- Explain why any finite Abelian group is a module over the integers. Explain why every finite Abelian group is the direct sum of cyclic subgroups.
- Let R be a commutative ring which has 1. Show it is a field if and only if the only ideals are
and.
- Let R be a commutative ring and let I be an ideal of R. Let
Show that N

is an ideal containing I. Also show that N= N. This is called the radical of I. Hint: If x^{n}∈ I for some n, what of yx? Is it true that^{n}∈ I? For the second part, it is clear that N⊇ N. If x ∈ N, then x^{n}∈ Nfor some n. Is it true that^{k}= x^{kn}∈ I for some k? - Let F be a field and p∈ F. Consider R ≡ F∕. Let
Show that N is an ideal and that it equals

if and only if pis not divisible by the square of any polynomial. Hint: Say N =. Then by assumption, if q^{n}∈, we must have qa multiple of p. Say p= ∏_{i=1}^{m}p_{i}^{ki}. Argue that a contradiction results if any k_{i}> 1 by replacing k_{i}in the product with l_{i},1 ≤ l_{i}< k_{i}. This is q. Explain why q^{n}∈but q. Therefore, p= ∏_{i=1}^{m}p_{i}, the p_{i}distinct noninvertible primes. Explain why this precludes having pdivisible by q^{2}. Conversely, if pis divisible by q^{2}for some polynomial, then p= ∏_{i=1}^{m}p_{i}^{ki}where some k_{i}> 1 and so N≠. Explain. - Recall that if pis irreducible in F, then F∕is a field. Show that if pis not irreducible, then this quotient ring is not even an integral domain.
- Find a polynomial pof degree 3 which is irreducible over ℤ
_{3}. Thus is a prime in ℤ_{3}. Now consider ℤ_{3}∕. Recall that this is a field. How many elements are in this field? Hint: Show that all elements of this field are of the form a_{0}+ a_{1}x + a_{2}x^{2}+where there are three choices for each a_{i}. - Let F be a field and consider F∕where pis prime in F(irreducible). Then this was shown to be a field. Let α be a root of pin a field G which contains F. Consider Fwhich means all polynomials in α having coefficients in F. Show that this is isomorphic to F∕and is itself a field. Hint: Let θ≡ k. Show that since pis irreducible, it has the smallest possible degree out of all polynomials qfor which q= 0 . You might use this to show that θ is one to one. Show that θ is actually an isomorphism. Thus Fmust be a field.
- Letting ℤ be the integers, show that ℤis an integral domain. Is this a principal ideal domain? Show that in ℤ, if you have f,ggiven, then there exists an integer b such that
where the degree of R

is less than the degree of g. Note how in a field, you don’t need to multiply by some integer b. Hint: Concerning the question about whether this is a p.i.d.,suppose you have an ideal in ℤcalled I. Say l∈ I and lhas smallest degree out of all polynomials in I. Let k∈ I. Then in ℚ, you have r,q∈ ℚsuch thatbut all the degrees are the same. Let the result on the left be. Then it is in I. Obtain a contradiction if r≠0. - Now consider the polynomials x
^{3}+ x and 2x^{2}+ 1. Show that you cannot writewhere the degree of r

is less than the degree of 2 x^{2}+ 1 and both r,qare in ℤ. Thus, even though ℤis a p.i.d. the degree will not serve to make ℤinto a Euclidean domain. - The symbol Fdenotes the polynomials in the indeterminates x
_{1},,x_{n}which have coefficients in the field F. Thus a typical element of Fis of the formwhere this sum is taken over all lists of nonnegative integers i

_{1},,i_{n}and only finitely many a_{i1,⋅⋅⋅ ,in}∈ F are nonzero. Explain why this is a commutative ring which has 1. Also explain why this cannot be a principal ideal ring in the case that n > 1. Hint: Consider Fand show that, denoting all polynomials of the form ax + by where a,b ∈ F cannot be obtained as a principal ideal. - Suppose you have a commutative ring R and an ideal I. Suppose you have a morphism h : R →whereis another ring and that I ⊆ ker. Also let
Here R∕I is as described earlier in the chapter. The entries are of the form r + I where r ∈ R.

such that h = θ ∘ fHint: It is clear that f is well defined and onto and is a morphism. Define θ

≡ h. Now show it is well defined. - The Gaussian integers ℤare complex numbers of the form m + in where m,n are integers. Show that this is also the same as polynomials with integer coefficients in powers of i which explains the notation. Show this is an integral domain. Reviewing the Definition 9.1.1 show that the Gaussian integers are also a Euclidean domain if δ≡ a . Hint: For the last part, if you have a,b ∈ ℤ, then as in [20]
where μ,ν ∈ ℚ the rational numbers. There are integers u,l such that

≤ 1∕2,≤ 1∕2. Then letting ε ≡ μ − u,η ≡ λ − l, it follows thatThus

It follows that r ∈ ℤ

. Why? Now considerVerify that δ

< δ. Explain why r is maybe not unique. - This, and the next several problems give a more abstract treatment of the cyclic decomposition
theorem found in Birkhoff and McClain. [6]. Suppose B is a submodule of A a torsion module over a
p.i.d.D. Show that A∕B is also a module over D, and that annequalsfor some α ∈ D. Now suppose that A∕B is cyclic
Say

= ann. Now define I ≡. Explain why this is an ideal which is at least as large as. Let it equal. Explain why α∕β. In particular, α≠0. Then let θ : D → B ⊕D be given byHere we write B ⊕D to indicate B ×D but with a summation and multiplication by a scalar defined as follows:

+≡,β≡. Show that this mapping is one to one and a morphism meaning that it preserves the operation of addition, θ= θδ + θλ. Why is αλa_{0}∈ B? - ↑In the above problem, define η : B ⊕ D → A as η≡ b + αa
_{0}. Show that η maps onto A, that η ∘ θ = 0, and that θ= ker. People write the result of these two problems asand it is called a “short exact sequence”. It is exact because θ

= ker. - ↑Let C be a cyclic module with ann=and μA = 0. Also, as above, let A be a module with B a submodule such that A∕B is cyclic, A∕B ≡ D,ann=, C = Dc
_{0}. From the above problem, ν≠0. Suppose there is a morphism τ : B → C. Then it is desired to extend this to σ : A → C as illustrated in the following picture where in the picture, i is the inclusion map.You need to define σ. First explain why μ = νλ for some λ. Next explain why νa

_{0}∈ B and then why this implies λτ= 0. Let τ= βc_{0}. The β exists because of the assumption that τ maps to C = Dc_{0}. Then since λτ= 0 , it follows that λβc_{0}= 0. Explain why λβ = μδ = νλδ. Then explain whyThus

From the above problem,

where θ

≡,η≡ b + κa_{0}and the above is a short exact sequence. So now extend τ to τ^{′}: B ⊕ D → C as follows.The significance of this definition is as follows.

⊆ ker. But also, θ= kerfrom the above problem. Defineas

This is well defined because if

−∈ θ, then this difference is in kerand so τ^{′}= τ^{′}. Now consider the following diagram.where

be given by

≡ η≡ b + κa_{0}. Then choose σ to make the diagram commute. That is, σ = i ∘^{′}∘^{−1}. Verify this gives the desired extension. It is well defined. You just need to check that it agrees with τ on B. IfNow recall that it was shown above that τ

^{′}extends τ. This proves the existence of the desired extension σ.This proves the following lemma: Let C = Dc

_{0}be a cyclic module over D and let B be a submodule of A, also over D. Suppose A = spanso that A∕B is cyclic and suppose μA = 0 where= annand letbe ann. Suppose that τ : B → C is a morphism. Then there is an extension σ : A → C such that σ is also a morphism. - ↑Let A be a Noetherian torsion module, let B be a sub module, and let τ : B → C be a morphism.
Here C is a cyclic module C = Dc
_{0}where ann=and μ= 0. Then there exists a morphism σ : A → C which extends τ. Note that here you are not given that A∕B is cyclic but A is a Noetherian torsion module. Hint: Explain why there are finitely manysuch that A = span. Now explain why if A_{k}= spanthen each of these is a module and A_{k}∕A_{k−1}is cyclic. Now apply the above result of the previous problem to get a succession of extensions. - ↑Suppose C is a cyclic sub module of A, a Noetherian torsion module and = ann. Also suppose that μA = 0. Then there exists a morphism σ : A → C such that
To so this, use the above propositon to get the following diagram to commute.

Then consider a = σa +

which is something in C added to something else. Explain why σ^{2}a = σ= σ. Thus the second term is in ker. Next suppose you have b ∈ kerand c ∈ C. Then say c + b = 0. Thus σc = c = 0 when σ is done to both sides. This gives a condition under which C is a direct summand. - ↑ Let M be a non zero torsion module for a p.i.d. D and suppose that
so that it is finitely generated. Show that it is the direct sum of cyclic submodules. To do this, pick a

_{1}≠0 and consider Da_{1}a cyclic submodule. Then you use the above problem to writefor a morphism σ. Let M

_{2}≡ kerσ. It is also a torsion module for D. If it is not zero, do the same thing for it that was just done. The process must end because it was shown above in the chapter that M is Noetherian, Proposition 9.5.4. - The companion matrix of the polynomial q= x
^{n}+ a_{n−1}x^{n−1}++ a_{1}x + a_{0}isShow that the characteristic polynomial, det

of C is equal to x^{n}+ a_{n−1}x^{n−1}++ a_{1}x + a_{0}. Hint: You need to takeTo do this, use induction and expand along the top row.

- Letting C be the n × n matrix as in the above where qis also given there, explain why
Thus e

_{k}= C^{k−1}e_{1}. Also explain why