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?
Best Answer
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 $j \geq i$.