# Module generated by $n$ elements, containing $n+1$ independent elements

Let $A$ be a ring and $M$ a $A$-module such that $M$ is generated by some system of $n$ elements, but contains some other system of $n+1$ linearly independent elements. I want to show that $M$ must contain an infinite system of linearly independent elements. In fact I've been able to prove it under the additional assumption that there is no zero divisor of $A$. Indeed it is easy then to prove by induction that there are two ideals in A, $I$ and $J$, both nonzeros and such that their intersection is trivial ; and then we can construct the infinite system by another induction. But without this assumption I'm stuck. Could you help?

(Edit. I put the edited part in slanted font)

Here's my proof in case there is no zero divisor. First, $A$ contains two nonzero ideals $I$ and $J$ such that $I \cap J=\{0\}$. Indeed we shall argue by contradiction; let $n$ be minimal such that $A^n$ contains some independent family of $n+1$ elements (it is easy to lift the hypothesis to $A^n$). If $n=1$ then the claim is obvious (just pick $Ax$ and $Ay$ where $x,y$ are independent). Otherwise, let $(x_1,\cdots,x_{n+1})$ be some independent family of $A^n$. Let us project $A^n$ onto $A^{n-1}=A^{n-1} \times \{0\}$. It gives us a family $(y_1,\ldots,y_{n+1})$. Since $n$ is minimal, we know $(y_1,\ldots,y_n)$ is not independent. So we can find some nontrivial $\lambda_1,\ldots,\lambda_n$ such that $$\sum_{i=1}^n \lambda_i x_i = (0,\ldots,0,\lambda)$$ where $\lambda \in A$ (but $\lambda \neq 0$). We can assume for example that $\lambda_1 \neq 0$. Then by the same token there is some nontrivial sequence $\mu_2,\ldots,\mu_{n+1}$ and some $\mu \neq 0$ such that $$\sum_{i=2}^{n+1} \mu_i x_i = (0,\ldots,0,\mu)$$ We assumed that any two nonzeros ideals of $A$ intersect nontrivially. Consequently there must exist $\alpha,\beta$ such that $\alpha \lambda = \beta \mu \neq 0$. Then we can prove that $(x_1,\ldots,x_{n+1})$ is not independent, a contradiction.

So let $I,J$ be nonzero ideals of $A$ such that $I \cap J=\{0\}$. We can build the infinite independent family by induction. We just put $x_{n+1}'=\alpha x_{n+1}$ and $x_{n+2}'=\beta x_{n+1}$ ($\alpha \in I,\beta \in J$ and $\alpha,\beta \neq 0$). Then the family $(x_1,\ldots,x_n,x_{n+1}',x_{n+2}')$ is independent if the family $(x_1,\ldots,x_{n+1})$ is independent (because $\lambda \alpha + \mu \beta=0$ can happen only if $\lambda=\mu=0$).

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Is the system of $n$ generators linearly independent? In other words: is $M\cong A^n$? –  Quimey Feb 4 '13 at 16:29
No. But but we can lift everything to $A^n$. For if there is a morphism $A^n \to M$ that is onto, and if $(x_1,\ldots,x_{n+1})$ is some independent system of $M$, then let $y_i$ be some lifting of $x_i$ to $A^n$ (i.e. the morphism from $A^n$ to $M$ maps $y_i$ to $x_i$). It is clear that the $y_i$ are independent. Now if we have some independent system (maybe infinite) $(y_i)_i$ in $A^n$, we get an independent system of same cardinality in $A^{n+1}$ (because there is an injective morphism $A^n \to A^{n+1}$) and we can push it back in $M$ since there is an injective morphism $A^{n+1} \to M$. –  timofei Feb 5 '13 at 17:26
For commutative $A \neq 0$ there is no embedding $A^{n+1} \hookrightarrow A^n$. So I assume $A$ is not assumed to be commutative? –  Martin Brandenburg Apr 26 '13 at 18:13

To say that $M$ contains $k$ linearly independent elements $\{x_1, \cdots, x_k\}$ is to say that the homomorphism $R^k \hookrightarrow M$, defined by sending the $i$th basis element to $x_i$, is injective. As you mention above, if $M$ is generated by $n$ elements and contains $n+1$ linearly independent elements, it follows that we may "lift" to $R^n$ so that we have an injective homomorphisms $R^{n+1} \hookrightarrow R^n$.

I claim that if there is such an embedding $R^{n+1} \hookrightarrow R^n$, then $R^n$ contains an infinite linearly independent set. Once this is proved, it will follow that $M$ contains an infinite linearly independent set, because $R^n \hookrightarrow R^{n+1} \hookrightarrow M$.

So suppose we have an embedding $R^{n+1} \hookrightarrow R^n$. The image of this injection can be written as $X_1 \oplus Y_1$, where $X_1 \cong R$ and $Y_1 \cong R^n$. Now we may proceed inductively to find submodules $(X_{k+1} \oplus Y_{k+1}) \subseteq Y_k$ where $X_k \cong R$ and $Y_k \cong R^n$. We obtain a chain of inclusions $$R^n \supseteq X_1 \oplus Y_1 \supseteq X_1 \oplus (X_2 \oplus Y_2) \supseteq X_1 \oplus X_2 \oplus (X_3 \oplus Y_3) \cdots$$ As a result, there is a submodule that is an infinite direct sum $$X_1 \oplus X_2 \oplus X_3 \oplus \cdots \subseteq R^n$$ where all $X_i \cong R$. Clearly it follows that $R^n$ has an infinite linearly independent subset.

Notes: (1) A relevant search term might be "strong rank condition" for rings. (2) The proof above generalizes to say that if $A$ and $B$ are modules such that $A \oplus B \hookrightarrow B$, then there is an embedding $(A \oplus A \oplus \cdots) \hookrightarrow B$.

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This is crystal clear. Thank you very much. Your proof makes it obvious why the result holds. –  timofei Apr 30 '13 at 16:07

It depends on what you mean by "linearly independent"; if what you mean is "minimal set of generators for the module they generate", then the statement is incorrect, even under the assumption that $A$ is an integral domain.

For instance, let $k$ be a field, let $A = k[X,Y]$ (where $X$ and $Y$ are indeterminates), and let $M$ be $A$, as an $A$-module. Then the set $\{1\}$ of course generates $M$ as an $A$-module, but so does the set $\{X, 1-XY\}$. Moreover, both are minimal generating sets for $M$. That is, if you remove either $X$ or $1-XY$ from the latter set, you generate a proper ideal of $A$.

For what it's worth, the set $\{X, 1-XY\}$ is also linearly independent over $k$ if you think of $A$ as a $k$-vector space, in case that was what you meant..

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Why would «linearly independent» mean «minimal w.r.t. genereting»? That would be very odd! –  Mariano Suárez-Alvarez Feb 2 '13 at 3:00
By "linearly independent" I mean "there is no nontrivial relation of the form $\sum \lambda_i x_i=0$". –  timofei Feb 2 '13 at 17:58
Okay; there are after all several different ways to generalize the notion of a linearly independent set in a vector space to the context of modules over a ring. Mine is one; timofei's is another. I guess that before answering, I should have just asked timofei which one (s)he meant. –  neilme Feb 2 '13 at 18:48

Assume that there is a linearly independent set of cardinal $n+1$ in $A^n$. In this case we get a monomorphism $\iota:A^{n+1}\to A^n$. Now, the morphism $\iota\oplus Id:A^{n+2}\to A^{n+1}$ is also a monomorphism, so that $\iota \circ (\iota\oplus Id):A^{n+2}\to A^n$ gives us a linearly independent set of cardinal $n+2$ in $A^n$. The end of the proof follows from an easy induction and the comment of timofei on his question. As pointed out in the comments this only shows that there is a linearly independent set of size $m$ for all $m>n$.

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I think there is a problem with the "easy induction" part. I had the same idea as you but (unless I'm wrong) you don't get an infinite independent family because at each step you change the whole family. In fact we need to build nested independent families of increasing size. –  timofei Feb 5 '13 at 19:11
I see your point. I will try to fix the proof. Thanks! –  Quimey Feb 5 '13 at 19:16

I think I have a proof.

I showed supra that there must exist some independent family $(\alpha,\beta)$ in $A$. Then we check by induction on $n$ that the family $$(\alpha,\alpha \beta,\ldots,\alpha \beta^n)$$ (where $n \geq 1$) is independent. It is easy to finish the proof.

Maybe it is possible do adapt this reasoning and construct an independent family without resorting to the case $n=1$?

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