Comaximality of ideals

Question

Suppose $P \in \mathbb{C}[X]$, and let $\prod_{i=1}^r(X-\alpha_{i})^{k_i}$ be its factorization. I'm working on an exercise where I need to prove that a ring $A$ is isomorphic to $$\prod_{i=1}^r\frac{\mathbb{C}[X]}{((X-\alpha_{i})^{k_i})},$$ where $((X- \alpha_i)^{k_i})$ is the ideal generated by $(X- \alpha_i)^{k_i}$.
I already have that $$\frac{\mathbb{C}[X]}{(P)}=\frac{\mathbb{C}[X]}{\prod_{i=1}^r((X-\alpha_{i})^{k_{i}})}\cong A.$$ If I can show that the ideals $((X- \alpha_i)^{k_i})$ are two by two comaximal, then I can apply the Chinese Remainder Theorem. But I how do I prove that they are comaximal?

Answer 1

Say $(X- \alpha_i)^{k_i}$ and $(X- \alpha_j)^{k_j}$ are coprime. So there exists $s, t\in \Bbb{C}[X]$ that $$ s(X)(X- \alpha_i)^{k_i}+t(X)(X- \alpha_j)^{k_j}=1 $$ Thus for any $f(X)\in \Bbb{C}[X]$ $$ f(X)= f(X)s(X)(X- \alpha_i)^{k_i}+f(X)t(X)(X- \alpha_j)^{k_j}\in \langle(X- \alpha_i)^{k_i}\rangle+\langle(X- \alpha_j)^{k_j}\rangle $$ This shows that $$ \Bbb{C}[X]\subset\langle(X- \alpha_i)^{k_i}\rangle+\langle(X- \alpha_j)^{k_j}\rangle $$ The other inclusion is obvious. So $$ \Bbb{C}[X]=\langle(X- \alpha_i)^{k_i}\rangle+\langle(X- \alpha_j)^{k_j}\rangle $$ i.e. they are comaximal.

Answer 2

The $\,\, P_i = X-\alpha_i\,$ are nonassociate primes so $\,i\neq j\,\Rightarrow\, \gcd(P_i,P_j)=1\,\Rightarrow\, \gcd(P_i^m,P_j^n)=1.\,$ But in a PID, $\ (a,b) = (c),\,$ for $\,c = \gcd(a,b),\,$ which yields the result (since here $\,c = 1)$.