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Let $(R,\mathfrak{m})$ be a local Noetherian ring with an ideal $I$ and $M$ and $N$ be finitely-generated modules over $R$. Then $\text{depth}_I (M \oplus N) = \min\{\text{depth}_I M,\text{depth}_I N \}$.

Proof: First, $\le$ is given for free since a sequence $(x_1,\dots,x_r)$ is $M\oplus N$-regular implies $(x_1,\dots,x_r)$ is both $M$ and $N$ regular (as the nonzerodivisors on $M \oplus N$ are nonzerodivisors on both $M$ and $N$), and we may use maximality of depth.

Then suppose $<$ is true, and now note it suffices to consider the depth $0$ case (i.e., $\text{depth}_I(M\oplus N) = 0$ implies $\text{depth}_I(N) =0$ or $\text{depth}_I(M) = 0$) by modding out by $(x_1,\dots,x_r)$ a maximal $M\oplus N$-regular sequence. But if $M\oplus N$ is depth $0$, $I$ consists entirely of zerodivisors on $M\oplus N$ and thus is contained in one of the associated primes of $M\oplus N$ by prime avoidance. But then since $\text{Ass}(M\oplus N) = \text{Ass}(M) \cup \text{Ass}(N)$, we have $I$ is in an associated prime of either $M$ or $N$, and thus consists of only zerodivisors on either $M$ or $N$, so one of them must have depth $0$.

I'm a little worried about the validity of only considering the depth $0$ case.

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    Do you know the Koszul complex? – 2017-02-09
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    @Bernard I know of it, but is it necessary to use in this proof? – 2017-02-09
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    I don't think it's necessary, but it makes the assertion almost obvious. – 2017-02-09
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    @Bernard The subject of the question is "proof verification". Alternative methods are welcome, but only after answering what the OP wants to know. (I also suggest the Ext characterization of grade/depth.) – 2017-02-09
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    I'll look at both the Koszul complex and using Ext. Thanks! – 2017-02-09

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