Exercise 3.5

Proof. Suppose $A$ has a nilpotent element, i.e., $𝔑\ne 0$. Then, by AM Corollary $3.12$, the nilradical of $A_{𝔭}$ is ${𝔑}_{𝔭}$. But $𝔑\ni x\ne 0⟹x∕1\in {𝔑}_{𝔭}$, which is a contradiction.

Now consider $A=k\times k$; clearly this is not an integral domain, for $(a,0)\cdot (0,b)=0$. Note $𝔭=k\times 0$ and $𝔮=0\times k$ are the only non-trivial prime ideals of $A$. Now consider ${𝔭}_{𝔭}$; for any $a∕b\in {𝔭}_{𝔭}$, we claim $a∕b\equiv 0∕1=0$. For, $a(0,1)=0$ and $(0,1)\in A\setminus 𝔭$ and by definition of the equivalence relation. But recall from AM Example $1$ from p.  $38$ that this ideal $0$ is maximal in $A_{𝔭}$, and so $A_{𝔭}$ is a field. Since the same argument applies for $A_{𝔮}$, we see that each $A_{𝔭}$ being an integral domain does not imply $A$ is an integral domain.

One another simple example would be $A=\mathbb{Z}∕6\mathbb{Z}=\mathbb{Z}∕2\mathbb{Z}\times \mathbb{Z}∕3\mathbb{Z}$. Prime ideals of $A$ are copies of $\mathbb{Z}∕(2)$ and $\mathbb{Z}∕(3)$ (i.e. ${𝔭}_1=\{0,3\}$ and ${𝔭}_2=\{0,2,4\}$) ,and $A_{𝔭}=\mathbb{Z}∕(3)$ or $\mathbb{Z}∕(2)$. Hence, $A_{𝔭}$ is a domain for all of its prime ideals $𝔭$, but $A$ itself is obviously not a domain as $2\cdot 3=0$.

In general Let $R$ be a ring with two nonzero prime ideals $𝔭$ and $𝔮$ such that each is not included in the other, and that $\mathrm{Spec}<!--\; FUNCTION\; APPLICATION\; -->R=\{𝔭,𝔮\}.$ Now let $A=R∕\mathit{𝔭𝔮}$. Then $A$ is not a domain because the image of any two element $p\in 𝔭∕𝔮$ and $q\in 𝔮∕𝔭$ are zero divisors of $A$. However, $A_{𝔭}$ and $A_{𝔮}$ are domains. □