Rings and modules problem sheet 4

$\DeclareMathOperator{lcm}{lcm}$

IV.1. Let$$A=\pmatrix{3&0&0\\0&2+2\sqrt{-5}&0\\0&0&4}$$Explain why A is in Smith normal form over ℂ, but is not in Smith normal form over $ℤ[\sqrt{-5}]$. Is A equivalent to a matrix in Smith normal form over $ℤ[\sqrt{-5}]$?

A is diagonal. Since ℂ is a field, $a|b∀a,b∈ℂ^*$, so A is in Smith normal form.
A is not in Smith normal form over $ℤ[\sqrt{-5}]$, since $3∤2+2\sqrt{-5}$.
$\gcd(3,4)=1,\lcm(3,4)=12⇒A$ is equivalent to $$\pmatrix{1&0&0\\0&2+2\sqrt{-5}&0\\0&0&12}$$This is in Smith normal form, since $12=(2+2\sqrt{-5})(1-\sqrt{-5})$.

IV.2. Find elements a₁ ∣ a₂ in ℤ[i], and a ℤ[i]-linear isomorphismℤ[i]/⟨a₁⟩ ⊕ ℤ[i]/⟨a₂⟩ → ℤ[i]/⟨2⟩ ⊕ ℤ[i]/⟨3 − 5i⟩.

Find GCD of $2,3-5i$ in $ℤ[i]$ by the extended Euclidean algorithm: \begin{align*} 3-5i&=(1-3i)2+1+i\\ 2&=(1-i)(1+i) \end{align*} $∴a_1=\gcd(2,3-5i)=1+i,a_2=\lcm(2,3-5i)=2+8i$.
Using $3-5i-(1-3i)2=1+i$ to perform elementary row and column operations \begin{align*}\pmatrix{2\\&3-5i}&\xrightarrow[c_1↦c_1+c_2]{r_2↦r_2-(1-3i)r_1}\pmatrix{2\\1+i&3-5i} \xrightarrow{r_1↔r_2}\pmatrix{1+i&3-5i\\2} \xrightarrow[c_2↦c_2+(1+4i)c_1]{r_2↦r_2-(1-i)r_1}\pmatrix{1+i\\&2+8i} \end{align*}Multiplying elementary matrices $$\pmatrix{1+i\\&2+8i}=\pmatrix{2-2 i & -1 \\-1-4 i & -1+i}\pmatrix{2\\& 3-5 i}\pmatrix{1 & -1-4 i \\1 & -4 i}$$The inverse of the matrix on the right is$$Q=\pmatrix{1 & -1-4 i \\1 & -4 i}^{-1}=\pmatrix{-4i&1+4i\\-1&1}$$The action of $Q$ on $ℤ[i]^2$ is\[\pmatrix{r_1&r_2}Q=\pmatrix{-4 ir_1-r_2,(1+4 i)r_1+r_2}\] By Thm 7.1 the isomorphism is\[(r_1+⟨1+i⟩,r_2+⟨2+8i⟩)↦(-4 ir_1-r_2+⟨2⟩,(1+4 i)r_1+r_2+⟨3-5i⟩)\]

A=ZZ[i]^1
A.span([[2]]).direct_sum(A.span([[3-5*i]]))

IV.3. Show that if A, B ∈ M_n(𝔽) are such that 𝔽ⁿ/𝔽ⁿA is 𝔽-linearly isomorphic to 𝔽ⁿ/𝔽ⁿB, then A and B are equivalent as matrices.

𝔽 is a field, so an 𝔽-module is a vector space,dim 𝔽ⁿ/𝔽ⁿA = dim 𝔽ⁿ/𝔽ⁿB ⇒ n − dim 𝔽ⁿA = n − dim 𝔽ⁿB ⇒ dim 𝔽ⁿA = dim 𝔽ⁿB ⇒ 𝔽ⁿA, 𝔽ⁿB are isomorphic ⇒ A is equivalent to B.

IV.4. Show that if R is the ring C(ℝ) with pointwise addition and multiplication then there are A, B ∈ M₁(R) with A ≁ B, and for which R¹/R¹A is R-linearly isomorphic to R¹/R¹B.

Proved in I.6.

IV.5. Determine the rational canonical form of the matrix$$A=\pmatrix{1&1&0&0\\0&1&0&0\\2&3&-1&4\\1&1&-1&3}$$

To transform a matrix on $ℚ[X]$ to SNF, we multiply it on both sides by invertible matrices.

The inverse operation of adding $f(x)$ times row $i$ to row $j$ is adding $-f(x)$ times row $i$ to row $j$, so $f(x)$ is allowed to be any element
The inverse operation of rescaling $i$th row/column by $f(x)$ is rescaling it by $f(x)^{-1}$, so $f(x)$ is only allowed to be units [$\deg f(x)>0$ disallowed]

Perform the SNF algorithm on $xI-A$

R.<x> = PolynomialRing(QQ)
A = matrix(QQ,4,4,[1,1,0,0,0,1,0,0,2,3,-1,4,1,1,-1,3])
M = x*identity_matrix(4)-A
print(latex(M))
M.swap_rows(0,3)
M.rescale_row(0,-1)
M.add_multiple_of_row(2,0,2)
M.add_multiple_of_row(3,0,-x+1)
M.add_multiple_of_column(1,0,-1)
M.add_multiple_of_column(2,0,1)
M.add_multiple_of_column(3,0,x-3)
print('∼',latex(M))
M.swap_rows(1,2)
M.rescale_row(1,-1)
M.add_multiple_of_row(2,1,-x+1)
M.add_multiple_of_row(3,1,x)
M.add_multiple_of_column(2,1,x-1)
M.add_multiple_of_column(3,1,-2*x+2)
print('∼',latex(M))
M.add_multiple_of_row(3,2,1)
M.add_multiple_of_column(3,2,2)
print('∼',latex(M))
print('We can verify the last invariant factor is equal to',A.minpoly())

\begin{multline*}\left(\begin{array}{rrrr} x - 1 & -1 & 0 & 0 \\ 0 & x - 1 & 0 & 0 \\ -2 & -3 & x + 1 & -4 \\ -1 & -1 & 1 & x - 3 \end{array}\right) ∼ \left(\begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & x - 1 & 0 & 0 \\ 0 & -1 & x - 1 & -2 x + 2 \\ 0 & -x & x - 1 & x^{2} - 4 x + 3 \end{array}\right)\\ ∼ \left(\begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & x^{2} - 2 x + 1 & -2 x^{2} + 4 x - 2 \\ 0 & 0 & -x^{2} + 2 x - 1 & 3 x^{2} - 6 x + 3 \end{array}\right) ∼ \left(\begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & x^{2} - 2 x + 1 & 0 \\ 0 & 0 & 0 & x^{2} - 2 x + 1 \end{array}\right)\end{multline*} We can verify the last invariant factor is equal to $m_A(x)=x^2 - 2x + 1$
$A$ have two invariant factors $x^2-2x+1$ with companion matrix $\pmatrix{ 0 & -1 \\ 1 & 2}$
So the rational canonical form of $A$ is\[\left(\begin{array}{cc|cc} 0 & -1 & 0 & 0 \\ 1 & 2 & 0 & 0 \\ \hline 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 2 \\ \end{array}\right)\]

Computing the Frobenius normal form

IV.6. Suppose that G is a finite commutative group. Show that there is an element x ∈ G such that the order of every y ∈ G divides the order of x. Assuming that every polynomial in 𝔽[X] of degree N has at most N roots show that U(𝔽) is a cyclic group when 𝔽 is a finite field.

By Corollary 7.2 G is isomorphic to $ℤ/⟨d_1⟩⊕⋯⊕ℤ/⟨d_n⟩$ for natural numbers $d_1∣⋯∣d_n$. Let x be the element of G corresponding to $(0,…,0,1)$ then x has the maximum order $d_n$, and the order of any element of G divides $d_n$.

Let |U(𝔽)| = n and x is an element of maximum order d, by Lagrange's theorem $d|n$.
We proved all elements of U(𝔽) have order dividing d, so they are roots of the polynomial $X^d-1$, which have$≤d$ roots in U(𝔽), so $n≤d$, so $n=d$, so U(𝔽) is generated by $x$.

Finite subgroups of the multiplicative group of a field are cyclic
sheet 3 and solutions Exercise 3

IV.7. Suppose that R is a commutative ring and for a₁, …, a_n ∈ Rⁿ_colwrite [a₁, …, a_n] for the matrix in M_n(R) with columns a₁, …, a_n in order. Show that the maps R_colⁿ→ R; a_i ↦ det[a₁, …, a_n] are R-linear for each 1 ≤ i ≤ n; and det[a₁, …, a_n] = 0 whenever a_i = a_j for some i ≠ j. By considering det[Ba₁, …, Ba_n] for B ∈ M_n(R) show that det BA = det B det A.

$$r\det[a_1,…,a_i,…,a_n]+s\det[a_1,…,a_i',…,a_n]=\det[a_1,…,ra_i+sa_i',…,a_n]$$So the maps a_i ↦ det[a₁, …, a_n] are R-linear.

If a = a_i = a_j for some i ≠ j. Since swapping columns negates the determinantdet[…, a, …, a, …] = −det[…, a, …, a, …] So it is zero.

Write $a_i$ as linear combination of basis $e_i$:$$a_i=\sum_{j=1}^na_{j,i}e_j$$In the linear expansion, using $\det[…,e_j,…,e_j,…]=0$, we get \[\left.\begin{aligned}\det[Ba_1,…,Ba_n]&=\det[Ba_{11}e_1,Ba_{22}e_2,…,Ba_{nn}e_n]\\&+\det[Ba_{21}e_2,Ba_{12}e_1,…,Ba_{nn}e_n]\\&+⋯\\&+\det[Ba_{n1}e_n,Ba_{n-1,2}e_{n-1},…,Ba_{1n}e_1]\end{aligned}\right\}n!\text{ terms}\]Let $B=[b_1,…,b_n]$, then $Be_j=b_j$, we get \[\left.\begin{aligned}\det[Ba_1,…,Ba_n]&=\det[a_{11}b_1,a_{22}b_2,…,a_{nn}b_n]\\&+\det[a_{21}b_2,a_{12}b_1,…,a_{nn}b_n]\\&+⋯\\&+\det[a_{n1}b_n,a_{n-1,2}b_{n-1},…,a_{1n}b_1]\end{aligned}\right\}n!\text{ terms}\] Factoring out the constants \[\left.\begin{aligned}\det[Ba_1,…,Ba_n]&=a_{11}a_{22}⋯a_{nn}\det B\\&-a_{21}a_{12}⋯+a_{nn}\det B\\&±⋯\\&+(-1)^{\small\frac{n(n-1)}2}a_{n1}a_{n-1,2}⋯a_{1n}\det B\end{aligned}\right\}n!\text{ terms}\]Factoring out $\det B$ we get $\det BA=\det B\det A$.

IV.8. Show that if$$A:=\pmatrix{2&0\\0&X}∈M_2(ℤ[X])\text{ and }B:=\pmatrix{2&X\\0&2}∈M_2(ℤ[X])$$then A is not equivalent to any matrix in Smith normal form, and B is not equivalent to any diagonal matrix.

Suppose $A∼A'$ in SNF, then $A'_{1,1}$divides both 2 and $X$, so $A'_{1,1}=±1$, but $±1∉⟨2,X⟩$.

Suppose $B∼\pmatrix{p(x)\\&q(x)}⇒p(x),q(x)∈⟨2,X⟩ $.

Lemma: $A∼B$ in $M_2(ℤ[X])$ then $\det A=±\det B$

$p(x)⋅q(x)|\det B=4⇒p(x)=q(x)=±2$ but $X∉⟨2⟩$, contradiction.

IV.9. Suppose that R is a PID and M is a finitely generated R-module. Show that there are elements a₁ ∣ ⋯ ∣ a_r in R such that M is R-linearly isomorphic to R/⟨a₁⟩ ⊕ ⋯ ⊕ R/⟨a_r⟩.

IV.10. Suppose that R is commutative or R is a division ring. Show that if φ : Rⁿ → R^m is an R-linear surjection and m > n, then R is trivial. Deduce that if R is non-trivial then any two bases of a finitely generated R-module have the same size.

PREVIOUSRings and modules problem sheet 3

NEXTTopologist's sine curve