PUMaC 2023 · 加试 · 第 3 题

Problem 3.0.1. Show that for any ring R and ideal I of R, I is an R − module under the addition and multiplication operations of the ring R. Definition 3.0.2. Given an R − module M, a linear combination of m , m , . . . , m ∈ M 1 2 n n P is an expression of the form r m , where r ∈ R. i i i i =1 A generating set of a module M is a set S such that every element can be written as a linear combination of some finite subset of S. We say that a module is finitely generated if it has a finite generating set. A set of elements of M is linearly independent if the only linear combination of these elements that equals zero is the linear combination where all of the coefficients r are 0 . i A set of elements of M is said to be a free basis of M if it is linearly independent and a generating set. In this case, if such a free basis exists, we say that M is a free module . Its rank is then the length of this free basis. These definitions should be reminiscent of definitions of linear independence and span from our discussions of linear algebra. We need to explicitly point out when our modules are free for the following reason. √ √ Example. Consider the ideal I = ⟨ 2 , 1 + − 5 ⟩ inside the ring R = Z [ − 5] , the set of √ integers of the form a + b − 5 for some integers a, b. Notice that this ideal is an R − module by Problem 3.0.1. We show that this is not a free module. To see this, suppose for the sake of contradiction that { r , r , . . . , r } was a free basis for 1 2 k I. If k ≥ 2 , then note that by linear independence that none of the elements can be zero. But then ( − r ) r + r r = 0 , meaning that this set is not linearly independent, contradiction. 2 1 1 2 Therefore, I would have to have a free basis with one element, say r . But then there 1 √ exist elements s , s ∈ R such that s r = 2 and s r = 1 + − 5 . But suppose that 1 2 1 1 2 1 √ √ √ s = a + a − 5 and r = b + b − 5 , then s r = ( a b − 5 a b ) + ( a b + a b ) − 5 . 1 1 2 1 1 2 1 1 1 1 2 2 1 2 2 1 For this to equal 2 , we need a b = − a b . Note however that multiplying this by ( a − 1 2 2 1 1 √ √ √ a − 5)( b − b − 5) , which equals ( a b − 5 a b ) − ( a b + a b ) − 5 = 2 , yields that 2 1 2 1 1 2 2 1 2 2 1 2 2 2 2 ( a + 5 a )( b + 5 b ) = 4 . For this to hold, as the a are integers, we need one pair of ( a , a ) i 1 2 1 2 1 2 to be ( ± 1 , 0) and the other to be ( ± 2 , 0) . However, note that r cannot be ± 2 , since that 1 √ √ 1+ − 5 implies that s = ± ∈ / Z [ − 5] , contradiction. 2 2 √ √ But if r = ± 1 , this means that 1 ∈ ⟨ 2 , 1 + − 5 ⟩ , meaning that there exist a + b − 5 1 √ √ √ √ √ and c + d − 5 in Z [ − 5] so 2( a + b − 5) + (1 + − 5)( c + d − 5) = 1 , or that (2 a + c − 5 d ) + √ (2 b + c + d ) − 5 = 1 . But c − 5 d would have to be odd and c + d even, which is impossible. Hence, no r can exist, and therefore I must not be a free module, which is what we 1 wanted to show. Observe that if we are given a free module, the following property from linear algebra does carry over to the module case. You may assume that this proposition holds without proof. Proposition 3.0.3. Suppose that F is a free module over a nonzero ring R that is finitely generated. Then, any two free bases of F have the same length. Similarly to the vector space case, we can also consider maps between modules, in the following way. Definition 3.0.4. A module homomorphism between R − modules M and N is a map ϕ : ′ ′ ′ M → N such that for all m, m ∈ M and r ∈ R, we have that ϕ ( m + m ) = ϕ ( m )+ ϕ ( m ) , M N and ϕ ( r · m ) = r · ϕ ( m ) . M N Definition 3.0.5. A module homomorphism is said to be an isomorphism if it is injective and surjective. Two modules are then isomorphic if there exists an isomorphism between them. We also have the kernel and the image, defined similarly to the vector space case. Definition 3.0.6. Let ker ϕ = { m ∈ M | ϕ ( m ) = 0 } , and im ϕ = { ϕ ( m ) | m ∈ M } . Just like before, we will omit which objects are being mapped if it is clear from context what objects we are mapping. Similarly to the vector space case, we can verify that the ker- nel and image of a module homomorphism are both modules; you may use this throughout the rest of the power round without proof. n