Can you tell me if my proof is right: Z / mZ and Z / nZ are both finite free Z -modules with the basis consisting of one single element {1}. So (Z / mZ) ⊗Z(Z / nZ) has the basis {1 ⊗ 1}. Therefore, any element in (Z / mZ) ⊗Z(Z / nZ) is of the form (ab)1 ⊗ 1 and any element in Z / gcd (m, n)Z is of the form k1 = k where k ∈ {0, …, gcd (n, m)}. I would like to construct an ...

Prove that Zm ×Zn ≅Zmn Z m × Z n ≅ Z m n implies gcd(m, n) = 1 gcd (m, n) = 1. This is the converse of the Chinese remainder theorem in abstract algebra. Any help would be appreciated. Thanks!

Finally Hom(Zn,Zm) =Zm[n] Hom (Z n, Z m) = Z m [n], i.e., the subgroup of elements of order dividng n n in a finite cylic group of order m m. I leave it to you to identify this subgroup explicitly.

Mar 27, 2017 · Use the exact same projective resolution and tensor with Z / nZ instead of taking hom.. then use the fact that Z / mZ ⊗ZZ / nZ = Z / dZ where d = gcd (m, n) (and if you weren't aware of this fact, try to prove it).

This is the answer for the "opposite direction". All tensor products and homs are over Z Z. The basic argument is based on the adjointness of Hom and tensor ( Adjointness of Hom and Tensor). For abelian groups (aka Z Z -modules) we get Hom(M ⊗ N, Q) ≅ Hom(M, Hom(N, Q)) H o m (M ⊗ N, Q) ≅ H o m (M, H o m (N, Q)). This is pretty clear from the universal property of tensor product: maps ...

Nov 7, 2020 · Note that the homomorphism could be zero. If the homomorphism is non zero, then as you remarked, everything is determined by where 1 1 goes. 1 1 can go to any of the generators of Zm Z m. Your confusion stems from the fact that you are confusing notations. For a group written in "multiplicative notation" 1G 1 G means the identity. For abelian groups like Zn Z n we use additive notation, and ...

I face the problem of finding how many non-trivial ring or group homomorphisms there are from Zm Z m to Zn Z n, where m <n m <n. Is there any general formula? At the moment, I want to know how many ring homomorphisms there are when m = 12, n = 28 m = 12, n = 28. Please help.

This amounts to counting all solutions to the congruence 2x = 0 2 x = 0 in Z/n. Clearly as gcd(2, n)|0 g c d (2, n) | 0 so solutions always exist and they will be gcd (2,n) (which may be either 1 or 2, depending upon whether n is odd or even repectively) in number. So Hom (Z/2,Z/n) will have either 1 or 2 elements.

Nov 28, 2017 · In general, the group Zmn Z m n under addition mod mn mod m n is isomorphic to the direct product Zm ×Zn Z m × Z n if and only if gcd(m, n) = 1. gcd (m, n) = 1.

May 15, 2014 · How to show that Zmn Z m n is isomorphic to Zm ×Zn Z m × Z n when m m and n n are coprime? It is easy to show that the natural map from Zmn Z m n to Zm ×Zn Z m × Z n is a ring homomorphism.