# A property of radical ideals

Let $A$ be a commutative ring with $1 \neq 0$.

Theorem (Atiyah-MacDonald 1.13 (v)). Let $\mathfrak{a, b} \subseteq A$ be ideals. Then $\sqrt{\mathfrak{a + b}} = \mathfrak{\sqrt{\sqrt{a} + \sqrt{b}}}$.

Question. Is the following generalization true?

For any finite collection of ideals $\mathfrak{a}_{1}, \dots, \mathfrak{a}_{n} \subseteq A$, we have $$\sqrt{\mathfrak{a}_{1} + \cdots + \mathfrak{a}_{n}} = \sqrt{\sqrt{\mathfrak{a}_{1}} + \cdots + \sqrt{\mathfrak{a}_{n}}}.$$

I'm pretty sure that it is true since if $x_{j} \in \sqrt{\mathfrak{a}_{j}}$ so that $x_{j}^{n_{j}} \in \mathfrak{a}_{j}$ for chosen $n_{j} \geq 1$, each term of $(x_{1} + \cdots + x_{n})^{N}$ is $x_{1}^{r_{1}} \cdots x_{n}^{r_{n}}$ and not all $r_{j} < n_{j}$ if we take $N$ large. (This is a recycled argument of Proposition 1.7 in Atiyah-MacDonald.) I just want to make sure that this statement is correct as I don't have a reference for it.

I am interested in this statement because I'd like to show that given finitely many elements $f_{j} \in A$, we have $\sqrt{(f_{1}, \dots, f_{n})} = \sqrt{0}$ if and only if $f_{j} \in \sqrt{0}$ for all $j$.

It is true when $n = 1, 2$, as follows.

Lemma. Let $f \in A$. We have $\sqrt{(f)} = \sqrt{0}$ if and only if $f \in \sqrt{0}$.

Proof. Since $f \in \sqrt{(f)}$, we only need to show one implication.

Let $f \in \sqrt{0}$. Then $(f) \subseteq \sqrt{0}$. Since taking radical does not change the order of inclusion, we have $\sqrt{(f)} \subseteq \sqrt{\sqrt{0}} = \sqrt{0} \subseteq \sqrt{(f)}$. We showed that $f \in \sqrt{0}$ implies $\sqrt{(f)} = \sqrt{0}$. Q.E.D.

Corollary. Let $f, g \in A$. We have $\sqrt{(f, g)} = \sqrt{0}$ if and only if $f, g \in \sqrt{0}$.

Proof. Since $f, g \in \sqrt{(f, g)}$, we only need to show one implication.

Suppose that $f, g \in \sqrt{0}$. Then by Lemma, we have $\sqrt{(f)} = \sqrt{(g)} = \sqrt{0}$, so $\sqrt{(f, g)} = \sqrt{(f) + (g)} = \sqrt{\sqrt{(f)} + \sqrt{(g)}} = \sqrt{0}$. Q.E.D.

• This was asked here a few days ago. Commented May 28, 2013 at 21:28
• My bad. I don't mind this being closed since I got what I wanted.
– Gil
Commented May 28, 2013 at 21:58

$$\sqrt{\mathfrak{a}_{1} + \cdots + \mathfrak{a}_{n}} = \sqrt{\sqrt{\mathfrak{a}_{1}} + \cdots + \sqrt{\mathfrak{a}_{n}}}$$ Take into account that the radical of an ideal is the intersection of all prime ideals containing it. Now let $\mathfrak p$ be a prime ideal of $R$ containing $\sqrt{\mathfrak{a}_{1}} + \cdots + \sqrt{\mathfrak{a}_{n}}$. This is equivalent to $\sqrt{\mathfrak{a}_{i}}\subseteq\mathfrak p$ for all $i$. But $\sqrt{\mathfrak{a}_{i}}\subseteq\mathfrak p$ iff $\mathfrak{a}_{i}\subseteq\mathfrak p$ and you are done. (You can also prove this by induction on $n\ge1$.)
$$\sqrt{(f_{1}, \dots, f_{n})} = \sqrt{0} \text{ if and only if } f_{j} \in \sqrt{0} \text{ for all } j=1,\dots,n$$ This is easy to prove without using the above equality for radicals. We have $(f_{1}, \dots, f_{n})\subseteq\sqrt{(f_{1}, \dots, f_{n})}=\sqrt 0$, so $f_j\in\sqrt 0$ for all $j$. Conversely, if $f_{j} \in \sqrt{0}$ for all $j$, then $(f_{1}, \dots, f_{n})\subseteq\sqrt0$ and therefore $\sqrt{(f_{1}, \dots, f_{n})}\subseteq\sqrt 0$. Since $(0)\subseteq (f_{1}, \dots, f_{n})$ we finally get an equality.
$V(\mathfrak{a}_{1} + \cdots + \mathfrak{a}_{n}) = V(\cup_{i} \mathfrak{a}_{i}) = V(\mathfrak{a}_{1}) \cap \cdots \cap V(\mathfrak{a}_{n}) = V(\sqrt{\mathfrak{a}_{1}}) \cap \cdots \cap V(\sqrt{\mathfrak{a}_{n}}) = V(\cup_{i}\sqrt{\mathfrak{a}_{i}}) = V(\sqrt{\mathfrak{a}_{1}} + \cdots + \sqrt{\mathfrak{a}_{n}}).$