# Direct limit of rings

I have a (perhaps silly) question motivated by the following exercise in Atiyah-MacDonald's book:

Say $(A_i,\alpha_{ij})$ is a direct system of rings with direct limit $A_\infty$ and associated homomorphisms $\alpha_i: A_i \to A_\infty$. Let $(\mathfrak{N}_i,\alpha_{ij}')$ denote the direct system of nilradicals $\mathfrak{N}_i \subseteq A_i$ with $\alpha_{ij}'$ the restriction of $\alpha_{ij}$ to $\mathfrak{N}_i$. Say $\mathfrak{N}_\infty$ is the direct limit of the nilradicals and $\nu_i: \mathfrak{N}_i\to \mathfrak{N}_\infty$ the associated homomorphisms.

We are to show that the nilradical of $A_\infty$ is the direct limit $\mathfrak{N}_\infty$.

The exercise is easy (based on previous exercises) if we identify each $\nu_i$ with the restriction of $\alpha_i$ to $\mathfrak{N}_i$. There is a solution online in which this is done. But what is the justification for this identification? Or, is this not even justified and I looked at a bad solution? It boils down to claiming that if $x\in \mathfrak{N}_i$, then $\alpha_i(x) \in \mathfrak{N}_\infty$.

When we look at an actual construction of the direct limit (e.g., start with a disjoint union then mod out by an appropriate relation), one sees that this identification is clearly not actual equality of maps. The collection of equivalence classes in $\mathfrak{N}_\infty$ is not actually contained in $A_\infty$. But perhaps there is some sort of identification that I may be missing that permits me to view $\nu_i$ as a restriction?

I tried showing that $\mathfrak{N}_\infty$ along with the restricted maps $\alpha_i' :=\alpha_i|\mathfrak{N}_i$ satisfied the universal mapping property of a direct limit in hopes of invoking uniqueness of direct limits, but that didn't seem to go anywhere...

Am I missing something?

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There are a lot of bad Atiyah–MacDonald solution sets out there. – Dylan Moreland Jul 19 '12 at 1:20
Indeed there are. Before I looked at the solution, I tried a different route: The collection of inclusions $e_i: \mathfrak{N}_i \to A_i$ form a map between the two systems. On passing to the limit, we obtain a map $e_\infty: \mathfrak{N}_\infty \to A_\infty$. It's then easy to see that the image of this map lands in the nilradical of $A_\infty$. I was hoping that this would turn out to be an isomorphism, but I couldn't show it. – John Myers Jul 19 '12 at 1:26
I'll have to think about whether I understand your objection. Note that by Ex. 19 in that same section, taking direct limits is exact. So the injections $\mathfrak N_i \to A_i$ induce an injection $\mathfrak N_\infty \to A_\infty$. So viewing $\mathfrak N_\infty$ as lying inside of this ring is really okay. Let me read again... – Dylan Moreland Jul 19 '12 at 1:29
I completely overlooked that fact. Ha! Got it. The map is also surjective. Start with $x_i$ in the nilradical, then use exercise 15 to pull back to an element $x\in A_i$. Then note then that we must have $\alpha_{ij}(x) \in \mathfrak{N}_j$. Apply $\nu_{j}$ and use a few relations among the maps and it's there. This is good, thanks for the help! – John Myers Jul 19 '12 at 1:34
So this is answered? – Martin Brandenburg Jul 19 '12 at 8:05

Just to collect our discussion in the comments: taking direct limits (over filtered index sets, as the book in question assumes) in the category of modules over a ring is exact. This is Exercise 19 of the same section. So the natural map $\mathfrak N_\infty \to A_\infty$ is an injection, and the composition of this map with $\nu_i$ is indeed just the restriction of $\alpha_i$.
As I think you've discovered, the rest of the exercise follows quickly from the facts (1) every element of $A_\infty$ is in the image of some $\alpha_i$ (2) if $\alpha_i(x) = 0$ then $\alpha_{ij}(x) = 0$ for some $j \geq i$.