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Let $R$ be a ring and $M$, $N$ finitely generated free modules modules over $R$. Let $A$ be a matrix representing a homomorphism $f: M \rightarrow N$. We know that the map $f$ is injective if and only if the columns of $A$ are linearly independent.

Now suppose $f$ is NOT injective. By the first isomorphism theorem, we know that the induced map $\bar{f}: M/ \ker f \rightarrow N$ is injective. But the matrix representing $f$ and $\bar{f}$ is the same, right? Then how does the above make sense?

user26857
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Artus
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  • Let me try again. A linear combination of things in $M/\text{ker} f$ can be zero when the same linear combination in $M$ is not zero. The linear combination in $M$ has to be zero proper while the same combination in $M/\text{ker} f$ just has to be in ker $f$. Meanwhile the actual matrix of $\bar f$ written in terms of a basis for $M/\text{ker} f$ cannot be the same as the matrix of $f$ because the domain has lower dimension, so it must have fewer columns. – Gregory Grant Jul 27 '16 at 02:14
  • @GregoryGrant We cannot be sure that $M/\text{ker} f$ is free since the quotient of a free module is not always free. So I guess we shouldn't be talking about a basis for $M/\ker f$ but maybe a generating set instead... – Artus Jul 27 '16 at 02:52
  • @GregoryGrant Since $M/\text{ker} f$ is not necessarily free, it does not make sense to talk about a matrix representing $\bar{f}$, right? So I guess my question makes no sense?? – Artus Jul 27 '16 at 02:55
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    This is wrong: "$f$ is injective if and only if the columns of $A$ are linearly independent". For matrices over a commutative ring the injectivity is related to its maximal minors. – user26857 Jul 27 '16 at 06:07
  • @user26857 Could you expand on that? – Artus Jul 27 '16 at 08:15
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    @Erciyes See this and this. – user26857 Jul 27 '16 at 17:48

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First of all, the quotient $M/\ker f$ may not be free -- consider $\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z}$. The criterion relating injectivity of $f$ to linear independence of the columns of $A$ works in the case where both source and target are free, but not necessarily otherwise.

If the quotient $M/\ker f$ is free, the matrix will not necessarily be the same -- for example, consider the projection $\mathbb{Z}^2 \to \mathbb{Z}$ onto the first factor: the number of columns of the matrix changes from $2$ to $1$.

Ashvin Swaminathan
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    First of all, injectivity is related to linear independence of columns for matrices over fields (and not necessarily for commutative rings). Second, I don't get what do you mean by "not necessarily otherwise": one can associate a matrix to a map of free modules, not in general. – user26857 Jul 27 '16 at 17:35