Submodules of a free module over a commutative ring

Question

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 $x_n\longmapsto x^ny$.

Answer 1

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.

Answer 2

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 $|J|\le|I|$.