Showing that $q=(z_1,z_2^2)$ is primary in $mathcal O_2 $

Question

Show that if $q$ is primary, then $sqrt{q}$ is prime.
Show that in the ring $mathcal O_2 = mathbb C{z_1, z_2}$, $q=(z_1,z_2^2)$ is primary.

original picture
I already did the first item. For the second, I tried writing two power series with variables $z_1$ and $z_2$, but I couldn't show how the product of them is a member of the ideal generated by $z_1, z_2^2$ (in fact, I don't know how can I re-write the series).
Again, I ONLY need help on the item $(2)$, and this is all that I did. I think this might be an easy problem, but I'm really stucked at this.

Rodrigo Torres · Answer

Remember that, by definition, the ideal $sqrt{q} =${$x in R / x^n in q$ for some $n$}.
It implies that, if $q$ is primary, for any product $ab in sqrt{q}$ you have that $a in sqrt{q}$ or $b in sqrt{q}$, because, for some $n$, $a^n in q$ or $b^n in q$, and that is the definition of being prime. There is no need of using power series. Just write down the definitions and go on with them!
For the second item, it is enough to show that every zero divisor in $C{z_1,z_2^2}/(q)$ is nilpotent, which implies that q is primary.
Good luck!

Showing that $q=(z_1,z_2^2)$ is primary in $mathcal O_2 $

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