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## Compute the powers of a given ideal in a polynomial ring

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.3 Exercise 7.3.36

Solution: We begin with a lemma.

Lemma: If $R$ is a commutative ring with 1 and $a,b \in R$, then $(aR)(bR) = abR$.

Proof: If $x \in (aR)(bR)$, then $x = \sum ay_i bz_i$, where $y_i,z_i \in R$. Then $$x = (ab)(\sum y_iz_i) \in abR.$$ If $x \in abR$, then $x = abr$ for some $r \in R$. Now $x = a1br \in (aR)(bR)$. $\square$

Now the problem at hand is merely to show that $$(x \mathbb{Z}[x])^n = x^n \mathbb{Z}[x],$$ which follows from the lemma by an easy induction.