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 $$\frac{a}{2}=a⋅k_0+\sum_{i=1}^na_i⋅k_i,$$for some $k_i∈ℤ$ and $a_i∈ T$. Then $$a=a⋅(2k_0)+\sum_{i=1}^na_i⋅(2k_i),$$ that is, $$a⋅m=\sum_{i=1}^na_i⋅(2k_i),$$ where $m=1-2k_0≠0$ as $k_0∈ℤ$.
$\frac{a}{m}∈⟨S⟩⇒\frac{a}{m}=a⋅r_0+\sum_{i=1}^lb_i⋅r_i$ for some $b_i∈ T$, $r_i∈ℤ$, thus $$a=a⋅mr_0+\sum_{i=1}^lb_i⋅mr_i=\sum_{i=1}^na_i⋅r_0(2k_i) +\sum_{i=1}^lb_i⋅mr_i.$$$⇒T=ℚ$, contradicting minimality of S.

Ⅲ.2. Let R be the ring $ℤ[\sqrt{-5}]$ and I be the ideal $⟨2,1+\sqrt{-5}⟩$. Show that$$Ψ:I⊕I→R⊕R;(x, y)↦(x, y)\small\left(\begin{smallmatrix}\frac{1-\sqrt{-5}}{2} & 1 \\ 1 &\small\frac{1+\sqrt{-5}}{2}\end{smallmatrix}\right)$$is a well-defined R-linear isomorphism. Is there an R-linear isomorphism from I to R?

\begin{matrix}Ψ(2,0)=\left(1-\sqrt{-5},2\right)&Ψ\left(1+\sqrt{-5},0\right)=\left(3,1+\sqrt{-5}\right)\\Ψ(0,2)=\left(2,1+\sqrt{-5}\right)&Ψ\left(0,1+\sqrt{-5}\right)=\left(1+\sqrt{-5},-2+\sqrt{-5}\right)\end{matrix}all ∈ R ⊕ R, so Ψ is well-defined.
Clearly Ψ is R-linear. det Ψ ≠ 0, so Ψ is injective. \[Ψ(1+\sqrt{-5},-2)=(1,0) Ψ(-2,1-\sqrt{-5})=(0,1)\] so Ψ is surjective, so Ψ is an isomorphism.

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$R$-module $R$ is cyclic, $R$-module $I$ is not cyclic, so $I$ and $R$ 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 $A$ be a R-submodule of a finitely generated R-module $B$. Since $B$ is finitely generated, there exists a surjective linear map $f:R^n→B$. Since $A$ is a submodule of $B$, there exists a surjective linear map $g:B→A$, then $g∘f:R^n→A$ is a surjective linear map, so $A$ 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 $α^n+r_1α^{n-1}+⋯+r_n=0$ for some $r_1,⋯,r_n∈R$, then $R[α]=⟨1,α,⋯,α^{n-1}⟩$ as proved in II.1.
Conversely, if $ℤ[α]=⟨x_1,⋯,x_n⟩$, there exists $A∈M_{n,n}(ℤ)$ such that\[(x_1,x_2,⋯,x_n)⋅α=(x_1,x_2,⋯,x_n)⋅A.\]so α is a root of the monic polynomial $\det(xI-A)$. ■

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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 $\{x_i\}_{i=1}^n$, ℤ[α]-module ℤ[α][β] is generated by $\{y_j\}_{j=1}^m$, then ℤ-module ℤ[α][β] is generated by $\{x_iy_j\}_{i,j}$ 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 $e_1,e_2,⋯,e_n$ be the standard basis of Rⁿ, since φ is surjective, there are $a_1,⋯,a_n∈R^m$ such that $φ(a_i)=ψ(e_i)$. Define $A=\pmatrix{a_1\\⋮\\a_n}$, then $φ(e_iA)=φ(a_i)=ψ(e_i)$, so φ(xA) = ψ(x) for all x ∈ Rⁿ.

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If ψ is also surjective, $∀y_0∈R^m:φ(y_0)∈M⇒∃x_0∈R^n:ψ(x_0)=φ(y_0)⇒φ(y_0-x_0A)=φ(y_0)-φ(x_0A)=ψ(x_0)-ψ(x_0)=0⇒y_0-x_0A∈\kerφ⇒Ψ(y_0-x_0A)=y_0-x_0A+R^nA=y_0+R^nA⇒Ψ$ is surjective.
For $y∈\kerφ$, if $y∈\kerΨ$ then $y∈R^nA⇒∃x∈R^n:y=xA⇒0=φ(y)=φ(xA)=ψ(x)⇒x∈\kerψ⇒y∈${vA : v ∈ ker ψ}. Therefore ker Ψ = {vA : v ∈ ker ψ}.

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Suppose R^m is generated by $\{y_i\}$, then R^m/RⁿA is generated by $\{y_i+R^nA\}$, so it is finitely generated.

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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.

$R/I=⟨1_R+I⟩$ so it is cyclic.
The quotient map $φ:R→R/I$ is $R$-linear and surjective.
Suppose $R/I$ is finitely presented, $∃m,n∈ℕ$ and $R$-linear, surjective map $ψ:R^m→R/I$ with $\kerψ=\operatorname{im}A$ for some linear map $A:R^n→R^m$, so $\kerψ$ is finitely generated, by Q5, $\kerφ=I$ is finitely generated as well, contradicting our assumption.

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

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 $2r+Xs=0$, then $X|2r$, since $X$ is prime in ℤ[X], we have $X|r⇒r=Xt$ for some $t∈ℤ[X]$, plugging in $2r+Xs=0$, we get $s=-2t⇒(r,s)=tA⇒\kerψ=ℤ[X]A$. 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 $\bigoplus_{i=1}^nN_i$ where $N_i$ 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 $α:ℚ[X]^2→ℚ[X]$ interweave the coefficients: $α(p,q)=p(x^2)+q(x^2)x$. It's easy to check α is a group homomorphism.

For $p(x)=\sum_{n=0}^∞a_nx^n$, let $β(x)=\sum_{n=0}^∞a_{2n}x^n,γ(x)=\sum_{n=0}^∞a_{2n+1}x^n$, it's easy to check that $p↦(β(p),γ(p))$ is the inverse of α.

It's easy to check\[E^2→E;(φ,ψ)↦(p↦φ(β(p))+ψ(γ(p)))\]is a well-defined homomorphism.

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To prove it's injective, suppose $(φ,ψ)↦0$, then $∀p:φ(β(p))+ψ(γ(p))=0$
Taking $p=x^0,x^1,⋯$ we have $0=φ(x^0)=ψ(x^0)=φ(x^1)=ψ(x^1)=⋯$, so $φ=ψ=0$.

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For all $r∈E$, \((r φ,r ψ)↦r φ(β(p))+r ψ(γ(p))\). So, it's \(E\)-linear.
To prove it's surjective, for any $f∈E$, let \[ϕ=\left(p↦\frac{1}{2}f(α(p,0))\right),ψ=\left(p↦\frac{1}{2}f(α(0,p))\right)\]Then $(ϕ, ψ)↦f$.