A case when $I \otimes_R M$ and $ IM$ are locally isomorphic at minimal primes?

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Let $(R,\mathfrak m)$ be a local Cohen--Macaulay ring of dimension $1$. Let $M$ be a finitely generated $R$-module of depth $1$. Let $I$ be an ideal of $R$ of height $1$.

Then my question is: Is it true that $I_P \otimes M_P \cong (IM)_P$ for every minimal prime ideal $P$ of $R$?

Of-course there is always a natural surjection $I \otimes_R M \to IM$ sending $x \otimes m\to xm$, but I am unable to see if this also becomes injective when localising at minimal primes.

Please help.

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As $R$ is Cohen-Macaulay, the set of minimal and associated primes are equal. Also, since $R$ is Cohen-Macaulay, the height of an ideal is equal to the grade.

Let $P$ be a minimal prime and $I$ a height 1 ideal. Since the grade of $I$ is $1$, $I \not\subset P$. Thus, $I_P \cong R_P$. This is sufficient to deduce the isomorphism $$ I_P \otimes_{R_P} M_P \cong R_P \otimes_{R_P} M_P \cong M_P $$ and $(IM)_P \cong I_P M_P \cong R_P M_P \cong M_P$.

If you prefer keeping track of the surjective map in your post and familiar with the Tor functor, one can show that $\operatorname{Tor}_1^R (R/I,M)$ becomes zero after localizing at a minimal prime $P$, since $(R/I)_P =0$.