Surjectivity of a Free Module Homomorphism implies injectivity

[caffeinemachine]

I am trying to prove the following:Let $R$ be a commutative ring and $M$ be a free $R$-module having a finite basis of $n$ elements.
Let $T:M\to M$ be a surjective $R$-module homomorphism.
Then $T$ is injective.Let $I$ be a maximal ideal of $R$.
Note that $I$ annihilates $\widehat M:=M/IM$ and thus $\widehat M$ is an $\bar (R/I)$-module.
Write $\bar R=R/I$.For each $r\in R$, write $\bar r$ to denote $r+I$ and for each $m\in M$ write $\widehat m$ to denote $m+IM$.Define $\widetilde T:\widehat M\to \widehat M$ as
$$\widetilde T(\widehat m)=\widehat{Tm},\quad \forall \widehat m\in \widehat M$$
It is easy to see that $\widetilde T$ is well defined.
Note that $\widetilde T$ is a surjective linear operator on a vector space and hence is also injective.Note that if $\mathcal B=\{e_1,\ldots,e_n\}$ is a basis of $M$, then $\widehat B=\{\widehat e_1,\ldots,\widehat e_n\}$ is a basis for $\widehat M$.Now suppose $T(r_1e_1+\cdots+r_ne_n)=0$. Then we get $\widetilde T(\bar r_1\widehat e_1+\cdots+\bar r_n\widehat e_n)=\widehat 0$.
Since $\widetilde T$ is injective, we get $\bar r_i=\bar 0$ for all $i$.
But this doesn't lead to $r_i=0$.Can anybody see how to complete the proof from here?

 

[Euge]

Hi caffeinemachine,

Let's start from the beginning, with your basis $\mathcal{B} = \{e_1,e_2,\ldots, e_n\}$. There are unique $R$ - scalars $a_{ij}$ ($i,j = 1,2,\ldots, n$) such that

\(\displaystyle T(e_i) = \sum_{j = 1}^n a_{ij}e_j.\)

Since $T$ is surjective, there exist $R$-scalars $b_{ki}$ ($i, k = 1, 2,\ldots, n$) such that

\(\displaystyle e_k = T\Bigl(\sum_{i = 1}^n b_{ki} e_i\Bigr).\)

Hence, for all $k$,

\(\displaystyle e_k = \sum_{i = 1}^n b_{ki} T(e_i) = \sum_{i = 1}^n \sum_{j = 1}^n b_{ki}a_{ij} e_j = \sum_{j = 1}^n c_{kj} e_j,\)

where the constants $c_{kj}$ are defined by the equation

\(\displaystyle c_{kj} = \sum_{i = 1}^n b_{ki}a_{ij}.\)

Since $\mathcal{B}$ is an $R$-basis for $M$, it follows that $c_{kj} = \delta_{kj}$, i.e.,

\(\displaystyle (*)\quad \sum_{i = 1}^n b_{ki}a_{ij} = \delta_{kj} \quad (k, j = 1, 2,\ldots, n)\)

Use the orthogonality relations $(*)$ to show that $T$ is injective.

 

[caffeinemachine]

Hi caffeinemachine,

Let's start from the beginning, with your basis $\mathcal{B} = \{e_1,e_2,\ldots, e_n\}$. There are unique $R$ - scalars $a_{ij}$ ($i,j = 1,2,\ldots, n$) such that

\(\displaystyle T(e_i) = \sum_{j = 1}^n a_{ij}e_j.\)

Since $T$ is surjective, there exist $R$-scalars $b_{ki}$ ($i, k = 1, 2,\ldots, n$) such that

\(\displaystyle e_k = T\Bigl(\sum_{i = 1}^n b_{ki} e_i\Bigr).\)

Hence, for all $k$,

\(\displaystyle e_k = \sum_{i = 1}^n b_{ki} T(e_i) = \sum_{i = 1}^n \sum_{j = 1}^n b_{ki}a_{ij} e_j = \sum_{j = 1}^n c_{kj} e_j,\)

where the constants $c_{kj}$ are defined by the equation

\(\displaystyle c_{kj} = \sum_{i = 1}^n b_{ki}a_{ij}.\)

Since $\mathcal{B}$ is an $R$-basis for $M$, it follows that $c_{kj} = \delta_{kj}$, i.e.,

\(\displaystyle (*)\quad \sum_{i = 1}^n b_{ki}a_{ij} = \delta_{kj} \quad (k, j = 1, 2,\ldots, n)\)

Use the orthogonality relations $(*)$ to show that $T$ is injective.

Hey Euge!

What you have done is absolutely correct and I thank you for helping out.

But the thing is, the equation $(*)$ is where I had started.

I am a beginner in module theory and yesterday I learned the fact that if $A$ and $B$ are square matrices with entries from a commutative ring $R$, then $AB=I$ if and only if $BA=I$.

If I use this fact then the equation $(*)$ does my job and solves the question I posted.

But the proof of the above which I know uses determinants quite extensively.
I do not want to use determinants.

This may seem unreasonable. But I think determinant-free proofs are much more elegant and enlightening.

So I was trying to prove that if you have a commutative ring $R$, and $R$-linear maps $T,S:R^n\to R^n$ such that $TS=\text{id}$, then $ST=\text{id}$.

From here it was a natural question that if $T:R^n\to R^n$ is surjective then it is also injective.

A determinant-free proof of this in the case where $R$ is a field is well-known. It uses rank-nullity theorem.

I doubt if there is an analog of the rank nullity theorem in the context of free modules of finite rank over a commutative ring without imposing some extra condition on the ring.

 

[Euge]

You do not need determinants to prove either injectivity of $T$ using $(*)$, or the statement $AB = I$ if and only if $BA = I$.

There are different, non-determinantal methods to prove injectivity of $T$. One way is to define a map $S : M \to M$ by the equation

\(\displaystyle S\Bigl(\sum_{i = 1}^n x_i e_i\Bigr) = \sum_{i = 1}^n \sum_{k = 1}^n x_k b_{ki}e_i\)

and show that $ST(e_i) = e_i$ for all $i$.

To show that $AB = I$ if and only if $BA = I$, first suppose $AB = I$. We want to show that $A$ is invertible. Suppose $A$ is not invertible. Then there is a sequence $\{E_i\}_{i = 1}^r$ of elementary matrices such that $E_1 E_2\cdots E_r A$ has a row of zeros. Then $E_1 E_2\cdots E_r AB$ has a row of zeros. Since $AB = I$, we deduce that the invertible matrix $E_1 E_2 \cdots E_r$ has a row of zeros, a contradiction. Therefore $BA = I$. A symmetric argument proves the converse.

 

[blue_raver22]

The proof can be completed by using the fact that $T$ is a surjective homomorphism. This means that for any element $m\in M$, there exists an element $m'\in M$ such that $T(m')=m$. In other words, every element in the image of $T$ has a preimage in $M$.

Since $T(r_1e_1+\cdots+r_ne_n)=0$, we can choose $m'=r_1e_1+\cdots+r_ne_n$ as the preimage of $0$ under $T$. This means that $T(m')=0$, which implies that $\widetilde T(\widehat{m'})=\widehat{T(m')}=\widehat{0}$. But since $\widetilde T$ is injective, this implies that $\widehat{m'}=\widehat{0}$, which in turn implies that $m'+IM=0$. This means that $m'\in IM$, which can only happen if all the coefficients $r_i$ are in the maximal ideal $I$. But since $I$ is a proper ideal, this means that $r_i=0$ for all $i$, and hence $T$ is injective.

In summary, since every element in the image of $T$ has a unique preimage, and the preimage of $0$ is uniquely determined by the coefficients $r_i$, we can conclude that $T$ is injective. Therefore, the surjectivity of $T$ implies its injectivity, and the proof is complete.