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Dummit & Foote explains the difference between "free module over a set $A$" and "direct sum" and uses the following example:

One should be careful to note the difference between the uniqueness property of direct sums, and the uniqueness property of free modules. Namely, in the direct sum of two modules, say $N_1 \oplus N_2$, each element can be writen uniquely as $n_1+n_2$ (how?); here the uniqueness refers to the module elements $n_1$ and $n_2$...... For example, if $R = \mathbb{Z}$, and $N_1 = N_2 = \mathbb{Z}_2$, then each element of $N_1 \oplus N_2$ has a unique representation in the form $n_1 + n_1$, where $n_1,n_2\in \mathbb{Z}_2$...

I really don't see why it is true. I suspect that the author is referring to the isomorphism between $\pi:N_1 \times N_2 \longrightarrow N_1 + N_2$ defined as $\pi(a_1,a_2) = a_1 + a_2$. But obviously this is not an isomorphism since $\pi(0,0) = \pi(1,1) = 0$.

Did I misunderstand anything?

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    $\begingroup$ There's some abuse of notation here. $n_1$ is presumably supposed to refer the respective element of $N_1 \oplus \{0\} \subset N_1 \oplus N_2$, and $n_2$ is the respective element of $\{0\} \oplus N_2 \subset N_1 \oplus N_2$. $\endgroup$ – Dustan Levenstein Nov 15 '16 at 5:07
  • $\begingroup$ @DustanLevenstein That makes more sense now. But still, in this case, $N_1 \oplus N_2$ is not isomorhpic to $N_1 + N_2$, right? Please see my edit. Thanks! $\endgroup$ – 3x89g2 Nov 15 '16 at 5:16
  • $\begingroup$ No, they are not isomorphic. $\endgroup$ – Dustan Levenstein Nov 15 '16 at 5:17
  • $\begingroup$ @DustanLevenstein Then I really don't know what point the author is trying to make here :( So he's trying to say that $N_1 \oplus N_2$ is not free on any set, and $N_1 + N_2$ is not $N_1 \oplus N_2$? $\endgroup$ – 3x89g2 Nov 15 '16 at 5:21
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    $\begingroup$ The author never made any mention of $N_1 + N_2$. That was all you. $\endgroup$ – Dustan Levenstein Nov 15 '16 at 5:22
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The direct sum $N_1\oplus N_2$ is not literally elements of $N_1$ added to elements of $N_2$: $N_1$ and $N_2$ need not both be submodules of the same module, in which case it doesn't make sense to add an element from $N_1$ to an element of $N_2$. We define $$ N_1\oplus N_2 := N_1\times N_2 $$ as a set, and give it the addition $(n,m) + (n',m') = (n + n', m + m')$ and $R$-module structure $r(n,m) = (rn,rm)$. I.e., $N_1\oplus N_2\cong N_1\times N_2$ as $R$-modules (the reason we have two different notation for the ``same thing" is that the definitions become more complicated when you look at the direct sum or direct product of infinitely many modules, and when you have infinitely many of them they will generally not be the same anymore). The isomorphism $\pi : N_1\times N_2\to N_1\oplus N_2$ is not quite right: it should send $(n_1,n_2)$ to $(n_1,n_2) = (n_1,0) + (0,n_2)$ (you can view $N_1$ and $N_2$ inside of $N_1\oplus N_2$ as $N_1\oplus \{0\}$ and $\{0\}\oplus N_2$).

If $N_1$ and $N_2$ are both submodules of a third module $M$, we may write $N_1 + N_2$ to mean the actual sums of elements of $N_1$ and $N_2$ within $M$; i.e., $$ N_1 + N_2 = \{n_1 + n_2\in M\mid n_1\in N_1, n_2\in N_2\}\subseteq M. $$ Again, if $N_1$ and $N_2$ are not considered as submodules of the same module, we may not write $N_1 + N_2$, as adding elements of two arbitrary modules doesn't make sense. If $N_1 + N_2\cong N_1\oplus N_2$ (i.e., if the homomorphism $\varpi : N_1\times N_2\to N_1 + N_2$ given by $\varpi(n_1,n_2) = n_1 + n_2$ is an isomorphism, which you've noticed it might not be), we might simply write $N_1+N_2 = N_1\oplus N_2$, and we say that $N_1 + N_2$ is an internal direct sum of $N_1$ and $N_2$. This happens if and only if every element of $N_1 + N_2$ can be written uniquely as $n_1 + n_2$ with $n_1\in N_1$ and $n_2\in N_2$. But in general, $N_1 + N_2\not\cong N_1\oplus N_2$. As you saw in your example: if you consider $\Bbb Z/2\Bbb Z$ as a submodule of itself, then $$ \Bbb Z/2\Bbb Z + \Bbb Z/2\Bbb Z = \Bbb Z/2\Bbb Z, $$ but on the other hand, $$ \Bbb Z/2\Bbb Z \oplus \Bbb Z/2\Bbb Z = \Bbb Z/2\Bbb Z^2. $$

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