divisibility in polynomial ring: if $(X - a)^t | f(X)$ and $(X - b)^t | f(X)$, then does $(X - a)^t \cdot (X - b)^t \mid f(X)$?

Question

let $K$ be a field an $A$ a commutative $K$-algebra with unit. fix an element $f(X) \in A[X]$, unequal elements $a$ and $b$ of $K \subset A$, and an integer $t > 0$.

Question. if $(X - a)^t \mid f(X)$ and $(X - b)^t \mid f(X)$, then does $(X - a)^t \cdot (X - b)^t \mid f(X)$ in $A[X]$?

this is "trivial" when $A \mathbin{/} K$ is itself a field, essentially by unique factorization in $A[X]$: $X - a$ and $X - b$ are unequal irreducibles in this ring.

i can prove this in general in the case $t = 1$: in this case the hypothesis implies that $f(a) = f(b) = 0$. by the Euclidean algorithm, $f(X) = (X - a) \cdot q(X) + r(X)$, where $\deg(R) < 1$, and in fact $r = 0$ must hold. now $0 = f(b) = (b - a) \cdot q(b)$, so that $q(X)$ also vanishes at $b$ (here we use the fact that $b - a \in K \subset A$ is a unit), and so $q(X) = (X - b) \cdot q'(X)$ by the same logic. we see that $f(X) = (X - a) \cdot (X - b) \cdot q'(X)$, as desired.

what about $t > 1$?

Over $\Bbb Z_8$ we have $x!-!1,x!-!3\mid x^2!-!1=x^2!-!9,$ but $,(x!-!1)(x!-!3)\nmid x^2-1,,$ see the Bifactor Theorem. $\ \ $ — Bill Dubuque, Mar 05 '24 at 00:41
indeed, the fact that $a$ and $b$ are in $K$ (and $b - a$ is a unit) will be essential if this is going to be true at all. — BD107, Mar 05 '24 at 00:53
As explained in the answer I linked, we need $,a-b,$ is cancellable (not a zero-divisor). Does that post satisfy you as an answer to your question? — Bill Dubuque, Mar 05 '24 at 00:57
no; essentially the key problem here is that we need to handle $t > 1$ too. (my proof above of the case $t = 1$ is essentially the content of that post.) — BD107, Mar 05 '24 at 01:11
$(x!-!a)^{k+1}!\mid (x!-!a)^k(x!-!b)^tg!\iff! x!-!a\mid (x!-!b)^tg!\iff! (a!-!b)^tg(a)!=!0!\iff! g(a)!=!0!\iff! x!-!a\mid g\ \ $ — Bill Dubuque, Mar 05 '24 at 01:37
i'm not sure i follow..... would you mind expanding this into an answer? — BD107, Mar 05 '24 at 01:40
Let $,h_k = (x!-!a)^k(x!-!b)^t.,$ By above $,h_k\mid f\Rightarrow h_{k+1}\mid f,$ if $,k<t.,$ Base case $h_0\mid f$ is true by hypothesis. — Bill Dubuque, Mar 05 '24 at 01:48
I appended the straightforward inductive extension to the linked proof of the Bifactor Theorem, so now the question is a dupe. — Bill Dubuque, Mar 05 '24 at 03:16

divisibility in polynomial ring: if $(X - a)^t | f(X)$ and $(X - b)^t | f(X)$, then does $(X - a)^t \cdot (X - b)^t \mid f(X)$?

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