Rank of a submodule less or equal to the rank of the module?

Question

Let $R$ be an integral domain and $M$ a module over $R$. Let $N$ be a submodule of $M$. I want to confirm my proof is correct and also am wondering why the argument doesn't work in general for any ring?

The definition of rank that my book uses is that the rank of a module $M$ is the supremum of the cardinalities of all linearly independent subsets of $M$.

My proof: Let $N$ be a submodule of $M$. Let $S= \{n_i: n_i \in N, i \in I\}$ (where $I$ is some index set) be a maximal linearly independent subset of $N$. Then $S$ is also a subset of $M$ and still a linearly independent set. So the supremum of the collection of cardinalities of linearly independent subsets of $M$ (which contains the cardinality of the subset $S$) must be bigger or equal to the cardinality of $S$, hence the rank of $M$ is bigger or equal to the rank of $N$.

Could someone check my proof and also explain how exactly I am using the fact that $R$ is an integral domain?

Answer 1

Since I cannot comment above... I believe it has to do with the fact that in the ring R, being an integral domain it has no zero divisors. R is an R-module. If R was not an ID, then the notion of a linearly independent set may run into trouble with zero-divisors causing a linear combination of supposedly independent elements summing to zero even when the ring elements acting on the module elements are all non-zero.

Answer 2

Your definition of rank doesn't immediately ensure that if $\alpha$ is the rank of $M$, then there exists a linearly independent set $S$ with $|S|=\alpha$.

Anyway, if $\Gamma$ and $\Delta$ are sets of cardinals and $\Gamma\subseteq\Delta$, then $\sup\Gamma\le\sup\Delta$, because any upper bound of $\Delta$ is obviously also an upper bound for $\Gamma$.

Thus your idea of proof is good, but it needs a slight modification: if $\Gamma(M)$ is the set of cardinals of the linearly independent subsets of $M$, then $\Gamma(N)\subseteq\Gamma(M)$, because a linearly independent subset of $N$ is also linearly independent as a subset of $M$.

So, no, the hypothesis that $R$ is an integral domain is not used. Of course, commutativity of $R$ is used.

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In the case $R$ is an integral domain, we can consider its field of fractions $Q$. For an $R$-module $M$, $M\otimes_RQ$ is a $Q$-vector space, say its dimension is $\gamma$.

Note that every element in $M\otimes_RQ$ can be written as $x\otimes q$, with $x\in M$ and $q\in Q$.

Let $(x_\alpha\otimes q_\alpha)_{\alpha\in I}$ be linearly independent in $M\otimes_RQ$. Then $(x_\alpha)_{\alpha\in I}$ is linearly independent. Indeed, suppose $$ \sum_{\alpha\in I} x_\alpha r_\alpha=0 $$ Then also $$ 0=\Bigl(\sum_{\alpha\in I} x_\alpha r_\alpha\Bigr)\otimes1= \sum_{\alpha\in I}(x_\alpha\otimes q_\alpha)\frac{r_\alpha}{q_\alpha} $$ and therefore $r_\alpha/q_\alpha=0$, implying $r_\alpha=0$, for $\alpha\in I$.

Conversely, if $(x_\alpha)_{\alpha\in I}$ is linearly independent in $M$, then $(x_\alpha\otimes 1)_{\alpha\in I}$ is linearly independent in $M\otimes_RQ$. Indeed, if $$ \sum_{\alpha\in I}(x_\alpha\otimes 1)t_\alpha=0 $$ and $F=\{\alpha\in I:t_\alpha\ne0\}$, then it's not restrictive to assume $t_\alpha\in R$, by using a common denominator and multiplying both sides by it. Therefore $$ \sum_{\alpha\in F}(x_\alpha t_\alpha)\otimes 1=0 $$ Since the set $\{x_\alpha:\alpha\in F\}$ is linearly independent, it generates a free module $N$ of $M$. The exact sequence $0\to N\to M\to M/N\to0$ transforms in the exact sequence $0\to N\otimes_RQ\to M\otimes_RQ\to (M/N)\otimes_RQ\to0$, because $Q$ is flat as $R$-module. Hence we get $$ \sum_{\alpha\in F}x_\alpha t_\alpha=0 $$ and the conclusion.

So, for modules over an integral domain, the rank is the same as the cardinality of a maximal linearly independent subset or, as well, the dimension of $M\otimes_RQ$ as a $Q$-vector space.