## In a commutative ring, prime ideals are radical

Let be a commutative ring. An ideal of is called *radical* if (since always holds, we can say equivalently that is radical if ).

- Prove that every prime ideal of is radical.
- Let be an integer. Prove that is radical in is radical if and only if is squarefree. Deduce that is radical in if and only if is squarefree.

We begin with a lemma.

Lemma: Let be a commutative ring and let be an ideal. Also let is an ideal containing . Then is radical in if and only if is radical in . Proof: Suppose is radical in , and let . Then for some . Now , so that . Thus , and is a radical ideal in . Conversely, suppose is radical in and let . Then for some , so that . Thus , and we have ; hence is radical.

- Let be a prime ideal. Let . Then for some , we have ; suppose that is minimal with this property. If , then . Since is prime, then either or , a contradiction since is minimal. Thus , and we have .
- Write the prime factorization of as . Suppose first that is radical in , and suppose some is at least 2. Now is nonzero in , and , a contradiction. Thus is squarefree. Conversely, suppose is squarefree and let with for some . If some prime divides but not , then no power of can be divisible by , a contradiction. Thus divides , and in fact .
Consider as an ideal of . By the lemma, is radical in if and only if is radical in , if and only if is squarefree.

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