Rings and modules problem sheet 3

Ⅲ.1. Show that the ℤ-module ℚ does not have a minimal generating set.

Let S be a minimal generating set of ℚ. Take a ∈ S. Let T ≔ ⟨S ∖ {a}⟩. We have for some and . Then that is, where as .
for some , , thus , contradicting minimality of S.

Ⅲ.2. Let R be the ring and I be the ideal . Show thatis a well-defined R-linear isomorphism. Is there an R-linear isomorphism from I to R?

all ∈ R ⊕ R, so Ψ is well-defined.
Clearly Ψ is R-linear. det Ψ ≠ 0, so Ψ is injective. so Ψ is surjective, so Ψ is an isomorphism.

---

-module is cyclic, -module is not cyclic, so and are not isomorphic.

Ⅲ.3. Suppose that R is a PID. Show that any submodule of a finitely generated R-module is finitely generated.

Let be a R-submodule of a finitely generated R-module . Since is finitely generated, there exists a surjective linear map . Since is a submodule of , there exists a surjective linear map , then is a surjective linear map, so is finitely generated.

Ⅲ.4. For α, β ∈ ℤ show that ℤ[α][β] is a finitely generated ℤ-module. Conversely, show that if ℤ[γ] is a finitely generated ℤ-module then γ ∈ ℤ. Deduce that ℤ is a ring.

R is a ring. α is integral over R iff R[α] is a finitely generated R-module.
Proof. If for some , then as proved in II.1.
Conversely, if , there exists such thatso α is a root of the monic polynomial . ■

---

Applying this result to R = ℤ[α], since β is integral over ℤ, it is integral over ℤ[α], we get ℤ[α][β] is a finitely generated ℤ[α]-module.
If ℤ-module ℤ[α] is generated by , ℤ[α]-module ℤ[α][β] is generated by , then ℤ-module ℤ[α][β] is generated by similar to the proof of Thm 2.40(Tower law)

By subring test α − β, αβ ∈ ℤ[α][β] ⊂ ℤ, so ℤ is a ring.

Ⅲ.5. Suppose that φ : R^m → M and ψ : Rⁿ → M are R-linear maps.
Show that if φ is surjective then there is A ∈ M_n,m(R) such that φ(xA) = ψ(x) for all x ∈ Rⁿ.
Show that if ψ is also surjective then the mapΨ : ker φ → R^m/RⁿA;   y ↦ y + RⁿAis surjective and ker Ψ = {vA : v ∈ ker ψ}. Why is R^m/RⁿA finitely generated? Deduce that if ker ψ is finitely generated then ker φ is finitely generated too.

Let be the standard basis of Rⁿ, since φ is surjective, there are such that . Define , then , so φ(xA) = ψ(x) for all x ∈ Rⁿ.

---

If ψ is also surjective, is surjective.
For , if then {vA : v ∈ ker ψ}. Therefore ker Ψ = {vA : v ∈ ker ψ}.

---

Suppose R^m is generated by , then R^m/RⁿA is generated by , so it is finitely generated.

---

If ker ψ is finitely generated, then ker Ψ = {vA : v ∈ ker ψ} is also finitely generated. Since both ker Ψ and im Ψ = R^m/RⁿA are finitely generated, by first isomorphism theorem, ker φ is finitely generated.

Ⅲ.6. Show that if R is a ring and I is an ideal that is not finitely generated then the R-module R/I is cyclic but does not have a finite presentation. Hence give an example of a module that is finitely generated but does not have a finite presentation.

so it is cyclic.
The quotient map is -linear and surjective.
Suppose is finitely presented, and -linear, surjective map with for some linear map , so is finitely generated, by Q5, is finitely generated as well, contradicting our assumption.

---

You could take any non-finitely generated ideal I and the cyclic module R/I as an example:
From II.4, has a non-finitely generated ideal I.
Also, continuous functions with compact support is a non-finitely generated ideal of continuous real functions: for of compact support , is a linear combination of , then the support of is a subset of , and ∃ function with compact support on .

Ⅲ.7. Show that the ℤ[X]-linear map ℤ[X]² → ⟨2, X⟩; (r, s) ↦ 2r + Xs has kernel equal to ℤ[X]A where A = (X, −2) ∈ M_1,2(ℤ[X]). Deduce that ⟨2, X⟩ has a finite presentation.

If , then , since is prime in ℤ[X], we have for some , plugging in , we get . By the First Isomorphism Theorem ⟨2, X⟩ ≅ ℤ[X]² / ℤ[X]A.

Ⅲ.8. Show that ⟨2, X⟩ is not ℤ[X]-linearly isomorphic to a direct sum of cyclic modules.

Suppose that I = ⟨2, X⟩ is ℤ[X]-linearly isomorphic to where are cyclic modules.
By Ex 2.12, I is non-cyclic, so n ≥ 2. By definition of direct sum, I contains a copy of N₁ and N₂ contradicting Prop 3.22(intersection of principal ideals in ℤ[X] contains a non-zero element). Let interweave the coefficients: . It's easy to check α is a group homomorphism.

For , let , it's easy to check that is the inverse of α.

It's easy to checkis a well-defined homomorphism.

---

To prove it's injective, suppose , then
Taking we have , so .

---

For all , . So, it's -linear.
To prove it's surjective, for any , let Then .