Let $R$ be a local ring with maximal ideal $I$. $M$ is a finitely generated module over $R$ generated by $a_1, \ldots, a_n$ and the generators are chosen such that their quotients in $M/IM$ form a basis. Then there is a surjective homomorphism $f: R^{n} \to M$. Suppose that $R^{m} = M \oplus \ker f$. How to show that $\ker f = I\ker f$?

Suppose $(p_1, \ldots, p_n) \in \ker f$, then $p_1a_1+\ldots+p_na_n = 0$. It then implies that $p_1a_1+\cdots+p_na_n = 0$ in $M/IM$. So $p_i \in I, \forall i$. But this is not enough to conclude that $\ker f = I\ker f$.

  • $\begingroup$ This implies that if $x\in R^m$, decomposing $x$ in $M\oplus \ker f$ shows that $R^m = M+IM$; so $R^m = M+IR^m$, so by Nakayama's lemma $R^m = M$, so $\ker f =0$, so $\ker f = I\ker f$ $\endgroup$ – Max Feb 8 at 7:57
  • $\begingroup$ Did you mean to write $R^{n} = M \oplus \ker(f)$ in the problem statement, or are there two distinct variables here, $m$ and $n$? $\endgroup$ – Alex Wertheim Feb 8 at 18:56

Put $N = \mathrm{ker}(f)$. Since $M$ is projective, the short exact sequence

$$0 \to N \to R^{n} \xrightarrow{~f~} M \to 0$$

is split, and so there is an isomorphism $\alpha \colon R^{n} \to M \oplus N$ witnessing this splitting. (This is all we need below, but it is worth noting that if $R^{m} \cong M \oplus N$, then $R^{m} \cong R^{n}$ via $\alpha$, whence $m = n$, since all commutative rings satisfy the invariant basis property.)

Now, let $k = R/I$ be the residue field of $R$. Then $\alpha \otimes \mathrm{Id}_{k} \colon R^{n} \otimes_{R} k \to (M \oplus N) \otimes_{R} k$ is an isomorphism of $k$-vector spaces. Since $R^{n} \otimes_{R} k \cong k^{n}$ as $k$-modules, it follows that $\dim_{k}((M \oplus N) \otimes_{R} k) = n$. But $(M \oplus N) \otimes_{R} k \cong M/IM \oplus N/IN$ as $k$-modules, and since $\dim_{k}(M/IM) = n$, we have $\dim_{k}(N/IN) = 0$. Hence, $N/IN = 0$, whence $N = IN$, as desired.

  • $\begingroup$ Why is $\dim_k (M/IM) = m$ ? It seems to me $\dim_k (M/IM) = n$ $\endgroup$ – Max Feb 8 at 10:01
  • $\begingroup$ @Max: yes, you are right. I had misread the problem thinking that there was only one variable used throughout (namely, $m$). I have asked the OP for clarification above, but if the problem was stated as intended, this answer is clearly wrong, and I will delete it. $\endgroup$ – Alex Wertheim Feb 8 at 18:59
  • $\begingroup$ @Max: on second thought, it seems to me that $m$ necessarily equals $n$, so there is no problem after all. If you see any additional difficulties with what I have written, I would appreciate your input. $\endgroup$ – Alex Wertheim Feb 10 at 8:05
  • $\begingroup$ I mean, since $N=0$ anyway in the end, you necessarily have $n=m$ (my comment below the OP is another way of seeing that). But yeah, your correction seems fine ! $\endgroup$ – Max Feb 10 at 10:11

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