# Proof nilradical is intersection of prime ideals.

It gives a proof of the above proposition that I don't fully understand. I'm sure I must have missed something simple, but can't see what it is. Wikipedia refers to the same proof in Chapter 1, Proposition 1.8 of Atyah, Mac Donald Introduction to Commutative Algebra.

Below the proof together with my 3 questions. If anyone can shed some light on this thanks in advance!

Proposition 1.8 : The nilradical $N$ of $A$ is the intersection of all prime ideals ($N'$) of $A$.

Proof : $\implies$ Let $N'$ denote the intersection of all prime ideals of $A$ (Question 1: doesn't $\{0\}$ count as a prime ideal? And if so doesn't that mean that the intersection of all prime ideals is always $\{0\}$?). If $f \in A$ is nilpotent and $P$ is a prime ideal then $f^m=0 \in P$ for some $m$. Hence $f \in P$ because $P$ is prime (Question 2: shouldn't it be: $f \in P$ OR $P=\{0\}$ ?). Hence $f \in N' \implies N \subseteq N'$.

Conversely: $\impliedby$ Let $x$ be a non-nilpotent. Let $\{I_0, I_1, \cdots \}$ be the set of ideals NOT containing any power of $x$. This set is non empty because $x$ is non-nilpotent so $I_0=\{0\}$ is an element. This is a partially ordered set with the inclusion $\subseteq$. By Zorn's lemma this set has a maximal element $I_{max}$. Now the proof goes on to prove $I_{max}$ must be prime. So we have constructed a prime ideal NOT containing $x$, this implies $N' \subseteq N$. Question 3: what if $I_{max}=\{0\}$? That could be right? But in that case we have proved that $x \notin \{0\}$ , which doesn't prove anything since we must exclude $\{0\}$ from the intersection of prime ideals somehow to get a result other than $\{0\}$.

$\{0\}$ is not a prime ideal if $A$ has divisor of zeros. This also answer the second question, because if $I_{max}=\{0\}$, this means that $A$ is integral.
The sentence "$\{0\}$ is a prime ideal in $A$" is equivalent to "$A$ is an integral domain", as both reduce to proving that $$\forall x,y\in A, xy=0 \Rightarrow x=0\, \text{or} \, y=0$$
Hence, it is not true in general that $\{0\}$ is a prime ideal in $A$.
For instance, in $\mathbb Z / 4 \mathbb Z$, we have the identity $$2\times 2=4=0$$ whereas $2$ surely is not $0$. In this ring, the zero ideal indeed is not prime.