(a) If Q is primary, then its radical is a prime ideal.

(b) If $Q_1$ and $Q_2$ are both P-primary, then their intersection is also P-primary.

(a)

Let $fg\in \sqrt{Q}$. Then $(fg)^N \in Q$ for some N and, since Q is primary, $f^N \in Q$ or $g^{MN}\in Q$ for some M. Hence, $f\in \sqrt{Q}$ or $g\in \sqrt{Q}$, ie $\sqrt{Q}$ is prime.

(b) Note: $Q_1, Q_2$ P-primary, ie $\sqrt{Q_1}=P=\sqrt{Q_2}$ where P is prime.

First, we show that the intersection is primary. Let $fg\in Q_1 \cap Q_2$. Then fg is an element of both, $Q_1$ and $Q_2$. Since they are primary, we get $f\in Q_1 \vee g^{N_1}\in Q_1$ and $f\in Q_2 \vee g^{N_2}\in Q_2$. In all cases, we get $f\in Q_1\cap Q_2$ or $g\in P$ and, hence, $f\in Q_1\cap Q_2$ or $g\in Q_1\cap Q_2$ for some M.

It remains to show that $\sqrt{Q_1\cap Q_2}=P$. If $f\in\sqrt{Q_1\cap Q_2}$, then $f^N\in Q_1\cap Q_2$ for some power N, ie $f\in P$.

For the reverse inclusion, let f be in P. Then some power of f is in $Q_1$ and in $Q_2$ (P is the radical ideal of both). Therefore, there is an N such that $f^N$ is in the intersection $Q_1\cap Q_2$ and f is in the radical of the intersection.