# I am at a very initial stage of commutative algebra.

I am at a very initial stage of commutative algebra. I want to know whether the power of a prime ideal in a commutative ring is prime ideal or not?
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

gutinyalk
An ideal $I\subset R$ is prime iff $R/I$ is an integral domain (that is, it contains no nontrivial zero divisors).
If $I$ is a prime ideal such that ${I}^{2}\ne I$, then in the quotient ring $R/{I}^{2}$ the elements of the form $x+{I}^{2}$, where $x\in I$, will be nilpotents. Thus, $R/{I}^{2}$ is not a domain and ${I}^{2}$ is not prime.
For a concrete and illustrative example, consider $I=\left(p\right)\subset \mathbb{Z}$ - the ideal of the ring of integers generated by a prime $p$. Then, ${I}^{2}=\left({p}^{2}\right)$, and is not prime (since it is generated by a non-prime integer).