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I'm taking a course on commutative algebra and we learn this theorem:
Every finitely generated module M over Dedekind domain A is direct sum of projective module P and torsion module T. T is direct sum of modules of the form $A/p^i$ where p is a prime ideal in A.

My questions (from HW): Find the composition as stated above for this cases:
1) Automorphism group of $C_{15}$ (cyclic group of order 15) as a module over $\mathbb{Z}$
2) The module $M=\mathbb{R}[x,y]/ (x^2 + y^2 -1)$ of real function $g(x,y)$ on the circle such that every point $r=(x,y)$ on the circle the vector $(g(r),0)$ tangent to the circle at point r.

I tried to solve (1). Denote by $M$ the module mentioned above, we know $M$ isomorphic to Euler group of order 15, which is $U_{15}=\{1,2,4,7,8,11,13,14\}$. Let $m \in M$, $0 \ne z\in\mathbb{Z}$, and we'll check when $z\cdot m=0$ to find the torsion. Here I got stuck. I don't understand how to continue from here, and how to find $P$ and $T$.

As for the second question, I'm helpless and don't know where to start.

Thanks in advance for any help.

Noy Rahmani
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  • $U_{15} = (\mathbb{Z}/ 15 \mathbb{Z})* \cong (\mathbb{Z}/ 3 \mathbb{Z})^* \times (\mathbb{Z}/ 5 \mathbb{Z})^*$ using the Chinese Remainder Theorem. Do you know how to continue from here ($3$ and $5$ are both prime!)? – M. Wang May 29 '20 at 15:13
  • no. I know that, but it seems I can't connect the dots. I want to say that, because prime number define prime ideals we get $\mathbb{Z}/(3)$ $\bigoplus$ $\mathbb{Z}/(5) = T$? and the rest is $P$? not sure how to explain this – Noy Rahmani May 29 '20 at 15:22
  • $(\mathbb{Z}/ 3 \mathbb{Z})^* \cong \mathbb{Z}/ 2 \mathbb{Z}$ because $3$ is prime. Similarly, $(\mathbb{Z}/ 5 \mathbb{Z})^* \cong \mathbb{Z}/ 4 \mathbb{Z} = \mathbb{Z}/ (2^2) \mathbb{Z}$. Does this answer the question? – M. Wang May 29 '20 at 15:25
  • no I can't see it. What you said is $M\cong\mathbb{Z}/ 2 \mathbb{Z} \times (\mathbb{Z}/ 2^2\mathbb{Z})$, How is it the composition? – Noy Rahmani May 29 '20 at 15:34
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    In this case it doesn't matter whether you take the direct sum or the cartesian product. So what I said is $M \cong \mathbb{Z}/2 \mathbb{Z} \oplus \mathbb{Z}/(2^2) \mathbb{Z}$. So in this case the Dedekind domain is $\mathbb{Z}$. The projective module $P$ is $0$. The torsion module $T$ is all of $M$. And it is indeed a direct sum of modules of the form $A/p^i$ with $p$ prime because the ideal $p = (2)$ is prime in $\mathbb{Z}$. $M = A / p^1 \oplus A / p^2 = T $ with $A = \mathbb{Z}$, $p = (2)$. – M. Wang May 29 '20 at 15:38
  • thank you! I understand the second part, but why is the direct sum equal to the cartesian product? – Noy Rahmani May 29 '20 at 15:42
  • that's always the case when you sum together finitely many modules. See: https://math.stackexchange.com/questions/39895/the-direct-sum-oplus-versus-the-cartesian-product-times – M. Wang May 29 '20 at 15:43
  • could you share a reference for this theorem? – Math_Day Nov 21 '24 at 21:35

0 Answers0