PUMaC 2023 · 加试 · 第 5 题
PUMaC 2023 — Power Round — Problem 5
题目详情
Problem 5.2.2. Show that Z /n Z is cyclic for every positive integer n. Problem 5.2.3. Show that Z [ x ] , the set of polynomials with coefficients in Z , is not a finitely generated abelian group under addition. (Hint: suppose you had a finite subset of Z [ x ] . What elements can be written as a sum of these elements?) Similarly to rings and modules, we have the notion of homomorphisms and isomorphisms of abelian groups, as follows. Definition 5.2.5. A group homomorphism is a map ϕ : G → G between groups G 1 2 1 and G such that for all g, h ∈ G, we have that ϕ ( g ) + ϕ ( h ) = ϕ ( g + h ) , where the first 2 addition is in group G and the second addition is in the group G . 2 1 An isomorphism is a bijective homomorphism. Example. As a simple example, we show that two cyclic groups that have the same finite order are isomorphic. To do this, suppose that our cyclic groups are G and G . Suppose that g is a generator 1 2 1 of G and g a generator of G . Let ϕ : G → G be the map that sends the sum of g with 1 2 2 1 2 1 itself n times to the sum of g with itself n times (for instance, ϕ ( g + g ) = g + g ). 2 1 1 2 2 Since g generates G , this specifies where ϕ sends every element of G . Furthermore, if 1 1 1 G , G both have order m, notice that g , g have order m as well. Indeed, if we consider 1 2 1 2 the sequence g , g + g , g + g + g , . . . , we will get a sequence where every element of G 1 1 1 1 1 1 1 appears, and that is periodic with period the order of g . But it will also need to be periodic 1 with the order of G , meaning that g has order m. The same logic applies for g . 1 1 2 In particular, this means that if the sum of g with itself k times equals the sum of g 1 1 with itself l times, then l − k is divisible by m. But then the sum of g with itself k times 2 equals the sum of g with itself l times (so are equal). This means that ϕ is well-defined. 2 Finally, if h is the sum of g with itself x times, then ϕ ( h + h ) equals the sum of g x 1 x y 2 with itself x + y times. But ϕ ( h ) equals the sum of g with itself x times, and ϕ ( h ) equals x 2 y the sum of g with itself y times, so ϕ ( h )+ ϕ ( h ) = ϕ ( h + h ) , and ϕ is our homomorphism. 2 x y x y We can also see that this map is bijective, since it an inverse, which we can construct using the same procedure as above but with G , G swapped. Thus, G , G are isomorphic. 1 2 1 2 In order to apply the PID Structure Theorem, we need to find an appropriate module structure.
解析
Problem 5.2.4. Show that every abelian group is a Z − module, where one of the operations is + . What is the scalar multiplication? Solution. Let G be an abelian group. We claim that this is a Z − module with scalar multiplication given by n · g = g + g + · · · + g, with n g s, if n > 0 , and n · g = − g + ( − g ) + · · · + ( − g ) , with − g the additive inverse of g and there being − n ( − g )s. Finally, we let 0 · g = 0 . We check the properties of being a Z − module. Properties 1, 3, 4, 5 follow from the fact that G is an abelian group, and the associativity of + also follows. For property 6 , this follows by definition of our module structure, since 1 · g = g. For the associativity of · , we first note that ( − n ) · g = − 1 · ( n · g ) = n · ( − 1 · g ) if n > 0 . Indeed, note that ( − n ) · g is equal to − g + ( − g ) + · · · + ( − g ) , where there are n − g s. Meanwhile, the right-hand side is equal to the additive inverse of n · g = g + g + · · · + g. However, ( n · g ) + ( − n · g ) = g + g + · · · + g + ( − g ) + ( − g ) + · · · + ( − g ) , where there are n g s and n ( − g )s. Each of these cancel out, and so ( n · g ) = ( − 1) · ( n · g ) . Finally, by definition, ( − n ) · g is equal to n · ( − 1 · g ) , which is the sum of n − 1 · g = ( − g )s. From here, note that for n = 0 that both sides are equal to 0 , and for n < 0 , we have that ( − 1) · (( − 1) · (( − n ) · g )) is equal to ( − n ) · g (since the two ( − 1)s corresponding to taking additive inverse and then applying additive inverse again, which cancel each other out). Thus, by the n > 0 case, this is also equal to ( − 1) · ( n · g ) , which is what we wanted. Again, note that ( − n ) · g is the sum of − n g s, which is also what n · ( − 1 · g ) is, as the additive inverse of the additive inverse of g is g. From here, given m, n ∈ Z , let s , s be the sign of m, n, respectively. Then, note that m · m n ( n · g ) = s · ( | m | · ( s · ( | n | · g ))) . But by the above, this is equal to s · ( s · ( | m | · ( | n | · g ))) . m n m n Furthermore, | m | · ( | n | · g ) is just the sum of | mn | g s (by having | m | sums of | n | g s), so this equals s · ( s · ( | mn | · g )) = s s | mn | · g = mn · g. This proves associativity, and hence property 2. m n m n For property 7, we note that ( − 1) · ( g + g ) is the additive inverse of g + g . But this is then equal 1 2 1 2 to ( − g ) + ( − g ) , since if the additive inverse is e, then e = e + g + g + ( − g ) + ( − g ) = − g + − g . 1 2 1 2 1 2 1 2 From here, we note that if s is the sign of n, then n · ( g + g ) = s · ( | n | · ( g + g )) = s · ( | n | · n 1 2 n 1 2 n g + | n | · g ) = n · g + n · g . 1 2 1 2 Furthermore, given m, n, we note that if both m, n are positive, then ( m + n ) · g = m · g + n · g, and the above argument also covers the case if m, n are both negative (by factoring out the − 1 first). If m + n > 0 but one of them is nonpositive, suppose without loss of generality that m > 0 , n ≤ 0 . Then, ( m + n ) · g is equal to the sum of m + n g s. But this is equal to m g s plus | n | ( − g )s, or m − | n | g s. The above observation, giving the distributivity of ( − 1) , gives us the case for when m + n < 0 as well, which proves property 7 and thus shows that G can be given a Z − module structure with this notion of scalar multiplication.