Atiyah-MacDonald Ch. 4 exercise 20: what's the module analogue of $\sqrt{\mathfrak{a}+\mathfrak{b}} = \sqrt{\sqrt{\mathfrak{a}}+\sqrt{\mathfrak{b}}}$? Atiyah-MacDonald exercises 20-23 in chapter 4 develop a theory of primary decomposition for modules, in analogy with the theory developed in the chapter for rings. Exercise 20 begins with this definition:
Definition: Given a (commutative, unital) ring $A$ and an $A$-module $M$, and a submodule $N\subset M$, the radical of $N$ in $M$ is
$$r_M(N) = \sqrt{\operatorname{Ann} M/N}$$
It then asks us to prove analogues to the formulas in exercise 1.13 for the radical of an ideal. Formula 1.13(v) is
$$\sqrt{\mathfrak{a}+\mathfrak{b}} = \sqrt{\sqrt{\mathfrak{a}}+\sqrt{\mathfrak{b}}}$$
This is true by taking radicals in the pair of inclusions $\mathfrak{a}+\mathfrak{b}\subset \sqrt{\mathfrak{a}}+\sqrt{\mathfrak{b}}$ and $\sqrt{\mathfrak{a}+\mathfrak{b}} \supset \sqrt{\mathfrak{a}}+\sqrt{\mathfrak{b}}$, the first of which is totally obvious and the second of which is because if $x^k\in\mathfrak{a}$ and $y^\ell\in\mathfrak{b}$ then $(x+y)^{k+\ell}\in\mathfrak{a}+\mathfrak{b}$.
It seems to me that the analogous formula in the module setting is 
$$r_M(N+N') = \sqrt{r_M(N)+r_M(N')}$$
While the inclusion $r_M(N+N') \supset r_M(N)+r_M(N')$ is true here for more or less the same reason as $\sqrt{\mathfrak{a}+\mathfrak{b}} \supset \sqrt{\mathfrak{a}}+\sqrt{\mathfrak{b}}$, and taking radicals gives $r_M(N+N')\supset \sqrt{r_M(N)+r_M(N')}$, the other inclusion has been confounding me and leading me to wonder if this is not the right generalization of formula 1.13(v). So, question:

Is this the right analogue of 1.13(v)? If so what's the proof of the inclusion $r_M(N+N') \subset \sqrt{r_M(N)+r_M(N')}$? If not, what's the right analogue?

Thoughts: if this is the right analogue, and the proof is also analogous, the desired inclusion should follow by taking radicals in the inclusion
$$\operatorname{Ann}\frac{M}{N+N'} \subset r_M(N)+r_M(N')$$
But I can't think of a reason for this. If $x\in A$ puts $M$ inside $N+N'$, to prove this statement I'd need to decompose it as something $y+y'$ where a power of $y$ puts $M$ inside $N$ and a power of $y'$ puts $M$ inside $N'$. How would I even start to seek such a decomposition?
 A: I don't think that your equation is true. Here is a different suggestion.
Let $M_1,M_2$ be finitely generated $A$-modules. Let $N_1$ (resp. $N_2$) be a submodule of $M_1$ (resp. $M_2$). Let $N$ denote the image of $N_1 \otimes M_2 \,\oplus\, M_1 \otimes N_2 \to M_1 \otimes M_2$. Hence, $(M_1 \otimes M_2)/N = M_1/N_1 \otimes M_2/N_2$. Using the Lemma below, it follows
$$\sqrt{\mathrm{Ann}((M_1 \otimes M_2)/N)} = \sqrt{\mathrm{Ann}(M_1/N_1 \otimes M_2/N_2)} = \sqrt{\mathrm{Ann}(M_1/N_1) + \mathrm{Ann}(M_2/N_2)}$$
i.e.
$$r_{M_1 \otimes M_2}(N) = \sqrt{r_{M_1}(N_1) + r_{M_2}(N_2)}.$$
Notice that for $M_1=M_2=A$ this gives back the case of ideals.

Lemma: If $M_1,M_2$ are finitely generated $A$-modules, then $$\sqrt{\mathrm{Ann}(M_1 \otimes M_2)}=\sqrt{\mathrm{Ann}(M_1)+\mathrm{Ann}(M_2)}.$$

Proof: From Nakayama one deduces that $\mathrm{supp}(M_1 \otimes M_2) = \mathrm{supp}(M_1) \cap \mathrm{supp}(M_2)$. Using $\mathrm{supp}(M)=V(\mathrm{Ann}(M))$ for finitely generated $M$, it follows that $$V(\mathrm{Ann}(M_1 \otimes M_2)) = V(\mathrm{Ann}(M_1)+\mathrm{Ann}(M_2)),$$
hence the claim.
Added. I think that it is important to assume that $M_1,M_2$ are finitely generated. To illustrate this with another property, consider the property that the radical commutes with localization, $r_{M_f}(N_f)=r_M(N)_f$ for $f \in A$. You will run into difficulties when $M$ is not finitely generated. Perhaps a more natural definition of the radical is the following:
$$r'_M(N) = \{x \in A : \forall m \in M \exists q \geq 0 (x^q m \in N)\} = \bigcap_{U \leq M \text{ f.g.}} r_{U}(N \cap U)$$
Then $r'$ commutes with localization.
