Let $R$ be a (commutative, otherwise the answer is easy, see the comment below) ring and let $M$ be a finitely generated $R$-module. Is it possible that $M$ admits an infinite linearly independent set? Cardinality of maximal linearly independent subset seems relevant, but it does not give an answer to my question. Thank you!
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4In the non-commutative (associative unital) free $K$-algebra $A_I$ on the variables $x_i$, $i\in I$, the $x_i$ freely generate a left-ideal (=left submodule) of $A_I$. So if $I$ is infinite this is an infinite linearly independent subset. – YCor Sep 09 '21 at 10:51
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Ah, nice! Thank you. – Ricky Sep 09 '21 at 11:18
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In a commutative ring $R$ this does not exist. Better for any $n\ge 0$, if $M$ is an $R$-module generated by $n$ elements, then $R^{n+1}$ doesn't embed into $M$.
Indeed, lifting if necessary, we can suppose $M=R^n$. So we get an $n\times (n+1)$ matrix $u$ over $R$ defining an injective operator $R^{n+1}\to R^n$. Let $B$ be the unital subring generated by entries of $u$. Then $B$ is finitely generated commutative, hence noetherian, and $u$ defines an injective operator $B^{n+1}\to B^n$, contradiction.
YCor
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Can't you just compose with the inclusion of $R^n$ into $R^{n+1}$ as the first n components to deduce you have an injective operator with determinant zero but the determinant of an injective linear map is a nonzero divisor? This way you don't need Hilbert's basis theorem – Benjamin Steinberg Sep 09 '21 at 12:30
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3The fact that for commutative rings there is no injective morphism $R^{n+1} \to R^n$ has been discussed several times in mathoverflow, see for example https://mathoverflow.net/questions/136/atiyah-macdonald-exercise-2-11. – Ricky Sep 09 '21 at 12:56