# finitely generated projective module and Nakayama's lemma

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 $$\ker f = I\ker f$$.

• 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 $\ker f = I\ker f$ – Max Feb 8 at 7:57
• Did you mean to write $R^{n} = M \oplus \ker(f)$ in the problem statement, or are there two distinct variables here, $m$ and $n$? – 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.
• Why is $\dim_k (M/IM) = m$ ? It seems to me $\dim_k (M/IM) = n$ – Max Feb 8 at 10:01
• @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. – Alex Wertheim Feb 8 at 18:59
• @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. – Alex Wertheim Feb 10 at 8:05
• 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 ! – Max Feb 10 at 10:11