Exercise 2.22

Proof. Suppose $x\in 𝔑$, the nilradical of $A=\underset{\text{→}}{\lim <!--\; FUNCTION\; APPLICATION\; -->}{A}_i$. Then, $x={\alpha }_i(x_i)$ for some $x_i\in A_i$ by AM Exercise $2.15$. Since these ${\alpha }_i$ are ring homomorphisms by AM Exercise $2.21$, ${\alpha }_i(x_i^n)=0$ for some $n$, and so by AM Exercise $2.15$ there exists $j$ such that ${\alpha }_{\mathit{ij}}(x_i^n)={\alpha }_{\mathit{ij}}{(x_i)}^n=0$, i.e., ${\alpha }_{\mathit{ij}}(x_i)\in {𝔑}_j$. Then, $x={\alpha }_j({\alpha }_{\mathit{ij}}(x_i))\in {\alpha }_j({𝔑}_j)\subseteq \underset{\text{→}}{\lim <!--\; FUNCTION\; APPLICATION\; -->}{𝔑}_i$.

In the other direction, suppose $x\in \underset{\text{→}}{\lim <!--\; FUNCTION\; APPLICATION\; -->}{𝔑}_i$; we can again write $x={\alpha }_i(x_i)$ for $x_i\in {𝔑}_i$. Then, $x_i^n=0$ for some $n⟹x^n=0⟹x\in 𝔑$. Thus, $𝔑=\underset{\text{→}}{\lim <!--\; FUNCTION\; APPLICATION\; -->}{𝔑}_i$.

Suppose $\mathit{xy}=0\in A$, and pick $i,j$ such that $x={\alpha }_i(x_i),y={\alpha }_j(y_j)$; by the well-definition of the ring structure on $A$, if we assume without loss of generality that $i\ge j$, we can find $y_i={\alpha }_{\mathit{ij}}(y_j)$ such that we have $\mathit{xy}={\alpha }_i(x_i){\alpha }_j(y_j)={\alpha }_i(x_i){\alpha }_i(y_i)={\alpha }_i(x_i{y}_i)=0$. By AM Exercise $2.15$, we can then find $k\ge i$ such that ${\alpha }_{\mathit{ik}}(x_i{y}_i)={\alpha }_{\mathit{ik}}(x_i){\alpha }_{\mathit{ik}}(y_i)=0$. But since $A_k$ is an integral domain, we have that ${\alpha }_{\mathit{ik}}(x_i)=0\vee {\alpha }_{\mathit{ik}}(y_i)=0$. Note that if we assume without loss of generality the former is true, ${\alpha }_k({\alpha }_{\mathit{ik}})(x_i)=x=0$. □