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Submodule of a free module over commutative ring with 1 is not free. For this statement, can $Z_n$ be the counter example for it?

When $n=8$, $\{0,2,4,6\}$ is an ideal of $Z_8$ and is not free.

My second question is as follows:

We can consider commutative ring $R$ with unity as a $R$-module. Then, it is finitely generated by the unity. For case of integral domain $D$, the unity generates $D$ and it is independent because of no zero divisor, so it is free.

Is it true?

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    $\begingroup$ Yes to the first part, but make sure to justify why the submodule (ideal) is not free. You will probably need to reformulate the second part, which appears unclear at the moment. Split in in two separate questions, if this is the case. $\endgroup$ – Andreas Caranti Feb 17 '17 at 22:05
  • $\begingroup$ I editted, so can you check again? $\endgroup$ – kswim Feb 18 '17 at 2:08
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Your counterexample is good. Any nonzero free $Z_8$-module must contain at least eight elements.

I don't understand the second part.

Edit. In answer to the second part as it's been rewritten, given any commutative ring $R$ with a unit element $1$, we have that $R$ is a free $R$-module. It isn't necessary to assume that $R$ is an integral domain.

As you said, $R$ is generated as a module by the element $1$. But the family $\{1\}$ is also linearly independent, because $a \cdot 1 = 0$ implies that $a = 0$. Therefore the family $\{1\}$ is a basis.

In an integral domain, for any $b \ne 0$, the family $\{b\}$ is linearly independent. This is not true in other rings with unit. However, when $b = 1$, it is always true.

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  • $\begingroup$ I editted. Can you see it again? $\endgroup$ – kswim Feb 18 '17 at 2:09
  • $\begingroup$ I've answered your second part. $\endgroup$ – user49640 Feb 18 '17 at 2:48

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