PUMaC 2023 · 加试 · 第 4 题
PUMaC 2023 — Power Round — Problem 4
题目详情
- Every element of G has an inverse with respect to + . Written out, for each g ∈ G, there exists an element h ∈ G such that g + h = 0 . A subgroup of a group G is a subset H of G that is a group under the same operation of H. Note that, by definition, every ring is also an abelian group, by discarding the · operation. Example. For instance, note that Z /n Z is an abelian group for any positive integer n, using the remark above, as well as Z / 2 Z × Z / 3 Z , under the addition operation. Note that Z is also an abelian group, with the set of even integers forming a subgroup. We also have the following useful notions. Definition 5.2.2. The order of an element g ∈ G is the smallest positive integer n so that g + g + . . . + g = 0 , if it exists. If it doesn’t, we say that g has infinite order. | {z } n times The order of a group G is the number of elements in the group.
解析
Problem 4.3.3. Suppose that the only nonzero entries of A are along the diagonal (that is, a ̸ = 0 i,j implies that i = j ). Show that M is then isomorphic to a module of the form k M r R ⊕ R/ ⟨ d ⟩ , i i =1 and describe how you obtain the values d and r. i n Solution. Suppose that e , e , . . . , e are the standard basis for R ; that is e is the tuple with the 1 2 n i only nonzero entry being a 1 in the i th coordinate. Then, since A only has nonzero entries along the diagonal, it follows that the kernel of ϕ is generated by a e , a e , . . . , a e . Let d = a . 1 , 1 1 2 , 2 2 m,m m i i,i L m n n − m Consider the map ψ : R → ( R/ ⟨ d ⟩ ) ⊕ R by sending the tuple ( r , r , . . . , r ) to i 1 2 n i =1 ( r , r , . . . , r , r , r , . . . , r ) , where r is the image of r under the quotient map R → R/ ⟨ d ⟩ . 1 2 m m +1 m +2 n i i i One can quickly verify that this is a surjective homomorphism: indeed, given ( r , r , . . . , r ) and 1 2 n ′ ′ ′ n ( r , r , . . . , r ) ∈ R , we have 1 2 n ′ ′ ′ ′ ′ ′ ′ ′ ψ (( r , r , . . . , r ) + ( r , r , . . . , r )) = ( r + r , r + r , . . . , r + r , r + r , . . . , r + r ) , 1 2 n 1 2 m m +1 n 1 2 n 1 2 m m +1 n which is equal to ′ ′ ′ ′ ′ ′ ′ ′ ( r , r , . . . , r , r , . . . , r )+( r , r , . . . , r , r , . . . , r ) = ψ (( r , r , . . . , r ))+ ψ (( r , r , . . . , r )) , 1 2 m m +1 n 1 2 n m m +1 n 1 2 n 1 2 and similarly for any r ∈ R, we have ψ ( r ( r , r , . . . , r )) = ( rr , rr , . . . , rr , rr , . . . , rr ) = rψ (( r , r , . . . , r )) . 1 2 n 1 2 m m +1 n 1 2 n L m n − m For surjectivity, given ( r , r , . . . , r , r , r , . . . , r ) ∈ ( R/ ⟨ d ⟩ ) ⊕ R , there exist r 1 2 m m +1 m +2 n i i i =1 for i = 1 , 2 , . . . , m so the image of r in R/ ⟨ d ⟩ is r , so then ( r , r , . . . , r ) maps to our above tuple. i i i 1 2 n Notice that the kernel of this homomorphism is precisely the set of tuples ( r , r , . . . , r ) where 1 2 n r ∈ ⟨ d ⟩ for i = 1 , 2 , . . . , m, and are 0 for the others. However, such tuples are precisely those i i m P of the form a d e , which is precisely the kernel of ϕ, as noted above. Therefore, we have the i i i i =1 isomorphisms ! m M n − m n n R/ ⟨ d ⟩ ⊕ R ≃ R / ker ψ = R / ker ϕ ≃ M. i i =1 Finally, noting that R/ ⟨ 0 ⟩ ≃ R (since the identity map on R has kernel 0 and is surjective) and that L k r ′ the map permuting the tuples is an isomorphism, it follows that M ≃ R ⊕ R/ ⟨ d ⟩ , where i i =1 ′ k is the number of nonzero d , the d are precisely the nonzero diagonal entries in A (with some i i ordering), and r = n − k.