Why is the power of maximal ideals comaximal in an Artinian ring?

Question

Let $R$ be an Artinian ring, commutative with 1.

We know :

1. there are only finitely many maximal ideals of $R$.
2. $Jac(R)^m = 0$ for some natural number $m$, using D.C.C.
3. Every prime ideal is maximal, using 2.

The backdrop of this question is we want to show $R$ is isomorphic to the product of some Artinian local rings. In the process of doing this, Dummit & Foote uses Chinese Remainder Theorem, which requires that the collection of powers of maximal ideals $\{M_1^m, ..., M_n^m\}$ is pairwise comaximal, where $m$ is the number from 2 above. I do not know why this comaximality is true.

Reference, Dummit & Foote 3ed, 16.1, page 753.

Answer 1

If $\mathfrak p + \mathfrak q = (1)$, then $r(\mathfrak p^n + \mathfrak q^m) = r(r(\mathfrak p^n) + r(\mathfrak q^m)) = r(\mathfrak p + \mathfrak q) = r((1)) = (1)$, so $\mathfrak p^n + \mathfrak q^m = 1$.

Answer 2

A more general fact. If $A, B$ are comaximal, write $1=a+b, a\in A, b\in B$. Given $A^n, B^m$, write $1 = (a+b)^{n+m-1} \in A^n + B^m$ because in the binomial expansion, one of the powers is always greater than or equal to n or m. Then $A^n + B^m = R$.