi }_i\alpha \otimes m_i\mapsto (x\mapsto \sum _i{\phi }_i(\alpha (x))m_i),\end{array}$$

where ${\sum }_i{\phi }_i(\alpha (x))m_i=\alpha (x)$ and so ${\tau }_{X,P}\circ \sigma ={\mathrm{id}}_{{\mathrm{Hom}}_R(X,P)}$. On the other hand, notice that

$$\begin{array}{rl}{X}^{\ast }\otimes P& \stackrel{{\tau }_{X,P}}{\to }{\mathrm{Hom}}_R(X,P)\stackrel{\sigma }{\to }{X}^{\ast }{\otimes }_{R}P,\\ \psi \otimes v& \mapsto {\tau }_{X,P}(\psi \otimes v)\mapsto \sum _i{\phi }_i{\tau }_{X,P}(\psi \otimes v)\otimes m_i.\end{array}$$

It remains to verify $\psi \otimes v={\sum }_i{\phi }_i{\tau }_{X,P}(\psi \otimes v)\otimes m_i$. Observe that ${\phi }_i(v)\psi ={\phi }_i\cdot {\tau }_{X,P}(\psi \otimes v)$, then

$$\psi \otimes v=\psi \otimes (\sum _i{\phi }_i(v)m_i)=\sum _i\psi \otimes {\phi }_i(v)m_i=\sum _i{\phi }_i(v)\psi \otimes m_i=\sum _i{\phi }_i{\tau }_{X,P}(\psi \otimes v)\otimes m_i$$

and we have done.

• iii)→iv)

iv)→v) It is trivial, by taking $X=P$.

v)→i) Assume $\{{\phi }_1,\cdots ,{\phi }_n\}\subseteq P^{\ast }$ and $\{x_1,\cdots ,x_n\}\subseteq P$ such that

$${\tau }_{P,P}\left(\sum _{i=1}^n{\phi }_i\otimes x_i\right)={\mathrm{id}}_P⟹\sum _{i=1}^n{\phi }_i(x)x_i=x.$$

It deduces that $P={⟨x_1,\cdots ,x_n⟩}_R$. Define $R$-module homomorphisms $\varphi :R^n\to P,(r_1,\cdots ,r_n)\mapsto {\sum }_{i=1}^n{r}_i{x}_i$ and $\psi :P\to R^n,x\mapsto ({\phi }_1(x),\cdots ,{\phi }_n(x))$. Then $\varphi \circ \psi ={\mathrm{id}}_P$ and so $R^n=\ker \varphi \oplus \mathrm{im}\psi \simeq \ker \varphi \oplus P$, i.e., $P$ is a direct summand of $R^n$. Therefore, $P$ is a finitely generated projective module. \end{proof}