PUMaC 2023 · 加试 · 第 2 题
PUMaC 2023 — Power Round — Problem 2
题目详情
- Show that M/ ker ϕ is isomorphic to im ϕ. 4 The PID Structure Theorem Although we’ve seen that modules in general can be quite complex, lacking the simplicity of vector spaces (in the sense that they aren’t generally classified by a single number), modules over PIDs are the next best case. Having built up the necessary ring and module theory over the previous sections, we are now ready to state and prove the main theorem of this power round! Theorem 4.0.1. Let R be a PID, and M be a finitely generated R − module. Then, there are positive integers k, r and nonzero elements d , d , . . . , d of R such that M is isomorphic 1 2 k to k M r R ⊕ R/ ⟨ d ⟩ . i i =1 Before we proceed onto the proof, we first need to understand some more important properties of PIDs and modules of PIDs. These properties arise in a more general class of rings, which are called Noetherian rings. The notion of being Noetherian will also be extended to modules. One of the main benefits of having this concept of Noetherian modules is that it de- scribes a certain “finiteness” condition that the module satisfies. Such a finiteness condition will allow us to conclude that certain algorithm stops, or that certain constructions (in par- ticular, having a finite set of generators) are possible, which we will find useful in the proof of the PID Structure Theorem. As such, before proceeding to the proof of the theorem, we begin by discussing Noetherian modules in general. 4.1 Noetherian Rings and Modules We first begin by introducing the notion of a Noetherian ring. Definition 4.1.1. A Noetherian ring is a ring R satisfying the following property: suppose we have a chain of ideals I ⊆ I ⊆ I ⊆ . . . . Then, there exists some n such that I = I 1 2 3 m n for all m ≥ n. We first consider some examples of a Noetherian ring. Example. For instance, Z is a Noetherian ring. Indeed, given any ideals I ⊆ I ⊆ I . . . , 1 2 3 notice that, by Problem 1.2.3, each ideal is of the form ⟨ i ⟩ for some i. Say that I = ⟨ i ⟩ . j j Then, observe that for I ⊆ I , we need i ∈ I , or that i is divisible by i . j j +1 j j +1 j j +1 We finally observe that if no such m exists, then we can pick a subset of the ideals I , I , . . . that forms a strictly increasing chain, namely where I ⊊ I ⊊ I ⊊ . . . . To do j j j j j 1 2 1 2 3 this, one can proceed inductively: pick one ideal to start with, I . Then, by assumption, j 1 not all the ideals are equal to I ; let one such ideal be I . We can repeat this logic for I j j j 1 2 2 to get some I , and so on. j 3 But then we have an infinite sequence of integers i , i , i , . . . , where i /i is an j j j j j 1 2 3 k k +1 integer. Furthermore, since our ideal inclusions are proper, this integer cannot be 1 or − 1 , as otherwise the ideals would be equal (since − 1 ∈ Z , and an integer is a multiple of x if and only if it is a multiple of − x ). In particular, we have that | i | > | i | for each k. j j k k +1 But then | i | , | i | , | i | , . . . is an infinite sequence of decreasing positive integers, which is j j j 1 2 3 impossible. Therefore, Z is a Noetherian ring. We also have the following equivalent way of defining a Noetherian ring, as follows, which is more reminiscent of the definition of a PID.
解析
- We show that the set of polynomials with complex coefficients form a ring, by checking each of n P i the conditions. For the first condition, for any polynomials f, g ∈ C [ x ] , we can write f ( x ) = f x i i =0 m P i and g ( x ) = g x , where f , g ∈ C . We may furthermore suppose that n = m, by adding more i i i i =0 n P i i terms of the form 0 x to these polynomials. Then, we observe that f ( x ) + g ( x ) = ( f + g ) x and i i i =0 n n n i 2 n n 2 n X X X X X X X X i + j i i i f ( x ) · g ( x ) = f g x = ( f g ) x + ( f g ) x = f g x i j j i − j j i − j j i − j i =0 j =0 i =0 j =0 i = n +1 j = i − n i =0 0 ≤ j ≤ n 0 ≤ i − j ≤ n which are both polynomials with complex coefficients. We could have stopped at the expression n n P P i + j f g x , but it is useful to have the explicit formula for the coefficients written out. i j i =0 j =0 1 n n P P i i For associativity, suppose we are given three polynomials f ( x ) = f x , g ( x ) = g x , h ( x ) = i i i =0 i =0 n P i h x (again, by the above argument, we can suppose that the sums are over the same range by i i =0 i adding additional 0 x terms). Then, n n n X X X i i i ( f ( x ) + g ( x )) + h ( x ) = ( f + g ) x + h x = (( f + g ) + h ) x i i i i i i i =0 i =0 i =0 n n n X X X i i i = ( f + ( g + h )) x = f x + ( g + h ) x i i i i i i i =0 i =0 i =0 = f ( x ) + ( g ( x ) + h ( x )) . Similarly, we can check that 2 n n X X X i i ( f ( x ) · g ( x )) · h ( x ) = f g x · h x j i − j i i =0 0 ≤ j ≤ n i =0 0 ≤ i − j ≤ n 3 n X X X i = ( f g ) h x k j − k i − j i =0 0 ≤ j ≤ 2 n 0 ≤ k ≤ n 0 ≤ i − j ≤ n 0 ≤ j − k ≤ n 3 n X X X i = f g h x k j − k i − j i =0 0 ≤ j ≤ 2 n 0 ≤ k ≤ n 0 ≤ i − j ≤ n 0 ≤ j − k ≤ n 3 n X X i = f g h x , k j − k i − j i =0 0 ≤ k ≤ n 0 ≤ i − j ≤ n 0 ≤ j − k ≤ n the last equality coming from the fact that 0 ≤ k, j − k ≤ n implies that 0 ≤ j ≤ 2 n. But then we can rewrite this as 3 n 3 n X X X X X X i i f g h x = f ( g h ) x k j − k i − j k j − k i − j i =0 0 ≤ k ≤ n 0 ≤ i − j ≤ n i =0 0 ≤ k ≤ n 0 ≤ i − j ≤ n 0 ≤ i − k ≤ 2 n 0 ≤ j − k ≤ n 0 ≤ i − k ≤ 2 n 0 ≤ j − k ≤ n n 2 n X X X i i = f x · ( g h ) x = f ( x ) · ( g ( x ) · h ( x )) . i i − j j i =0 i =0 0 ≤ j ≤ n 0 ≤ i − j ≤ n This gives us associativity. 2 Next, for commutativity, we can quickly check that, given any polynomials f, g, which we can write in the form above, that n n X X i i f ( x ) + g ( x ) = ( f + g ) x = ( g + f ) x = g ( x ) + f ( x ) i i i i i =0 i =0 and 2 n 2 n X X X X i i f ( x ) · g ( x ) = f g x = g f x = g ( x ) · f ( x ) , j i − j j i − j i =0 0 ≤ j ≤ n i =0 0 ≤ j ≤ n 0 ≤ i − j ≤ n 0 ≤ i − j ≤ n where the second-to-last equality arises from replacing the index j with i − j. Using these formulas, we can check that the polynomial 0 is the additive identity, as n n n X X X i i i 0 + f x = (0 + f ) x = f x , i i i i =0 i =0 i =0 2 n P P i and 1 is the multiplicative identity, as 1 · f ( x ) = a f x , where a is 1 for j i − j j i =0 0 ≤ j ≤ n 0 ≤ i − j ≤ n n P i j = 0 and 0 otherwise. But then this is just f x . i i =0 n n P P i i For additive inverse, we note that given f ( x ) = f x , the polynomial ( − f ) x is the additive i i i =0 i =0 n P i inverse, as adding f ( x ) to this yields ( f − f ) x = 0 . Finally, for the distributive law, we note i i i =0 that n 2 n X X X i i f ( x ) · ( g ( x ) + h ( x )) = f ( x ) · ( g + h ) x = f ( g + h ) x i i j i − j i − j i =0 i =0 0 ≤ j ≤ n 0 ≤ i − j ≤ n 2 n 2 n X X X X i i = f g ) x + f h ) x j i − j j i − j i =0 0 ≤ j ≤ n i =0 0 ≤ j ≤ n 0 ≤ i − j ≤ n 0 ≤ i − j ≤ n = f ( x ) · g ( x ) + f ( x ) · h ( x ) . Having verified all of the conditions for the set of polynomials with complex coefficients, we thus see that this forms a ring with the typical addition and multiplication operations.