Let $R$ be a commutative unital ring, $I$ a set, and $R^{(I)}$ the free module on $I$.

  1. Can there be a submodule $R^{(J)}\cong M\leq R^{(I)}$ with $|J|\!>\!|I|$?

  2. Can $R^{(I)}$ be generated (as a $R$-module) by a subset $J$ with $|J|\!<\!|I|$?

Slightly related: does there exist an embedding of $R$-algebras $R[x_1,x_2,\ldots]\longrightarrow R[x,y]$?

I know that there exist an embedding of free groups $\langle x_1,x_2,\ldots|\emptyset\rangle \longrightarrow \langle x,y\|\emptyset\rangle$ and an embedding of free algebras $R\langle x_1,x_2,\ldots\rangle\longrightarrow R\langle x,y\rangle$, namely $x_n\longmapsto x^ny$.

  • $\begingroup$ The answers for (1) and (2) are no in the finitely-generated case, by Nakayama's lemma. $\endgroup$ – Zhen Lin Mar 12 '13 at 8:15
  • $\begingroup$ @Zhen Lin: Can you explain this in more detail? $\endgroup$ – Martin Brandenburg Mar 12 '13 at 9:31
  • $\begingroup$ I guess only (2) is really proved using Nakayama's lemma. In my mind, the Cayley-Hamilton theorem over a general ring and Nakayama's lemma are almost interchangeable. $\endgroup$ – Zhen Lin Mar 12 '13 at 11:02
  • $\begingroup$ And why Nakayama when you can go directly to fields? $\endgroup$ – Martin Brandenburg Mar 12 '13 at 15:07
  • 2
    $\begingroup$ The question about $R$-algebras is now a separate question. $\endgroup$ – user26857 Mar 16 '13 at 17:31

For $R=0$ the answer is yes in each case. So assume $R \neq 0$ in the following.

1) No, see MO/136. It has also appeared several times on this site (math.SE/106786, math.SE/132729, math.SE/310166).

2) No, tensor with $R/\mathfrak{m}$ for some maximal ideal $\mathfrak{m}$ and then use Linear algebra.

3) No if $R$ is an integral domain. Because this would yield an embedding $K(x_1,x_2,\dotsc) \to K(x,y)$ of extension fields of $K$, the field of fractions of $R$, which contradicts the transcendence degree. I am not sure what happens with general $R$. I hope that someone else can explain the general case.

Here is a small observation: If there is an embedding of $R$-algebras $R[x_1,x_2,\dotsc] \to R[x,y]$, then there is a countable subring $S \subseteq R$ and an embedding $S$-algebras $S[x_1,x_2,\dotsc] \to S[x,y]$. Namely, let $S$ be the subring generated by the coefficients of the image of $x_i$, where $i \in \mathbb{N}$. Thus, we may restrict our attention to countable rings.

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  • $\begingroup$ 1) Do your answers also cover the case where $I$ and $J$ are arbitrarily large sets? 2) Ok, if $J$ is the generating set of the $R$-module $R^{(I)}$, then $J\otimes_RR/\mathfrak{m}$ is the generating set of the $R/\mathfrak{m}$-module $R^{(I)}\otimes_RR/\mathfrak{m} \cong (R\otimes_RR/\mathfrak{m})^{(I)} \cong (R/\mathfrak{m})^{(I)}=:(\ast)$, hence $J\otimes_R\{1\}$ is a generating set of $(\ast)$, but $(\ast)$ has dimension $|I|$, hence $|I|\leq|J\otimes_R\{1\}|=|J|$. Correct? 3) Could you please elaborate, I've forgotten most of what I knew from field theory. $\endgroup$ – Leo Mar 12 '13 at 20:00
  • $\begingroup$ 1) Yes. 2) Yes, $J \otimes 1$ not $J \otimes R/\mathfrak{m}$. 3) Please consult any text on transcendence bases. If, after that, there are still problems, I can help. $\endgroup$ – Martin Brandenburg Mar 12 '13 at 20:11
  • $\begingroup$ Ok, will do, thank you. After I get through, I may have additional questions, so I'll accept your answer in the following weeks. Just one more question: the rank of an $R$-module $M$ is defined as $rk\,M:=\sup\{|B|;\, B\!\subseteq\!M\text{ is }R\text{-linearly independent}\}$. If $R$ is a domain, then $rk\,M=\dim_{Q(R)}M\otimes_RQ(R)$, where $Q(R)$ is the field of fractions. When $R$ is not a domain, is $$rk\,M=\sup_{\mathfrak{p}\in Spec R}\dim_{Q(R/\mathfrak{p})}M \otimes_RQ(R/\mathfrak{p})?$$ What about $\sup_{\mathfrak{m}\in Max R}\dim_{R/\mathfrak{m}}M \otimes_RR/\mathfrak{m}$? $\endgroup$ – Leo Mar 12 '13 at 21:10
  • $\begingroup$ No. But this question can be asked separatedly. In that case, spend some time before and google "rank of a module" etc. $\endgroup$ – Martin Brandenburg Mar 13 '13 at 0:28
  • $\begingroup$ I did google it before, but I only found odd (seemingly inequivalent) definitions (via skew fields) or unrelated ones (rank of free modules), etc. I asked as a comment because I just wondered if it's true. I don't really need the result at this point, hence I don't want to waste too much of people's time solving problems I just wondered about. I asked the question hoping that you've seen this somewhere before; it wasn't meant as problem solving. Also, I don't know where in the literature this rank can be found. Anyway, thank you! $\endgroup$ – Leo Mar 13 '13 at 1:02

Since all links given for problem $1.$ by Martin Brandenburg deal with the finite case, let me settle the infinite case.

If $R^{(J)}\cong M\leq R^{(I)}$, then $|J|\le|I|$.

If $I$ and $J$ are finite, this is well known. If $J$ is finite and $I$ is infinite there is nothing to prove. If $I$ is finite and $J$ is infinite we reach a contradiction as we can see below.

Therefore we can assume that $I$ and $J$ are infinite. Let $(e_j)_{j\in J}$ be a basis for $R^{(J)}$. This is a linearly independent system into the free module $R^{(I)}$ and denote by $(f_i)_{i\in I}$ the canonical basis of $R^{(I)}$. Then $e_j=\sum_{i\in I}a_{ij}f_i$ with $a_{ij}\in R$. Set $I_j=\{i\in I:a_{ij}\neq 0\}$. This is a finite subset of $I$ and it's uniquely determined by $e_j$. Now define a map $h:J\to\mathcal P_f(I)$ by $h(j)=I_j$, where $\mathcal P_f(I)$ denotes the set of finite subsets of $I$. This map has no reason to be injective, but let's see how many elements of $J$ can be sent to the same finite subset of $I$: suppose that $h(i_1)=\cdots=h(i_t)=I'$, where $I'\subset I$ is a finite set. This means that $e_{i_1},\dots,e_{i_t}$ belong to the free submodule of $R^{(I)}$ generated by $(f_i)_{i\in I'}$. Since $e_{i_1},\dots,e_{i_t}$ are linearly independent, from the finite case we get $t\le |I'|$. On the other side, $$J=\bigcup_{I'\subset I,\ I'\text{ finite}}h^{-1}(I')$$ with $|h^{-1}(I')|\le |I'|$. (In particular this shows that we can't have $I$ finite and $J$ infinite.) Now it's easy to deduce that $|J|\le|I|$.

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