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 watson December 25th, 2011 12:23 PM

Free modules .. does this solution work?

Hi all, could someone take a look at my claim and solution below and see if it makes sense? Thanks!

Claim: Let R be a commutative ring with $1\not=0$, and let $f:R^m\to R^n$ be a surjective homomorphism of free $R$-modules. Then $m\geq n$.

Proof:
Take $\beta_1,...,\beta_n$ to be any $n$ linearly independent elements of $R^n$. Then there exists $\alpha_1, \alpha_2,...,\alpha_n$ in $R^m$ such that $f(\alpha_1)=\beta_1,...,f(\alpha_n)=\beta_n$. But since $\beta_1,...,\beta_m$ are linearly independent, we have that if $c_1 \beta_1+\cdots+c_n\beta_n=0$, then $c_1=\cdots=c_n=0$. Also, there exists some $\delta\in R^m$ such that $f(\delta)=c_1\beta_1+\cdots +c_n\beta_n=0$. So, as $f$ is a ring homomorphism, we have the following:
\begin{align*}
f(c_1\alpha_1+\cdots+c_n\alpha_n)&=c_1f(\alpha_1)+ \cdots+c_nf(\alpha_n)\\
&=f(\delta)\\
&=c_1\beta_1+\cdots+c_n\beta_n\\
&=0=f(0),
\end{align*}

but in order for this to be true, again $c_i=0$ for $i\leq n$, since $\beta_i$ are linearly independent. But this means that $\alpha_1,...,\alpha_n$ must be linearly independent in $R^m$, and so $m\geq n$.

 Peter December 30th, 2011 06:21 AM

Re: Free modules .. does this solution work?

Your approach works, provided, that your last implication is known: "If there are n linear independent elements in $R^m$ then $n\leq m$." But unless this is a theorem in a book or your course, the last assertion seems similarly hard to prove as the original problem.

An alternative approach would be: Recall that every commutative ring with $0\neq 1$ has at least one maximal ideal, say M. Then check $R^i/M(R^i)=(R/M)^i$ for all positive integers i. Then you get a surjective map $(R/M)^m\rightarrow (R/M)^n$ and these are vector spaces of dimiesion m and n respectively over the field $R/M$. Thus the problem is reduced to the case where R is a field, i.e., the case well-known from linear algebra.

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