Possible mistake in Atiyah-MacDonald (Radicals of arbitrary subsets)

Question

On page 9, Atiyah-MacDonald defines $r(E)$ (the radical of an arbitrary subset $E$ or a commutative ring $A$) to be the intersection of all prime ideals containing $E$. They then assert that

$r(\bigcup_{\alpha} E_{\alpha})=\bigcup_{\alpha} r(E_{\alpha})$.

However, if we take $A=\mathbb{Z}$, $E_1=\{ 6 \}$, and $E_2= \{ 10 \}$, then

$r(E_1 \cup E_2)=r(\{6, 10 \})=(2)$

and

$r(E_1) \cup r(E_2)=r(\{6 \}) \cup r(\{ 10 \})=(6) \cup (10)$.

Is this counterexample valid, or did I make a mistake somewhere? If the book is wrong, is there a simple way to correct it?

Answer 1

Proposition 1.19, which is that the radical of an ideal is the intersection of all containing prime ideals, is not a definition. The definition of radical is on the bottom of page 8, and reads, for a general subset $E\subseteq A$:

$$r(E)=\{x\in A\mid x^n\in E\text{ for some }n>0\}$$

Answer 2

In my copy of the book, $$ r(\mathfrak{a})=\{x\in A:x^n\in \mathfrak{a},\text{ for some }n>0\} $$ This is the intersection of the prime ideals containing $\mathfrak{a}$ under the assumption that $\mathfrak{a}$ is an ideal.

The book is careful to say that $r(E)$ is not necessarily an ideal if $E$ is not an ideal.

In your example, $r(\{6\})=\{6\}$ and $r(\{10\})=\{10\}$.