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I am a student of a masters course and this question was asked in my quiz of commutative algebra.

Question: Prove that $\mathbb{Z}/ 3\mathbb{Z}$ is a projective $\mathbb{Z} / 6\mathbb{Z}$ module which is not free.

By web surfing I found 2 similar questions here:Showing that $\mathbb{Z}/3\mathbb{Z}$ is a projective $\mathbb{Z}/12\mathbb{Z}$-module but not a free $\mathbb{Z}/12\mathbb{Z}$-module and here :Prove that every $\mathbb{Z}/6\mathbb{Z}$-module is projective and injective. Find a $\mathbb{Z}/4\mathbb{Z}$-module that is neither.

But unfortunately I am not able to use these 2 questions to answer my question ( maybe my concepts are lacking). I read my notes again and found out some relevant results here:

(1) Every free module $F$ over a ring $R$ with $1$ is projective.

(2) Let $R$ be a ring. Then the following conditions on an $R-$ module P are equivalent:

(1) $P$ is projective. (2) Every short exact sequence $0\to A\to B \to P\to 0$ is split exact .

(3) there is a free module $F$ and an $R-module K$ such that $F\approx K \oplus P$.

But I am not sure how to use these to prove what is asked in question ( unfortunately,I am not able to prove either of the question asked).

Can you please tell how to prove the question not just hints?

I really want to learn how to solve this problem.

3 Answers3

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The fact that $\mathbb{Z}/3\mathbb{Z}$ is not a free $\mathbb{Z}/6\mathbb{Z}$-module is obvious, as any finite free $\mathbb{Z}/6\mathbb{Z}$-module has cardinality a multiple of $6$.

Why is it projective? Well, use the definition. Take any $\mathbb{Z}/6\mathbb{Z}$-module homomorphism $f \colon \mathbb{Z}/3\mathbb{Z} \to M$ and any surjective $\mathbb{Z}/6\mathbb{Z}$-module homomorphism $g \colon N \to M$. We want to see that there exists a $\mathbb{Z}/6\mathbb{Z}$-module homomorphism $h \colon \mathbb{Z}/3\mathbb{Z} \to N$ such that $g \circ h = f$.

Consider $1 \in \mathbb{Z}/3\mathbb{Z}$, and $m = f(1) \in M$:

  • If $m=0$, then take $h$ to be the zero homomorphism and you are done.
  • $m$ cannot have order $2$ in $M$.
  • If $m$ has order $3$ in $M$, take any $n \in N$ such that $g(n) = m$ (which exists, by surjectivity of $g$). Now, the order of $n$ in $N$ is either $3$ or $6$. If it is $3$, define $h(1) = n$ and extend. If it is $6$, then define $h(1) = -2n$ and extend.
Diego Artacho
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  • Just a quick question: Is there a particular reason why $h(1)=-2n$ and not $h(1)=2n$? – Prism Mar 16 '23 at 13:11
  • Yes, because you need $g \circ h = f$. If you take $h(1)=2n$, you get $g(h(1))=g(2n)=2g(n)=2f(1)=f(2 \cdot 1) = f(-1) = -f(1)$. – Diego Artacho Mar 16 '23 at 13:40
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    Thank you very much for the response. That makes sense! – Prism Mar 16 '23 at 14:20
  • @DiegoArtacho "The fact that Z/3Z is not a free Z/6Z -module is obvious, as any finite free Z/6Z-module has cardinality a multiple of 6. " It is not at all obvious to me. "as any finite free Z/6Z-module has cardinality a multiple of 6." Which result you used to deduce this? –  Mar 16 '23 at 18:37
  • Let $R$ be a commutative ring with unit. Any free $R$-module is isomorphic to a direct sum $\oplus_{i \in I} R$, for some index set $I$. In particular, if $R$ is finite and the free module is finite and non-trivial, the cardinality of the direct sum of $n$ copies of $R$ is $n \cdot card(R)$. – Diego Artacho Mar 16 '23 at 20:12
  • @DiegoArtacho I don't understand why free $\mathbb{Z}_6$ have $6n$ elements, why not $6^n$, can you give me an example of free $\mathbb{Z}_6$ module which has $12$ elements? – MGIO May 30 '24 at 09:06
  • @MGIO Thanks for your comment, you are obviously right. In my previous comment, I should change $n \cdot card(R)$ to $card(R)^n$. – Diego Artacho May 30 '24 at 10:25
  • @DiegoArtacho Thank your answer! I saw many answers written $6n$, so I thought I was wrong, and I thought about it for a long time, but I still don't know where I made a mistake, so I ask you. Thank you so much! – MGIO May 30 '24 at 10:43
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In addition to Diego's answer, another approach is to use the fact that a module $P$ is projective $R$-module if it is a direct summand of a free $R$-module $F$ (i.e. $P$ is an $R$-submodule of $F$ and there exists a "complementary" $R$-submodule $Q$ of $F$ such that $F = P \oplus Q$).

In your example, $R = \mathbb Z / 6\mathbb Z$. Consider the $F = \mathbb Z / 6\mathbb Z$, which is clearly a free module over $R$. Define $P = \{ 0, 2, 4 \}$ and $Q = \{0, 3 \}$. $P$ and $Q$ are both $R$-submodules of $F$, and $F = P \oplus Q$. Thus, by the criterion in the previous paragraph, $P$ is projective. Now observe that $P$ is isomorphic to $\mathbb Z / 3\mathbb Z$ as an $R$-module.

Kenny Wong
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We have $\mathbb{Z}/6\mathbb{Z} \cong \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/3\mathbb{Z}$ as rings. In $\mathbb{Z}/6\mathbb{Z}$, ideal generated by $\overline{2}$ (resp. $\overline{3}$) is isomorphic to $\mathbb{Z}/3\mathbb{Z}$ (resp. $\mathbb{Z}/2\mathbb{Z}$). Try finding an "easy" ring isomorphism.
(We know that for a ring $R$ (lets say commutative with unity), an ideal $I$ of $R$ can be considered an $R$-module.)
Verify that the "easy" ring isomorphism is infact an $\mathbb{Z}/6\mathbb{Z}$-module isomorphism. So, $\mathbb{Z}/3\mathbb{Z}$ (as an $\mathbb{Z}/6\mathbb{Z}$-module) is a direct summand of a free $\mathbb{Z}/6\mathbb{Z}$-module (namely $\mathbb{Z}/6\mathbb{Z}$ itself). This is one of the equivalent criteria for a projective module (see here). It is clear that $\mathbb{Z}/3\mathbb{Z}$ is not a free $\mathbb{Z}/6\mathbb{Z}$-module.
$\therefore$ $\mathbb{Z}/3\mathbb{Z}$ is projective (yet not free) $\mathbb{Z}/6\mathbb{Z}$-module.

Ajin Shaji Jose
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