Proposition

$r(a) \supset a$.
$r(r(a)) = r(a)$.
$r(ab) = r(a \cap b) = r(a) \cap r(b)$.
$r(a) = (1) \iff a = (1)$.
$r(a + b) = r(r(a) + r(b))$.
if $p$ is prime, $r(p^n) = p$ for all $n > 0$.

Solution

1

If $x \in a$, then $x^1 \in a$, so $x \in r(a)$.

2

By 1, $r(a) \subset r(r(a))$. Let $x \in r(r(a))$. Then $x^m \in r(a)$, and $(x^m)^n \in a$ for some $m, n \geq 1$. Therefore, $x^{mn} \in a$, so $x \in r(a)$.

3

Since $ab \subset a \cap b$, $r(ab) \subset r(a \cap b)$. Let $x \in r(a \cap b)$. Then $x^m \in a \cap b$. Thus $x^m \in a$ and $x^m \in b$, so $x^{2m} \in ab$. Therefore, $x \in r(ab)$.

Let $x \in r(a \cap b)$. Then $x^m \in a \cap b$, so $x^m \in a$ and $x^m \in b$. Therefore, $x \in r(a)$ and $x \in r(b)$, so $x \in r(a) \cap r(b)$. Let $x \in r(a) \cap r(b)$. Then $x^m \in a$ and $x^n \in b$ for some $m, n \in \mathbb{Z}$. Then $x^{m + n} \in a \cap b$, so $x \in r(a \cap b)$. Thus $r(a \cap b) = r(a) \cap r(b)$.

4

If $r(a) = (1)$, then $1 \in r(a)$, so $1^m = 1 \in a$, so $a = (1)$. If $a = (1)$, then $a \subset r(a)$, so $r(a) = 1$.

5

By 1, $a \subset r(a)$ and $b \subset r(b)$. Then $a + b \subset r(a) + r(b)$, so $r(a + b) \subset r(r(a) + r(b))$.

Let $x \in r(r(a) + r(b))$. Then $x^m \in r(a) + r(b)$ for some $m \in \mathbb{N}$. Let $y \in r(a)$ and $z \in r(b)$ such that $x^m = y + z$. Let $n, l \in \mathbb{N}$ be chosen such that $y^n \in a$ and $z^l \in b$. Then $x^{n + l} = (y + z)^{n + l} = \sum_{i=0}^{n + l} \binom{n + l}{i} y^{i}z^{n + l - i} \in a + b$ because for each $i$, $i \geq n$ or $n + l - i \geq l$. Therefore, $x \in r(a + b)$. This implies $r(r(a) + r(b)) \subset r(a + b)$.

Thus $r(a + b) = r(r(a) + r(b))$.

6

By 1, $p \subset r(p)$. Let $x \in r(p)$. Then $x^m \in p$ for some $m \geq 1$. Since $p$ is prime, $x \in p$. Thus $p = r(p)$.

By 3, $r(p^n) = r(p) \cap r(p^{n - 1}) = p \cap r(p^{n - 1})$, so by induction, $r(p^n) = p$.