I just came up with a geometric interpretation for Nakayama's Lemma, and I'm surprised that no one here has already mentioned it!

In the geometric picture, we see $A$ as functions on a space $X$ (the spectrum $\mathrm{Spec}(A)$), with $I$ corresponding to functions that vanish on a closed subset $Z$ (the subset $V(I)$); $M$ is the module of global sections for a "vector bundle" $\mathcal F$ (the sheaf of modules $\widetilde M$). The elements of $IM$ represent those global sections vanishing along $Z$, so the condition $IM=M$ says that these are all the sections possible. In other words, $\mathcal F$ is identically zero along $Z$, and so its support $\mathrm{Supp}(\mathcal F)$ must be disjoint from $Z$. Now the conclusion of Nakayama's Lemma simply affirms the existence of a "bump function" $a$, that is, a function being $0$ along $Z$, and $1$ along the support of $\mathcal F$: i.e. it lies in $I$ and it acts on $M$ like $1$!

If one works out the details of commutative algebra, without any finite generation condition on $M$, we have

• $IM=M$ is equivalent to $V(I)\cap \mathrm{Supp}(M)=\emptyset$;
• $\mathrm{Supp}(M)$ is contained in $V(\mathrm{Ann}(M))$;
• $\exists\ a\in I$ s.t. $(a-1)M=0$ is equivalent to $V(I)\cap V(\mathrm{Ann}(M))=\emptyset$ (the "partition of unity").

Now the key issue is that $\mathrm{Supp}(M)$ is not necessarily closed: the addition of each generator in $M$ adds a new component to $\mathrm{Supp}(M)$; if the number is infinite we might get an infinite union of closed subsets, which is no longer closed. Also the closure of $\mathrm{Supp}(M)$ might be smaller than the expected $V(\mathrm{Ann}(M))$. So the finite generation condition on $M$ is given to ensure that $\mathrm{Supp}(M)$ is indeed a closed subset and equals to $V(\mathrm{Ann}(M))$, so that we have the two disjoint closed subsets as wanted. Also this allows us to easily construct counterexamples in non-finitely generated cases: $A=\mathbf Z$, $I=2\mathbf Z$, and $M=\mathbf Q$ or $M=\bigoplus_{p\ge3}\mathbf F_p$, since they both have non-closed support, whose closure intersects $Z=V(I)$; and $M=\bigoplus_{n\ge1}\mathbf Z/3^n\mathbf Z$, in which case the support is closed but not equal to $V(\mathrm{Ann}(M))$.

So, to conclude, Nakayama's Lemma (in the above form) says that the support of a finitely generated $A$-module $M$ can be identified with the closed subset $V(\mathrm{Ann}(M))$, and given any closed subset $Z=V(I)$ disjoint from it, we can find a "bump function" separating the two.

In real life, however, we are usually just using another interpretation of the closedness property: the support is closed under specialization. Since for an $A$-module $M$ to be non-zero, it must at least have some support, and by specialization, it must be supported at least at one closed point (corresponding to a maximal ideal). Therefore, if we can verify that $M$ is not supported at any of the closed points, (which is provided, say, by the condition $IM=M$ for $I$ contained in the Jacobson radical $J(A)$), then $M$ must be zero.

I just came up with a geometric interpretation for Nakayama's lemma, and I'm surprised that no one here has already mentioned it!

In the geometric picture, we see $A$ as functions on a space $X$ (the spectrum $\mathrm{Spec}(A)$), with $I$ corresponding to functions that vanish on a closed subset $Z$ (the subset $V(I)$); $M$ is the module of global sections for a "vector bundle" $\mathcal F$ (the sheaf of modules $\widetilde M$). The elements of $IM$ represent those global sections vanishing along $Z$, so the condition $IM=M$ says that these are all the sections possible. In other words, $\mathcal F$ is identically zero along $Z$, and so its support $\mathrm{Supp}(\mathcal F)$ must be disjoint from $Z$. Now the conclusion of Nakayama's lemma simply affirms the existence of a "bump function" $a$, that is, a function being $0$ along $Z$, and $1$ along the support of $\mathcal F$: i.e. it lies in $I$ and it acts on $M$ like $1$!

If one works out the details of commutative algebra,

• the condition "$\exists\ a\in I$ s.t. $(a-1)M=0$" is equivalent to $V(I)\cap V(\mathrm{Ann}(M))=\emptyset$ (the "partition of unity");
• the support $\mathrm{Supp}(M)$ is contained in $V(\mathrm{Ann}(M))$;
• for finitely generated module $M$, the condition $IM=M$ implies $V(I)\cap\mathrm{Supp}(M)=\emptyset$, moreover $\mathrm{Supp}(M)$ can be shown to coincide with the closed subset $V(\mathrm{Ann}(M))$.

When $M$ is not finitely generated, $\mathrm{Supp}(M)$ does not behave well, so we have courterexamples

• $IM=M$ does not imply $V(I)\cap\mathrm{Supp}(M)=\emptyset$: the classical counterexample $A=\mathbf Z_{(p)}$ with its maximal ideal, and $M=\mathbf Q$;
• $\mathrm{Supp}(M)$ can be non-closed: $A=\mathbf Z$, $I=2\mathbf Z$, and $M=\bigoplus_{p\ge3}\mathbf F_p$;
• $\mathrm{Supp}(M)$ can be closed yet different from $V(\mathrm{Ann}(M))$: $A=\mathbf Z$, $I=2\mathbf Z$, and $M=\bigoplus_{n\ge1}\mathbf Z/3^n\mathbf Z$.

So, to conclude, Nakayama's lemma (in the above form) says that the support of a finitely generated $A$-module $M$ can be identified with the closed subset $V(\mathrm{Ann}(M))$, and given any closed subset $Z=V(I)$ disjoint from it, we can find a "bump function" separating the two.

In real life, however, we are usually just using another (weaker) property of the closedness: the support is closed under specialization. Since for an $A$-module $M$ to be non-zero, it must at least have some support, and by specialization, it must be supported at least at one closed point (corresponding to a maximal ideal). Therefore, if we can verify that $M$ is not supported at any of the closed points, (which is provided, say, by the condition $IM=M$ for $I$ contained in the Jacobson radical $J(A)$), then $M$ must be zero.