Matrix Roots (12.3.37-39)

Dummit and Foote Abstract Algebra, section 12.3, exercises 37-39:

37. Let $J$ be a Jordon block of size $n$ with eigenvalue $λ∈\mathbb{C}$.
(a) If $λ \neq 0$, prove the Jordan canonical form of $J^2$ is the Jordan block of size $m$ with eigenvalue $λ^2$.
(b) If $λ = 0$, prove the Jordan canonical form of $J^2$ is two Jordan blocks of sizes $\dfrac{m}{2}$, $\dfrac{m}{2}$ if $m$ is even and of sizes $\dfrac{m-1}{2}$, $\dfrac{m+1}{2}$ if $m$ is odd.
38. Determine the necessary and sufficient conditions for a matrix over $\mathbb{C}$ to have a square root.
39. Let $J$ be a Jordan block of size $m$ with eigenvalue $λ$ over a field $F$ with characteristic 2. Determine the Jordan canonical for for the matrix $J^2$. Determine the necessary conditions for a matrix over $F$ to have a square root.

Proof: We consider a more general case - determining when a matrix over $\mathbb{F}$ has an $n^{th}$ root for any natural $n$, first when $n \neq 0$ in $F$, and when $n = 0$ in $F$ and $n$ is prime in $\mathbb{Z}$.

(37)(a) Let $T$ be the linear transformation of $J$ with regard to some basis $e_1,...,e_m$. We shall show$$J^n=\begin{bmatrix}λ^n{n \choose 0} & λ^{n-1}{n \choose 1} & \cdots & {n \choose n}λ^0 & 0 & \cdots \\ 0 & λ^n{n \choose 0} & λ^{n-1}{n \choose 1} & \ddots & 0 & \cdots \\ 0 & 0 & λ^n{n \choose 0} & \ddots & 0 & \cdots \\ \ddots & & & \cdots & \cdots & \cdots \\ & \ddots & & & \cdots & \cdots \end{bmatrix}$$In the language of linear transformations, this is equivalent to showing $T^n$ is the linear transformation acting on the basis by$$T^n(e_k)=\sum_{j=0}^{k-1}λ^{n-j}{n \choose j}e_{k-j}$$This is evident for $n=1$, so we prove the proposition by induction:$$T^n(e_k)=T \circ T^{n-1}(e_k) = T(\sum_{j=0}^{k-1}λ^{n-1-j}{n-1 \choose j}e_{k-j}) =$$$$\sum_{j=0}^{k-1}λ^{n-1-j}{n-1 \choose j}T(e_{k-j})$$We note at this point that $T$ acts on the basis by $T(e_k)=λe_k+e_{k-1}$ if $k > 1$ and $T(e_1)=λe_1$, so we may continue$$\sum_{j=0}^{k-1}λ^{n-1-j}{n-1 \choose j}T(e_{k-j})=$$$$[\sum_{j=0}^{k-2}{n-1 \choose j}(λ^{n-j}e_{k-j}+λ^{n-j-1}e_{k-j-1})]+{n-1 \choose k-1}λ^{(n-1)-(k-1)+1}e_1=$$$$λ^ne_k+[\sum_{j=1}^{k-2}({n-1 \choose j} λ^{n-j}+{n-1 \choose j-1}λ^{n-1-(j-1)})e_{k-j}]+$$$$({n-1 \choose k-1}λ^{n-k+1}+{n-1 \choose k-2}λ^{n-1-(k-2)})e_1=$$$$λ^ne_k+[\sum_{j=1}^{k-2}λ^{n-j}{n \choose j}e_{k-j}]+λ^{n-(k-1)}e_1=\sum_{j=0}^{k-1}λ^{n-j}{n \choose j}e_{k-j}$$And so our claim is proven. Now, since $J^n-λ^n$ is an upper triangular matrix, it is nilpotent and thus $m_J(x) | (x-λ^n)^m$. As well, note the nullity of $T^n-λ^n$ on $V$; we have $(T^n-λ^n)V$ is generated by $λ^{n-j}{n \choose j}e_{m-j}$ for $1≤j≤m-1$, which arises as a linear combination of the linearly independent elements $e_1,...,e_{m-1}$ when $λ \neq 0$, ${n \choose 1} = n \neq 0$ and so $(T^n-λ^n)V$ is of rank $m-1$ and $T^n-λ^n$ has nullity of $1$. Thus there is one Jordan block in the Jordan canonical form of $J$ with eigenvalue $λ^n$ and by the minimal monomial this must be the only block, and hence the Jordan canonical form of $J^n$ for $J$ a Jordan block of size $m$ with eigenvalue $λ \neq 0$ is a Jordan block of size $m$ with eigenvalue $λ^n$.

(b) When $λ=0$, by the above we have $J^n$ is the matrix where the $i,j$ entry is $1$ if $j=i+n$ and all other entries are $0$. Thus $T^n$ acts on the basis by $T(e_k)=0$ if $k ≤ n$ and $T(e_k)=e_{k-n}$ when $k > n$. Write $m = an+b$ where $0 ≤ b < n$. Assume $m < n$ so $a = 0$: Then $J^n=0$ and there are $m=b$ blocks of size $1=a+1$ so the form is of the same pattern as follows when $a ≥ 1$: We may see $r_k = m-nk$ when $k ≤ a$ and $r_k = 0$ when $k > a$, where $r_k = \text{dim}(T^n)^kV$. We count the Jordan blocks by 12.3.30:$$r_{a-1}-2r_a+r_{a+1}=m-n(a-1)-2(m-na)=$$$$m-na-2(m-na)+n=n-b$$$$r_a-2r_{a+1}+r_{a+2}=m-na=b$$Thus there are $n-b$ blocks of size $a$ and $b$ blocks of size $a+1$ (all with eigenvalue $0$, remember). We count their combined dimension over $F$ by $(n-b)a+b(a+1)=na+b=m$ so this describes the Jordan form entirely.

38. We prove a matrix $A$ over $F$ wherein $n \neq 0$ and which contains all the eigenvalues of its matrices (e.g. $\mathbb{C}$) has an $n^{th}$ root if and only if:

1) Each of its nonzero eigenvalues has an $n^{th}$ root in $F$, and
2) It is possible to divide its list of exponents of its invariant factors $x^k$ into regions with generation rules:
     i) Size $m$ containing only $1$s for $m < n$, and
     ii) Size $n$ containing elements which all differ from each other by $1$ or $0$.

Lemma 1: Let $G$ be a ring. An element $g∈G$ has a multplicative $n^{th}$ root in $G$ if and only if each of the elements in its similarity class has an $n^{th}$ root. Proof: ($⇐$) is clear, and ($⇒$) follows by $f^n = g ⇒ (hfh^{-1})^n = hf^nh^{-1}=hgh^{-1}$.$~\square$

Lemma 2: Let $A$ be the block matrix with blocks $A_1,...,A_n$ and let $B$ be the block matrix with blocks $B_1,...,B_n$ where $A_i$ is the same size as $B_i$. Then $AB$ is the block matrix with blocks $A_1B_1,...,A_nB_n$. Proof: Expanding $AB=(A_1'+...+A_n')(B_1'+...+B_n')$ where $A_i'$ is the block matrix in position of a matrix of size $A$, and similar for $B_i'$, by viewing these individual blocks as collapses onto disjoint subspaces followed by a particular linear transformation, we see $A_i'B_j'=0$ for $i \neq j$ and $A_i'B_i'=(A_iB_i)'$. Therefore $(A_1'+...+A_n')(B_1'+...+B_n')=(A_1B_1)'+...+(A_nB_n)'$ is the block matrix described above.$~\square$

By the above lemmas and our knowledge of the Jordan canonical forms of powers of Jordan blocks, the logical equivalence now holds. $A$ has an $n^{th}$ root if and only if there is a matrix for which $B^n$ is of the same Jordan canonical form as $A$ if and only if there is such $B$ that is in Jordan canonical form itself (since $(P^{-1}BP)^n=P^{-1}B^nP$ and $B^n$ are in the same similarity class), and we see $B^n$ has blocks each of which can be separately conjugated into Jordan canonical form giving rise to a block matrix $P$ of conjugating blocks conjugating $B^n$ into its Jordan canonical form as the block sum of the Jordan canonical forms of its component blocks. By the nature of how these canonical forms arise in these blocks by (37), this completes the proof.

39. The second requirement is the same as above but applies to each class of eigenvalues as well as $0$, as when $A_λ$ is the matrix in Jordan canonical form with eigenvalue $λ$ of some fixed size, then $A_λ^n-λ^n=A_0^n$, since when $n$ is prime $n | {n \choose k}$ for all $0 < k < n$.$~\square$

General Canonical Forms (12.3.25-28)

Dummit and Foote Abstract Algebra, section 12.3, exercises 25-28:

25. Determine the Jordan canonical form of the $n \times n$ matrix over $\mathbb{Q}$ whose entries are all equal to $1$.
26. Determine the Jordan canonical form of the $n \times n$ matrix over $\mathbb{F}_p$ whose entries are all equal to $1$.
27. Determine the Jordan canonical form of the $n \times n$ matrix over $\mathbb{Q}$ whose diagonal entries are all equal to $0$ and other entries are all $1$.
28. Determine the Jordan canonical form of the $n \times n$ matrix over $\mathbb{F}_p$ whose diagonal entries are all equal to $0$ and other entries are all $1$.

Proof: When $n=1$ all these matrices are already in Jordan canonical form, so assume $n > 1$.

(25) Let $B=A-n$ and $C=AB$. We see$$c_{ij}=\sum_{k=1}^n a_{ik}b_{kj}=\sum_{k=1}^nb_{kj}=b_{jj}+\sum_{k=1,k \neq j}b_{kj}=1-n+n-1=0$$so now $m_A(x) \mid x(x-n)$ and since $A,B \neq 0$ we have $m_A(x)=x(x-n)$ and $A$ is diagonalizable with $n$s and $0$s.

Assume there are multiple $n$s down the diagonal of the Jordan form of $A$, so there are multiple $x-n$ factors in the invariant decomposition. Letting $1_a,1_b∈\mathbb{Q}^n$ be the $\mathbb{Q}[x]$ generators of these summands we see $x(1_a)$ and $x(1_b)$ are linearly independent over $\mathbb{Q}$. But we see $A \begin{bmatrix}f_1 \\ ... \\ f_n \end{bmatrix}=\begin{bmatrix}\sum f_i \\ ... \\ \sum f_i \end{bmatrix}$ so that all elements under the image of $A$ are either zero or associate to $\begin{bmatrix}1 \\ ... \\ 1 \end{bmatrix}$, a contradiction. Therefore the Jordan canonical form of this matrix is the $n \times n$ matrix with $a_{1,1}=n$ and all other entries $0$.

(26) Assume $p \not \mid n$. Then as before we can conclude $m_A(x)=x(x-n)$ and that there is only one $x-n$ invariant factor and the Jordan canonical form is as above. Assume $p \mid n$. Then $x=x-n$ and since $A \neq 0$ we have $m_A(x)=x^2$. Once again we see that assuming there is more than invariant factor not dividing $x$ leads to linearly independent elements in the image of $A$, so that we may conclude the invariant decomposition is $x,...,x,x^2$ and the Jordan canonical form is simply the matrix with $a_{1,2}=1$ and all other entries $0$.

(27) Letting $A$ be this matrix and $A'$ the matrix of (25), we see $A+1=A'$ and thus $((x+1)-n)(x+1)=(x-n+1)(x+1)=0$. As before we note there is one associativity class in the image of $A'=A+1$ so that there is exactly one invariant factor of $x-n+1$ in the decomposition, and so the Jordan canonical form is the matrix with $a_{1,1}=n-1$, the other diagonal entries $-1$ and the other entries $0$.

(28) When $p \not \mid n$ we have $n \not ≡ 0$ and so $x-n+1 \neq x+1$ and the Jordan canonical form is as above. When $p \mid n$ we again notice there is one invariant factor of $(x+1)^2$ and thus the Jordan canonical form is the matrix with $a_{1,2}=1$, the diagonal entries all $-1$ and the other entries $0$.$~\square$

Diagonalization of Special Matrices (12.3.21-22)

Dummit and Foote Abstract Algebra, section 12.3, exercise 21-22:

21. Let $A$ be a matrix such that $A^2=A$. Show that $A$ can be diagonalized with $1$s and $0$s down the diagonal.

22. Let $A$ be a matrix such that $A^3=A$. Show that $A$ can be diagonalized over $\mathbb{C}$. Is this true over any field $F$?

Proof: (21) Let $f_i(x)^{α_i}$ be the $i^th$ invariant factor in the variant factor decomposition of $V$ over $F[x]$. Let $1$ be viewed as the $F[x]$ generator of this direct summand. Since $x^2(1)=x(1)$ we have $x(x-1)=0$. Since $-1 \neq 0$ and $f_i(x)^{α_i}$ is a power of a single prime power dividing $x(x-1)$ we must have $f_i(x)^{α_i}∈\{x,x-1\}$. Thus in the Jordan form of $A$ it is diagonal with $1$s and $0$s down the diagonal.

(22) Restarting as above we can see $f_i(x)^{α_i}$ divides $x^3-x=x(x-1)(x+1)$. When $F$ has characteristic greater than $2$ these are all distinct prime factors and thus $A$ is diagonalizable with $-1$s, $0$s, and $1$s down the diagonal. When $F$ has characteristic $2$ we note $x-1=x+1$ so $x^3-x=x(x-1)^2$ and thus by choosing the Jordan canonical form of the matrix with invariant factors $(x-1)^2$ we obtain a matrix that is not diagonalizable and is seen to satisfy$$\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}^3=\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}~\square$$

Prime Cyclotomic Polynomials Modulo p (12.2.20)

Dummit and Foote Abstract Algebra, section 12.2, exercise 20:

Let $n$ be prime, and let $Φ_n(x) = \dfrac{x^n-1}{x-1} = x^{n-1}+...+1$ be the $n^{th}$ cyclotomic polynomial, which is irreducible over $\mathbb{Q}$. Fix a prime $p$. This exercise determines when $Φ_n(x)∈\mathbb{F}_p[x]$ is irreducible and thus provides information about the conjugacy classes of matrices of prime order over prime fields.

(a) Let $p = n$. Show $x-1$ divides $Φ_n(x)$.
(b) Let $p \neq n$. Let $f$ be the multiplicative order of $p$ in $\mathbb{F}_n^\times$. Show that $m=f$ is the smallest integer for which $\text{GL}_m(\mathbb{F}_p)$ contains a matrix $A$ of order $n$.
(c) Show that no polynomial in $\mathbb{F}_p[x]$ of degree less than $f$ divides $Φ_n(x)$, and that $m_A(x)$ is an irreducible polynomial of degree $f$ dividing $Φ_n(x)$.
(d) In particular, prove $Φ_n(x)$ is irreducible over $\mathbb{F}_p$ if and only if $p$ is a primitive root modulo $n$, i.e. $p$ generates $\mathbb{F}_n$.

Proof: (a) Since $Φ_n(1)≡0$, $x-1$ divides $Φ_n(x)$.

(b) We have $|GL_m(\mathbb{F}_p)|=(p^m-1)...(p^m-p^{m-1})$. By Lagrange's and Cauchy's theorems, $GL_m(\mathbb{F}_p)$ has a matrix of order $p$ if and only if its order is divisible by $p$, so - since $\mathbb{F}_p$ is an integral domain - if and only if $n \mid p^m-p^k=p^k(p^{m-k}-1)$ for some $0≤k≤m$. Since this first occurs when $m=f$ and $k=0$, the claim is proven.

(c) Assuming $g(x)~|~Φ_n(x)~|~x^n-1$ implies $m_B(x)~|~c_B(x) = g(x)~|~x^n-1$ where $B$ is the companion matrix of $g(x)$. Thus $B^n=1$ and by (b) $g(x)$ is of degree $≥f$. Letting $A$ be as in (b) we see $m_A(x)~|~x^n-1=(x-1)Φ_n(x)$ and by the restriction of polynomials dividing $Φ_n(x)$ and the size of $A$ we have $m_A(x)$ is of degree $f$. Since $m_A(x)$ is a monomial of minimal degree dividing $Φ_n(x)$ it must be irreducible, and since $A \neq 1$ we have $m_A(x) \neq x-1$ and now $m_A(x)~|~Φ_n(x)$.

(d) The case is clear when $p=n$. Otherwise, let $p$ be a primitive root. Then $f=n-1$ and no polynomial of degree smaller than $n-1$ divides $Φ_n(x)$, and thus $Φ_n(x)$ is irreducible. If $Φ_n(x)$ is irreducible, then a polynomial of degree smaller than $n-1$ divides $Φ_n(x)$ and thus the order of $p$ in $\mathbb{F}_n$ is smaller than $n-1$.$~\square$

Matrix Similarity Classes Over Extension Fields of Q (12.2.13)

Dummit and Foote Abstract Algebra, section 12.2, exercise 13:

Show that there are the same number of similarity classes of $3 \times 3$ matrices over $\mathbb{Q}$ for a given characteristic polynomial over $\mathbb{Q}[x]$ as there are for when the entries are over any extension field of $\mathbb{Q}$. Give an example to show this is not true in general for $4 \times 4$ matrices.

Proof: Consider all the cases of the decomposition of $c_A(x)∈\mathbb{Q}[x]$. Let $a \neq b \neq c$.

$(x+a)(x+b)(x+c)$: The only choice for minimal polynomial is $c_A(x)$, and there is one similarity class over $\mathbb{Q}$ and $F$.

$(x+a)^2(x+b)$: Whether viewed over $\mathbb{Q}$ or $F$, there are two similarity classes.

$(x+a)^3$: As before, there are invariably three similarity classes. These are all a result of $x+a$ being irreducible in both $\mathbb{Q}[x]$ and $F[x]$.

$(x^2+ax+b)(x+c)$: There is only one similarity class over $\mathbb{Q}$. If $x^2+ax+b=(x+v_1)(x+v_2)$ for $v_1 \neq v_2$, then there is again only one similarity class. If $x^2+ax+b=(x+v_1)^2=x+2v_1x+v_1^2$, then $2v_1 ∈ \mathbb{Q}$ so $v_1 ∈ \mathbb{Q}$, even though $x^2+ax+b$ doesn't factor and thus doesn't have zeros in $\mathbb{Q}$.

$x^3+ax^2+bx+c$: If this polynomial decomposes in $F[x]$ to three distinct linear factors, or an irreducible quadratic and a linear factor, or doesn't decompose further, then the minimal polynomial remains the same. If it decomposes into $(x+v)^3$, then comparing coefficients of $x^2$ we obtain $3v∈\mathbb{Q}$ so $v∈\mathbb{Q}$. Therefore $x^3+ax^2+bx+c = (x+v_1)^2(x+v_2)$ and we obtain the relations$$2v_1+v_2 = a$$$$v_1^2+2v_1v_2 = b$$$$v_1^2v_2 = c$$We first observe $v_2 = -2v_1 + a$ to manipulate the second equation$$v_1^2 = \dfrac{2}{3}av_1-\dfrac{1}{3}b$$Substituting these both into the third equation yields a rational quadratic expression over $v_1$, so employing the first observation again and rearranging yields$$(6b-2a^2)v_1 = 9c-ab$$Since the original polynomial can't have zeros in $\mathbb{Q}$, $-v_1 \not ∈ \mathbb{Q}$ so $v_1 \not ∈ \mathbb{Q}$ implying $6b-2a^2=0$ implying $b = \dfrac{1}{3}a^2$ and $c = \dfrac{1}{27}a^3$. Now $x^3+ax^2+bx+c=(x+\dfrac{1}{3}a)^3$ decomposes in $\mathbb{Q}[x]$, a contradiction.

Observe the polynomial $x^4+2x^2+1=(x^2+1)^2$ in $\mathbb{Q}[x]$. There are two lists of invariant factors for matrices with this characteristic polynomial, and thus there are two similarity classes over $\mathbb{Q}$. In $\mathbb{C}$ this polynomial decomposes to $(x-i)^2(x+i)^2$, and there are seen to be four lists of invariant factors and thus four similarity classes over $\mathbb{C}$.$~\square$

Complex Computations and Inequalities (1.13-15)

Walter Rudin Principles of Mathematical Analysis, chapter 1, exercises 13-15:

13. For complex $x,y$ show$$||x|-|y||≤|x-y|$$ 14. If $z$ is complex and $|z|=1$, compute$$|1+z|^2+|1-z|^2$$ 15. Under what conditions does equality hold in the Schwarz inequality? I.e.,$$|\sum a_j\overline{b_j}|^2 = \sum |a_j|^2 \sum |b_j|^2$$for complex $a_j,b_j$.

Proof: (13) Assume $|x|≥|y|$. Then we must show $|x|-|y| > |x-y|$ is impossible. Multiply both sides by (positive) $|x|+|y|$ to obtain $|x|^2-|y|^2 > |x-y|(|x|+|y|) ≥ |x-y||x+y| = |x^2-y^2|$ so $|x^2| > |x^2-y^2|+|y^2| ≥ |x^2|$, a contradiction. The case is parallel when $|y|≥|x|$.

(14) Then $z=a+bi$ such that $a^2+b^2=1$. We see $|1-z|=|z-1|$ and now $|z+1|^2+|z-1|^2=(a+1)^2+b^2+(a-1)^2+b^2=2(a^2+b^2)+2=4$.

(15) Let $v_a,v_b∈\mathbb{C}^k$ be the the vectors of the $a_j$ and $b_j$. Let $A = \sum |a_j|^2$,$B = \sum |b_j|^2$, and $C = \sum a_j\overline{b_j}$. We claim $|C|^2=AB$ if and only if $v_b$ is associate to $v_a$ (i.e. $v_b=cv_a$ for some $c∈\mathbb{C}$) or at least one of $v_a$ or $v_b$ is zero. ($⇒$) Assume $|C|^2=AB$ and $v_a,v_b \neq 0$ so $A,B > 0$. As we have seen, $∑|Ba_j-Cb_j|^2 = B(AB-|C|^2)$. When $|C|^2=AB$ we then thus have $a_j=\dfrac{C}{B}b_j$ for all $j$, and thus $v_b = \dfrac{C}{B}v_a$. ($⇐$) When either one of $v_a$ or $v_b$ is zero the equality clearly holds, so assume $v_a,v_b \neq 0$ and $v_b = cv_a$ for some complex $c$. We can thus manipulate$$B=\sum |b_j|^2 = \sum |ca_j|^2 = |c|^2\sum |a_j|^2 = |c|^2A$$and$$\overline{c}A=\overline{c}\sum a_j\overline{a_j} = \sum a_j\overline{b_j} = C$$so for all $j$ we have$$Ba_j = |c|^2Aa_j = \dfrac{|c|^2a_j\overline{a_j}A}{\overline{a_j}} = \dfrac{|ca_j|^2}{\overline{a_j}}A = b_j \dfrac{\overline{b_j}}{\overline{a_j}} A = b_j \overline{c} A = Cb_j$$so that $∑ |Ba_j - Cb_j|^2 = B(|C|^2 - AB) = 0$. Since $v_b \neq 0$ and thus $B > 0$, we have $|C|^2 = AB$.$~\square$

Real Logs (1.7)

Walter Rudin Principles of Mathematical Analysis, chapter 1, exercise 7:

Fix $b > 1, y > 0$ and prove that there is a unique real $x$ such that $b^x = y$ by completing the following outline. Say $x = \text{log}_b y$.
(a) For natural $n$, $b^n - 1 ≥ n(b-1)$.
(b) Hence $b - 1 ≥ n(b^{1/n}-1)$.
(c) If $t > 1$ and $n > (b-1)/(t-1)$ then $b^{1/n} < t$.
(d) If $w$ is such that $b^w < y$ then $b^{w+1/n} < y$ for sufficiently large $n$.
(e) If $b^w > y$ then $b^{w-1/n} > y$ for sufficiently large $n$.
(f) Let $A = \{w~|~w < y\}$ and show $b^{\text{sup }A} = y$.
(g) Show $x$ is unique.

Proof: (a) We show $b^n ≥ n(b-1)+1$. This is clear when $n=1$, so by induction$$b^n = bb^{n-1} ≥ b((n-1)(b-1)+1)) = (n-1)b(b-1)+b$$and we must show $(n-1)b(b-1)+b ≥ n(b-1)+1$. Collecting terms on the left and manipulating we must thus show $(n-1)(b-1)^2 ≥ 0$, which is clear as all these terms are nonnegative.
(b) This is clear from the previous, as $b^{1/n} > 1$ by noting $c^n ≤ 1$ for $c ≤ 1$ by induction.
(c) Assume $b^{1/n} ≥ t$. Then $b - 1 < n(t-1) ≤ n(b^{1/n}-1)$, a contradiction by (b).
(d) We have $1 < \dfrac{y}{b^{w}}$. Observe $b^{1/n} ≤ \dfrac{b-1}{n}+1$ by (b) so we may choose $n$ such that $\dfrac{b-1}{n}+1 < \dfrac{y}{b^{w}}$ by conditioning $\dfrac{b-1}{n} < \dfrac{y}{b^w}-1 > 0$ and now $n > \dfrac{b-1}{\dfrac{y}{b^w}-1}$ so that $b^{w+1/n} < y$.
(e) As before, since we showed $b^{1/n}$ tends toward 1 we can choose $n$ such that $b^{1/n} < \dfrac{b^w}{y}$ to fulfill.
(f) By (a) we have $b^w$ is divergent so $A$ has $x = \text{sup }A$. Suppose $b^x < y$; then by (d) we have some $b^{x+1/n} < y$ so $x$ is not an upper bound. Suppose $b^x > y$; then by (e) choose $n$ for $b^{x-1/n} > y$ and $x$ is not a minimal upper bound. Therefore $b^x = y$.
(g) If $x' \neq x$ then $b^x - b^{x'} = b^{x'}(b^{x-x'}-1) = 0$ is a contradiction.$~\square$

Real Exponents (1.6)

Walter Rudin Principles of Mathematical Analysis, chapter 1, exercise 6:

Fix $b>1$.
(a) Let $m,n,p,q$ be integers, $n,q > 0$ with $m/n=p/q$. Prove$$(b^m)^{1/n}=(b^p)^{1/q}$$Hence it makes sense to define $b^{m/n}=(b^{m})^{1/n}$.
(b) Prove $b^{r+s}=b^rb^s$ for $r,s∈\mathbb{Q}$.
(c) For $x∈\mathbb{R}$, define $B(x) = \{b^t~|~t∈\mathbb{Q},~t≤x\}$. When $r∈\mathbb{Q}$, prove$$b^r=\text{sup }B(r)$$Hence it makes sense to define$$b^x=\text{sup }B(x)$$(d) For $x,y∈\mathbb{R}$, prove $b^{x+y}=b^xb^y$.

Proof: (a) Recall $(z_1z_2)^{1/n}=z_1^{1/n}z_2^{1/n}$ for natural $n$. For natural $n_1,n_2$ it is also clear that $(z^{n_1})^{n_2}=z^{n_1n_2}$, and when $n_1,n_2$ may be negative the equality holds by case analysis together with the definition $z^{-n}=(z^n)^{-1}$. Therefore, we derive$$(b^m)^{1/n}=(b^{1/n})^m=(((b^{1/n})^m)^q)^{1/q}=((b^{1/n})^{mq})^{1/q}=$$$$((b^{1/n})^{pn})^{1/q}=(((b^{1/n})^n)^p)^{1/q}=(b^p)^{1/q}$$ (b) The case is clear for integer $r,s$. As well, we note $z^{1/n_1n_2}=(z^{1/n_1})^{1/n_2}$ as $((z^{1/n_1})^{1/n_2})^{n_1n_2}=(((z^{1/n_1})^{1/n_2})^{n_2})^{n_1}=z$. Otherwise let $r=r_1/r_2$ and $s=s_1/s_2$. We have$$b^{r+s}=b^{r_1/r_2+s_1/s_2}=b^{(r_1s_2+r_2s_1)/(r_2s_2)}=(b^{r_1s_2+r_2s_1})^{1/r_2s_2}=$$$$(b^{r_1s_2}b^{r_2s_1})^{1/r_2s_2}=(((b^{r_1})^{s_2})^{1/s_2})^{1/r_2}(((b^{s_1})^{r_2})^{1/r_2})^{1/s_2}=b^rb^s$$ (c) Lemma 1: When $a = a_1/a_2 > 0$ is rational we have $b^a > 1$. This implies $b^a > b^c$ for rationals $a > c$. Proof: Rewrite $a_2 > 0$ so that $a_1 > 0$ so by induction $b^{a_1} > 1$ and it suffices to prove $b^{1/a_2} > 1$ for $b > 1$. Here it suffices to prove for $c ≤ 1$ then $c^n ≤ 1$ for natural $n$, which is clear by induction $c^n = cc^{n-1} ≤ c ≤ 1$.

Now, for $t≤r$ we show $b^r ≥ b^t ⇔ b^t(b^{r-t}-1) ≥ 0$, so since $b^t$ is nonnegative and by the lemma $b^{r-t}-1$ is nonnegative we have $b^r$ is an upper bound of $B(r)$. As well, if $z < b^r$ then since $b^r∈B(r)$ we have $z$ is not an upper bound of $B(x)$ and thus $b^r = \text{sup }B(r)$.

(d) Lemma 2: Let $X⊆\mathbb{R}$ be bounded above such that when $x∈X$ and $y≤x$, then $y∈X$. We see $\text{sup }X = \text{sup }(X \setminus \{\text{sup }X\})$. Proof: $z = \text{sup }X$ bounds the latter, so assume $y < z$ also bounds. But $y$ doesn't bound $X$ and thus $y < x$ for some $x∈X$ and we may choose some $y < x_1 < x ≤ z$ so $x_1 ∈ X \setminus \{\text{sup }X\}$ yet $y < x_1$ and $z$ is the smallest bound.

Let $B^*(x) = \{v ≤ b^t~|~t∈\mathbb{Q},~t≤x\}$ so that $\text{sup }B^*(x) = \text{sup }B(x)$ and it serves as an equivalent definition for $b^x$. Now if $b^x∈B^*(x)$ then by the above lemma we may remove it, and as well $b^t$ for $t < x$ to define $B^{**}(x)$ with an equivalent supremum. Now we must show that when $z < \text{sup }B^{**}(x)$ then $z < b^v$ for some rational $v < x$. We already see this is the case when for some rational $v ≤ x$, so we must construct rational $v' < v$ such that $z < b^{v'} < b^v$; this may be done by observing lemma 1 and parts (a),(b), and (g) to show $b^{1/n}$ tends to 1 and setting $v' = v - 1/n$ for $1 < b^{1/n} < b^v/z$.

Now, suppose $b^xb^y < b^{x+y}$; then $b^xb^y < b^v$ for some rational $v < x+y$ by the above discussion. Set rational $v - y < v_1 < x$ and $v - v_1 < v_2 < y$ so $v_1+v_2 > v$ and $b^xb^y ≥ b^{v_1}b^{v_2} = b^{v_1+v_2} > b^{v}$.

Suppose $b^{x+y} < b^xb^y$. Then $(b^x)^{-1}b^{x+y} < b^{v_2}$ for some rational $v_2 < y$. Choose rational $v_2 < v_2 < y$ and set $v_4 = v_3 - v_2 > 0$, and then choose $x-v_4 < v_1 < x$. We have $b^{v_2} > (b^x)^{-1}b^{x+y} ≥ (b^x)^{-1}b^{v_1+v_2+v_4}$ implying $1 > (b^x)^{-1}b^{v_1}b^{v_4}$ then $b^x > b^{v_1+v_4}$. But $v_1+v_4 > x$ and choosing rational $x < x' < v_1+v_4$ we see $b^{x'}$ is an upper bound of $b^x$ and $b^{v_1+v_4} > b^{x'} ≥ b^x$ is a contradiction.$~\square$

Special Modules over PIDs (12.1.20-22)

Dummit and Foote Abstract Algebra, section 12.1, exercise 20-22:

20. Let $R$ be an integral domain with field of fractions $F$. Prove $\text{dim }F ⊗_R M = \text{rank }M$ for $F ⊗_R M$ a vector space over $F$ and $M$ any $R$-module.

21. Let $R$ be a PID. Prove any finitely generated projective $R$-module is free.

22. Let $R$ be a non-field PID. Prove no finitely generated $R$-module is injective.

Proof: (20) ($≤$) We see $\{f ⊗ m~|~f∈F,m∈M\}$ spans $F ⊗_R M$, so there is a basis $\{f_i ⊗ m_i\}_{i∈I}$. As well, we see $\{1⊗m_i\}$ is a basis. We claim $\{m_i\}$ is a set of linearly independent elements in $M$; otherwise, we have $∑r_im_i=0$ implying $∑r_i(1⊗m_i)=0$. ($≥$) Let $\{m_i\}$ be a set of linearly independent elements. We claim $\{1⊗m_i\}$ are linearly independent over $R$. We have $∑r_i(1⊗m_i)=∑1⊗r_im_i=1⊗∑r_im_i$; since the $m_i$ are linearly independent, we may assume that if some $r_i \neq 0$ then $∑r_im_i \neq 0$ and as well $∑r_im_i$ is nontorsion; now by 10.4.8 $1⊗∑r_im_i \neq 0$ and the $1⊗m_i$ are linearly independent over $R$. Consequently, collecting the denominators of any linear independence over $F$ leads to an equation of the form $\dfrac{1}{d}(∑r_i(1⊗m_i))=0$ so that $∑r_i(1⊗m_i)=0$ and now $r_i=\dfrac{r_i}{d}=0$ for all $i$ and the $1⊗m_i$ are linearly independent over $F$.

(21) $M$ is projective if and only if it it is a direct summand of a free module only if it is a submodule of a free module, which would imply it has no torsion elements. By the fundamental theorem this eliminates any elementary divisors of $M$ and forces $M$ free.

(22) $M≅R^n⊕R/(a_1)~⊕~...~⊕~R/(a_k)$ is injective if and only if each of its summands is injective by 10.5.4. By Baer's criterion, $M$ is injective over a PID if and only if $rM=M$ for all nonzero $r∈R$ so necessarily $R/(a_i)$ is not injective and we must only prove $R$ is not injective; indeed, for $p∈R$ a prime we see $1∉pR$.$~\square$

Proof of Finitely Generated Module Theorem Over EDs by Matrices (12.1.19)

Dummit and Foote Abstract Algebra, section 12.1, exercise 19:

By the previous two exercises we may perform elementary row and column operations on a given relations matrix by choosing different generators for $R^n$ and $\text{ker }φ$. If all relation matrices are the zero matrix then $\text{ker }φ=0$ and $M=R^n$. Otherwise let $a_1$ be a nonzero g.c.d. of all the entries in a fixed initial relations matrix for $M$.

(a) Prove that by elementary row and column operations we may assume $a_1$ occurs in a relations matrix $(a_{ij})$ where $a_1 \mid a_{ij}$ for all $i,j$.
(b) Prove there is a relations matrix of the form $(a_{ij})$ where $a_{1,1} = a_1 | a_{ij}$ for all $i,j$ and $a_{1k}=a_{k1}=0$ for all $k > 1$.
(c) Let $a_2$ be a g.c.d. of all the entries except the element $a_1$ in the relations matrix in (b). Prove there is a relations matrix of the form above where additionally $a_{2,2} = a_2$ and $a_{2k} = a_{k2} = 0$ for $k \neq 2$.
(d) Prove there is a relations matrix diagonalized from the top left $a_1,a_2,...,a_k$ for some $k ≤ n$ such that $a_1 \mid a_2 \mid ... \mid a_k$. Derive the theorem for finitely generated modules over Euclidean Domains.

Proof: (a,b) Let $N : R → \mathbb{N}$ be the norm on $R$. For a matrix $M$ with entries in $R$ define $N(M)$ to be the minimum norm taken over the nonzero entries over $M$. For $A$ the initial relations matrix, let $A'$ be a minimal matrix attainable from $A$ by elementary row and column operations with regard to $N(A')$, and let $a_1$ be the nonzero entry of $A'$ possessing such minimum norm. Move $a_1$ to position $(1,1)$ by a row and column interchange. We claim $a_1$ divides all the entries of $A'$, and as such is the g.c.d. of all the entries of $A$ (since any divisor of all the entries of $A$ still acts as a divisor after an operation, and the operation is reversible). First, we show $a_1 \mid a_{1i}$ and $a_1 \mid a_{i1}$. By definition of the norm write $a_{1i} = qa_1 + r$, and subtract $q$ times the first column from the $i^{th}$ column to set $r$ in the $(1,i)$ position; since $N(r) < N(a_1)$ or $r = 0$, we must have the latter by the minimality of $A'$, and thus $a_1 \mid a_{1i}$. The case for the first column is parallel.

Now attain a matrix like the one in (b) by subtracting appropriate multiples of the first row and column. We show $a_1 \mid a_{ij}$ for all $i,j$ and we will be done. Add the $i^{th}$ row to the first to set $a_{ij}$ in the $(1,j)$ position while still retaining $a_1$ in the $(1,1)$ position (as the entry in the $(i,1)$ is zero), and as above prove the remainder of the division is zero. This is matrix $B$.

(c) Simply iterate the process of (a,b) by focusing on the matrix $B_{1,1}$. These operations will not disturb the outer part of the matrix because of the zeros.

(d) Continue iterating to obtain the matrix in (d). By considering the new basis for $R^n$, we have$$R^n/\text{ker }φ = (R~⊕~...~⊕~R) / (Ra_1~⊕~...~⊕~Ra_k~[⊕~0~⊕~...~⊕~0]) ≅$$$$R^{n-k}~⊕~R/(a_1)~⊕~...~⊕~R/(a_k)~\square$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Though the above proof is nonconstructive, a process for obtaining such a matrix with minimal norm can be described as follows: Given $A$, move the nonzero element with minimal norm to the top left. Use Euclid's algorithm to obtain the smallest possible norm among the elements of the first row, which is a g.c.d. of the row. Subtract the other elements of the row to zero and repeat with the first column. Now if $a_{1,1}$ doesn't divide $a_{ij}$ then add row $i$ to row $1$ and use Euclid's algorithm between $a_{1,1}$ and $a_{1,j}$ to obtain an element of smaller norm. Repeat until completion.

Irreducible Modules Over PIDs (12.1.14)

Dummit and Foote Abstract Algebra, section 12.1, exercise 14:

Letting $M$ be a torsion module over $R$ a P.I.D., prove $M$ is irreducible if and only if $M=Rm$ for some nonzero element $m∈R$ with prime annihilator $(p)$.

Proof: ($⇒$) $M$ is clearly generated by a single element $m$, as a partial generator would imply a nontrivial submodule. Assume $\text{Ann }(m)=(r)$ for composite, nonzero $r$, so write $r = ab$ for nonzero nonunits $a$ and $b$. We show $Rbm ⊂ Rm = M$. Assuming otherwise, we can write $xbm = m$ for some $x∈R$, and now $axbm = am ⇒ am = 0$ despite $(r) ⊂ (a)$, a contradiction. ($⇐$) Take nonzero $rm ∈ M$; we thus have $p \not \mid r$ and there are elements $a,b∈R$ providing $ar+bp = 1$. We have $arm = (ar+bp)m=m$ so that $Rrm = M$ and $M$ has no nontrivial submodules.$~\square$

Direct Sums and Rank (12.1.3)

Dummit and Foote Abstract Algebra, section 12.1, exercise 3:

Let $R$ be an integral domain and let $A$ and $B$ be $R$-modules of rank $m$ and $n$ respectively. Show $A ⊕ B$ is an $R$-module of rank $m+n$.

Proof: Letting $a_1,...,a_m$ and $b_1,...,b_n$ be the fulfilling linearly independent elements, we see $(a_1,0),...,(0,b_n)$ are linearly independent. Observing the homomorphism $(x,y) ↦ (x \text{ mod }R^m, y \text{ mod }R^n)$ admitted by these former elements we see the kernel is the submodule generated by the latter. By 12.1.2, we will show the quotient is torsion and thus $A ⊕ B$ is of rank $m+n$. Take $\overline{(a,b)}$. Also by 12.1.2 we have the respective images of $A$ and $B$ under the homomorphism are those of $A$ and $B$ under the homomorphisms provided by their own linearly independent elements, and as such there are nonzero $r_1,r_2$ such that $r_1\overline{a}=r_2\overline{b}=0$, so that $r_1r_2$ is nonzero and $r_1r_2\overline{(a,b)}=0$.$~\square$

Torsion Elements and Rank (12.1.1)

Dummit and Foote Abstract Algebra, section , exercise :

Let $M$ be a module over the integral domain $R$.
(a) Suppose $x \neq 0$ is a torsion element of $M$. Show $x$ and $0$ are "linearly dependent." Conclude the rank of $\text{Tor}(M)$ is $0$, so that in particular any torsion $R$-module has rank $0$.
(b) Show $\text{rank} M = \text{rank} M/\text{Tor}(M)$.

Proof: (a) By definition we have $rx=0$ for nonzero $r$. Therefore there exist no candidates for linearly independent sets.

(b) Let $n = \text{rank} M$ and $m = \text{rank} M/\text{Tor}(M)$. ($≥$) Let $\overline{x_1},...,\overline{x_m}∈M/\text{Tor}(M)$ be linearly independent. If $m > n$ then we can arrange for a linear dependence$$r_1x_1+...+r_mx_m = 0$$which clearly extends to give a linear dependence in the quotient. ($≤$) Let $x_1,...,x_n∈M$ be linearly independent. Assume that their images in the quotient are linearly dependent, so that$$r_1x_1+...+r_nx_n∈\text{Tor}(M)$$and by definition of $\text{Tor}(M)$ we obtain a nonzero annihilator to obtain a contradictory linear dependence.$~\square$

Tensor, Symmetric, and Exterior Subalgebras (11.5.14)

Dummit and Foote Abstract Algebra, section 11.5, exercise 14:

Prove that when the $R$-module $M$ is a direct summand of $N$ by $φ$, then $\mathcal{T}(M)$, $\mathcal{S}(M)$, and $\bigwedge M$ are $R$-subalgebras of $\mathcal{T}(N)$, $\mathcal{S}(N)$, and $\bigwedge N$, respectively.

Proof: For each of the above algebras from $M$, we have induced algebra homomorphisms$$Φ(m_1 ⊗ ... ⊗ m_k)=φ(m_1) ⊗ ... ⊗ φ(m_k)$$$$Φ(m_1 ⊗ ... ⊗ m_k \text{ mod }\mathcal{A}(M)) = φ(m_1) ⊗ ... ⊗ φ(m_k) \text{ mod }A(N)$$$$Φ(m_1 ∧ ... ∧ m_k) = φ(m_1) ∧ ... ∧ φ(m_k)$$As well, we have a natural homomorphism $φ' : N → M$ by collapsing onto the $M$ component which induces homomorphisms of algebras by $Φ'$. Since we see $φ'∘φ=1$, likewise we have $Φ'∘Φ=1$ so that $Φ$ is injective.$~\square$