# Finitely generated $R$-module in the product of quotient field.

Let $R$ be an integral domain with the quotient field $K$.

Let $M$ be a finitely generated $R$-submodule in $K^n$.

Is it true that $M$ is free $R$-module?

• Well, it seems to be that nice implies integral domain, which is a rather heavy assumption for a (general) ring: unitary, commutative without non-trivial zero divisors. – DonAntonio Oct 7 '12 at 13:27
• I just need to work in integral domain. Is this statement true? – Submodule Oct 7 '12 at 13:29
• Your question is true if there exists a unique homomorphism $\hat f:M\to N$ such that $\hat f \circ i= f:S \to N$, where $S$ is a set and $i$ is a set map $i: S \to M$. – user28511 Oct 7 '12 at 13:32

Non, it is almost never true. E.g. an finitely generated ideal $M$ in $R$ is free only if it is generated by one element because two elements $x_1, x_2\in M$ are always related by the linear relation $x_1.x_2+(-x_2).x_1=0$, thus any basis of $M$ can only have one element.

Example: $R=\mathbb C[X, Y]$, $M=XR+YR$. If $M$ was free, it would be generated by one element $P(X,Y)$. Then $P(X,Y)$ divides $X$ and $Y$,so $P(X,Y)$ is constant and generates the unit ideal $R$. But $M\ne R$.

• @navigetor23: you are right, $M$ don't have to be finitely generated. But I don't see why my argument would imply $M$ has only element ? It says two elements in $M$ are never linearly independent. – user18119 Oct 7 '12 at 18:56

Let $\,M=\langle\,m_1,...,m_s\,\rangle_R\leq K^n\,$ be a finitely generated $\,R-\,$submodule of $\,K^n\,$. Assuming the above is a minimal generating set for $\,M\,$, we thus clearly have $(1)\,\,\,s\leq n\,$ (why?) and also $(2)\,\,\,\{m_1,...,m_s\}\,$ are $\,K-\,$ linearly independent (why?). Thus

$$r_1m_1+...+r_sm_s=0\,\,,\,\,r_i\in R\Longrightarrow r_i=0\,\,\forall\,i$$

since otherwise we get a contradiction to (2) above.

• Why should one state (1) before (2)? – Marc van Leeuwen Oct 7 '12 at 13:56
• @DonAntonio, what's your conclusion? – user26857 Oct 7 '12 at 19:03