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Let $k$ be a commutative ring and let $M,N$ be two flat modules over $k$.

$\mathbf{EDIT}:$ A minimal generating set $X \subseteq M$ is a set which generates $M$ and no proper subset of $X$ generates $M$. There is no canonical notion of size for a minimal generating set, for example $\mathbb Z$ has generating sets $\{ 1 \}$ and $\{ 2, 3\}$ over $\mathbb Z$.

My question: is it true that

If $\{ m_i \mid i < \lambda\}$ is a minimal generating set for $M$ and $\{ n_j \mid j < \kappa \}$ is a minimal generating set for $N$ then $\{ m_i \otimes n_j \mid i < \lambda; j< \kappa\}$ is a minimal generating set for $M \otimes N$?

or equivalently,

If $\{ m_i \mid i < \lambda\}$ is a minimal generating set for $M$ and $\{ n_j \mid j < \kappa \}$ is a minimal generating set for $N$ then whenever a finite linear combination $ \displaystyle\sum_{(i,j)< \lambda \times \kappa}\alpha_{(i,j)} m_i \otimes n_j = 0$, each $\alpha_{(i,j)}$ is a non-unit?

I attempted to prove this for the case when $k$ is a field, thinking in terms of flatness : Let $V, W$ be vector spaces and let $\{ v_i \mid i< \lambda\}, \{ w_j \mid j < \kappa\}$ be bases for $V,W$ respectively. We can easily define maps $f \colon \oplus_{i < \lambda} k \to V$ and $g \colon \oplus_{j< \kappa} k \to W$ which send sequences $(\alpha_i )_{i< \lambda}$ to $\sum_{i < \lambda } \alpha_i v_i$ and $( \beta_j )_{j < \kappa }$ to $\sum_{j < \kappa }\beta_j w_j$. These maps are injective because of the linear independence property of the bases. By flatness the maps $$ \bigoplus_{i < \lambda}k \otimes \bigoplus_{j < \kappa}k \xrightarrow{f \otimes 1} V \otimes \bigoplus_{j < \kappa}k \quad \text{and} \quad V \otimes \bigoplus_{j < \kappa} k \xrightarrow {1 \otimes g} V \otimes W $$ are injective and so the composite $$ \bigoplus_{(i,j) < \lambda \times \kappa} k \cong \bigoplus_{i < \lambda}k \otimes \bigoplus_{j < \kappa}k \xrightarrow {f \otimes g} V \otimes W$$ is injective. This map being injective tells us that the proposed basis is actually linearly independent.

I know there are much easier ways to do that but as I said I wanted to think about the flatness of the modules more than their free-ness. However this proof does not generalise to modules because $f,g$ are not necessarily going to be injective, and the notion of linear independence doesn't really work for modules in general. Instead we have that non-unit condition as above. Is there a way to adapt this proof, or a totally different proof?

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However your remarks about vectorspaces extend easily over local rings. –  user26857 Oct 26 '12 at 12:33
    
@navigetor23 I tried to show this, but I couldn't get it to work... If $k$ is local then we have an ideal of non-units $I$ . If we define $f \colon k \to M$ by $f(\alpha) = \alpha m_i$ this map will only descend to a quotient map $\tilde f \colon k/ I \to M $ if $f(I) = 0$... but is it true that $\alpha$ is a non-unit implies that $\alpha m_i = 0$? The converse of that statement is true however by hypothesis. –  Paul Slevin Oct 26 '12 at 13:13

2 Answers 2

up vote 4 down vote accepted

No this is not true. Take $M=N=\mathbb Z$ over the ring $\mathbb Z$. Let $\{2, 3\}$ (resp. $\{2, 5\}$) be a minimal generating set of $M$ (resp. of $N$). Then $M\otimes N=\mathbb Z$ and the generating set obtained by tensoring the minimal ones is $\{4, 10, 6, 15\}$. As $10=4+6$, the subset $\{4, 10, 15\}$ generates $M\otimes N$ and $\{ 4, 10, 6, 15\}$ is not minimal.

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Perfect, thank you. –  Paul Slevin Oct 29 '12 at 12:42

This should be a comment...

Take $R$ to be a Dedekind domain and $I$ an ideal of $R$ which is not principal and which has order $2$ in the ideal class group of $R$, so that $I^2$ is a principal ideal. Then $I$ is projective, so flat, it cannot be generated by one element, and $I\otimes I\cong I^2$ is free of rank $1$. This shows that minimal in your question must be with respect to inclusion, and not minimal number of elemets.

A concrete example if $R=\mathbb Z[\sqrt{-5}]$, which is the ring of integers in $\mathbb Q(\sqrt{-5})$, so is a Dedekind domain, and $I=(3,2+\sqrt{-5})$, which is a non-principal ideal such that $I^2=(2+\sqrt{-5})$.

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Yeah minimal means that no smaller set generates $M$. I will edit my question. –  Paul Slevin Oct 26 '12 at 12:15

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