Tensor Products and Field Extensions (13.3.22)

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

Let $K/F$ be a finite field extension, and let $K_1,K_2⊆K$ be field extensions of $F$. Show the $F$-algebra $K_1 ⊗_F K_2$ is a field if and only if $[K_1K_2~:~F] = [K_1~:~F][K_2~:~F]$.

Proof: Let $A$ be the set of finite sums of elements of the form $k_1k_2$ for $k_1∈K_1,k_2∈K_2$, let $φ : K_1 × K_2 → A$ be the bilinear map defined by $φ(k_1,k_2)=k_1k_2$, and let $Φ : K_1 ⊗ K_2 → A$ be the corresponding $F$-linear transformation. We observe $$Φ(k_1⊗k_2)Φ(k_1'⊗k_2') = k_1k_1'k_2k_2' = Φ((k_1⊗k_2)(k_1'⊗k_2'))$$ allowing us to show $$Φ(\sum_i k_{i1} ⊗ k_{i2})Φ(\sum_j k_{j1}'⊗k_{j2}')=\sum_{i,j} Φ(k_{i1} ⊗ k_{i2})Φ(k_{j1}' ⊗ k_{j2}') =$$ $$\sum_{i,j} Φ((k_{i1} ⊗ k_{i2})(k_{j1}' ⊗ k_{j2}')) = Φ((\sum_i k_{i1} ⊗ k_{i2})(\sum_j k_{j1}' ⊗ k_{j2}'))$$ so that $Φ$ is an $F$-algebra homomorphism. Note $A = \text{img Φ}$. As well, let $K_1$ have for basis over $F$ $\{n_i\}$ and let $K_2$ have $\{m_j\}$.

($⇒$) We thus have $Φ$ is a nonzero field homomorphism, and is thus an isomorphism. Now $A$ is a field and by definition we have $K_1K_2 ⊆ A$ and by construction we observe $A ⊆ K_1K_2$ so that $K_1K_2 ≅ K_1 ⊗ K_2$ as $F$-algebras, the latter of which has for basis $\{n_i ⊗ m_j\}$ of order $[K_1~:~F][K_2~:~F]$. ($⇐$) We still have $A ⊆ K_1K_2$, and since we observe $[K_1~:~F][K_2~:~F]$ elements $n_im_j$ of $A$ linearly independent over $F$ by Proposition 13.2.21, we must have this is a basis for $K_1K_2$ and thus again $A = K_1K_2$. The $F$-algebra homomorphism $Φ$ above sends basis to basis and is thus an isomorphism, and now $K_1 ⊗ K_2$ is a field.$~\square$

Applications of Algebraic Extensions (13.2.16-17)

Dummit and Foote Abstract Algebra, section 13.2, exercises 16-17:

16. Let $K/F$ be an algebraic extension and let $R$ be a ring where $F ⊆ R ⊆ K$. Show $R$ is a field.

17. Let $f(x)∈F[x]$ be irreducible of degree $n$, and let $g(x)∈F[x]$. Prove that every irreducible factor of $f(g(x))$ has degree divisible by $n$.

Proof: (16) It suffices to show that every element $r∈R$ has a multiplicative inverse. Since $r∈K$ is algebraic over $F$, we observe $r^{-1}∈F(r) ⊆ R$.

(17) Let $h(x) \mid f(g(x))$ be irreducible of degree $k$, and let $α$ be a solution to $h(x)$, so that $h(α)=0$ and thus $f(g(α))=0$. Since $f(x)$ is irreducible, we must have $g(α)$ is of degree $n$. We thus have $$\text{deg }h(x) = k = [F(α)~:~F] =$$ $$[F(α)~:~F(g(α))] \cdot [F(g(α))~:~F] = [F(α)~:~F(g(α))] \cdot n$$ so that $n \mid k$.$~\square$

Convergence of Matrices (12.3.40-45)

Dummit and Foote Abstract Algebra, section 12.3, exercises 40-45:

40. Letting $K$ be the real or complex field, prove that for $A,B∈M_n(K)$ and $α∈K$:
(a) $||A+B|| ≤ ||A|| + ||B||$
(b) $||AB|| ≤ ||A|| \cdot ||B||$
(c) $||αA|| = |α| \cdot ||A||$

41. Let $R$ be the radius of convergence of the real or complex power series $G(x)$.
(a) Prove that if $||A|| < R$ then $G(A)$ converges.
(b) Deduce that for all matrices $A$ the following power series converge: $$\text{sin}(A)=\sum_{k=0}^∞ (-1)^k \dfrac{A^{2k+1}}{(2k+1)!}$$ $$\text{cos}(A)=\sum_{k=0}^∞(-1)^k \dfrac{A^{2k}}{(2k)!}$$ $$\text{exp(A)}=\sum_{k=0}^∞\dfrac{A^k}{k!}$$ 42. Let $P$ be a nonsingular $n \times n$ matrix, and denote the variable $t$ by the matrix $tI$ (in light of the theory of differential equations).
(a) Prove $PG(At)P^{-1}=G(PAtP^{-1})=G(PAP^{-1}t)$, so it suffices to consider power series for matrices in canonical form.
(b) Prove that if $A$ is the direct sum of matrices $A_1,...,A_m$, then $G(At)$ is the direct sum of the matrices $G(A_1t),...,G(A_mt)$.
(c) Show that if $Z$ is the diagonal matrix with entries $z_1,...,z_n$ then $G(Zt)$ is the diagonal matrix with entries $G(z_1t),...,G(z_nt)$.

43. Letting $A$ and $B$ be commuting matrices, show $\text{exp}(A+B)=\text{exp}(A)\text{exp}(B)$.

44. Letting $λ ∈ K$, show $$\text{exp}(λIt+M)=e^{λt}\text{exp}(M)$$ 45. Let $N$ be the $r \times r$ matrix with $1$s on the first superdiagonal and zeros elsewhere. Show $$\text{exp}(Nt)=\begin{bmatrix}1 & t & \dfrac{t^2}{2!} & \cdots & \cdots & \dfrac{t^{r-1}}{(r-1)!} \\ ~ & 1 & t & \dfrac{t^2}{2!} &~ & \vdots \\ ~ & ~ & \ddots & \ddots & \ddots & \vdots \\ ~ & ~ & ~ & \ddots & t & \dfrac{t^2}{2!} \\ ~ & ~ & ~ & ~ & 1 & t \\ ~ & ~ & ~ & ~ & ~ & 1 \end{bmatrix}$$ Deduce that if $J$ is the $r \times r$ elementary Jordan matrix with eigenvalue $λ$ then $$\text{exp}(Jt)=\begin{bmatrix}e^{λt} & te^{λt} & \dfrac{t^2}{2!}e^{λt} & \cdots & \cdots & \dfrac{t^{r-1}}{(r-1)!}e^{λt} \\ ~ & e^{λt} & te^{λt} & \dfrac{t^2}{2!}e^{λt} &~ & \vdots \\ ~ & ~ & \ddots & \ddots & \ddots & \vdots \\ ~ & ~ & ~ & \ddots & te^{λt} & \dfrac{t^2}{2!}e^{λt} \\ ~ & ~ & ~ & ~ & e^{λt} & te^{λt} \\ ~ & ~ & ~ & ~ & ~ & e^{λt} \end{bmatrix}$$ Proof: (40)(a) We have $$||A+B||=\sum_{i,j}|a_{ij}+b_{ij}| ≤ (\sum_{i,j}|a_{ij}|)+(\sum_{i,j}|b_{ij}|)=||A||+||B||$$ (b) We have $$||AB||=\sum_{i,j}|\sum_{k=0}^n a_{ik}b_{kj}| ≤ \sum_{i,j} \sum_{k=0}^n |a_{ik}| \cdot |b_{kj}| ≤$$ $$\sum_{i,j,i',j'} |a_{ij}| \cdot |b_{i'j'}| = (\sum_{i,j}|a_{ij}|)(\sum_{i,j}|b_{ij}|) = ||A|| \cdot ||B||$$ (c) We have $$||αA|| = \sum_{i,j}|αa_{ij}| = |α| \sum_{i,j} |a_{ij}| = |α| \cdot ||A||$$ (41)(a)(Method due to Project Crazy Project) For each entry $i,j$ entry $a_{(k)ij}$ of $A^k$, we note that $|a_(k){ij}| ≤ ||A^k|| ≤ ||A||^k < R^k$. Therefore, we note $\sum_{k=0}^N |α_ka_{(k)ij}| ≤ \sum_{k=0}^N |α_k| \cdot r^k$, where $r$ has been chosen $||A|| < r < R$. By the Cauchy-Hadamard theorem this last power series displays the same radius of convergence $R$ and thus converges for $r$ as $N$ approaches infinity. Thus the partial power series for $G(A)$ in the $i,j$ coordinate converges absolutely, and thus converges.

(b) Since these functions are shown to have radius of convergence $R = ∞$ over $K$, by the above they converge for matrices of arbitrary absolute value, i.e. for all matrices.

(42)(a) Lemma 1: Let $x_n → x$ and $y_n → y$ be convergent sequences of $m × m$ matrices over $K$. Then $x_ny_n → xy$. Proof: We see $x → x_n$ if and only if for any $ε > 0$ we have $||x-x_N|| < ε$ for sufficiently large $N$, since the forward is evident when the entries all converge within range of $ε/m^2$, and the converse forces all entries to converge within $ε$. For sufficiently large $N$ we have - for some asymptotically small matrices $||ε_{(N)1}||,||ε_{(N)2}||$ - the result $$||xy-x_Ny_N|| = ||xy-(x+ε_{(N)1})(y+ε_{(N)2})|| =$$ $$||xε_{(N)2}+ε_{(N)1}y+ε_{(N)1}ε_{(N)2}|| ≤ ||x|| \cdot ||ε_{(N)2}|| + ||y|| \cdot ||ε_{(N)1}|| + ||ε_{(N)1}|| \cdot ||ε_{(N)2}||$$ which vanishes to zero.$~\square$

Now we see $$PG(At)P^{-1}=(\text{lim }P)(\text{lim }G_N(At))(\text{lim }P)=$$ $$\text{lim }PG_N(At)P^{-1} = \text{lim }G_N(PAtP^{-1}) = G(PAtP^{-1}) = G(PAP^{-1}t)$$ (b) By considering lemma 2 of 12.3.38, we see that the power series' summands may be computed in blocks, leading to independent convergences as blocks.

(c) This is simply a special case of (b).

(43) Lemma 2: $\lim_{n → ∞}\dfrac{x^n}{\lfloor n/2 \rfloor !}=0$ for any real $x$. Proof: When $n > x^2$ $$\lim_{n → ∞}\dfrac{x^n}{\lfloor n/2 \rfloor !} ≤ \lim_{n → ∞}x\prod_{k=1}^{\lfloor n/2 \rfloor} \dfrac{x^2}{k} = \lim_{n → ∞}x \prod_{k = 0}^{\lfloor x^2 \rfloor}\dfrac{x^2}{k} \prod_{k = \lceil x^2 \rceil}^{\lfloor n/2 \rfloor}\dfrac{x^2}{k} =$$ $$x \prod_{k = 0}^{\lfloor x^2 \rfloor}\dfrac{x^2}{k} \lim_{n → ∞} \prod_{k = \lceil x^2 \rceil}^{\lfloor n/2 \rfloor}\dfrac{x^2}{k} = 0~~\square$$ Lemma 3: Let $x_n → x$ and $y_n → y$ be convergent sequences of $m × m$ matrices over $K$ such that $|x_n-y_n| → 0$. Then $x=y$. Proof: Assume $x \neq y$ so $|x-y| > 0$. Let $|x-x_n| < |x-y|/2 - ε$ for $n > n_1$ and some chosen $0 < ε < |x-y|/2$, let $|y-y_n| < |x-y|/2$ for $n > n_2$, let $|x_n-y_n| < ε$ for $n > n_3$, and choose $N > \text{max}(n_1,n_2,n_3)$. We have $$|x-y| = |(x-x_N)+(x_N-y_N)+(y_N-y)| ≤$$ $$|x-x_N|+|x_N-y_N|+|y-y_N| < |x-y|$$ a contradiction.$~\square$

Utilizing lemma 1, we have $\lim_{N → ∞}\text{exp}_n(A)\lim_{N → ∞}\text{exp}_n(B)=\lim_{N → ∞}\text{exp}_n(A)\text{exp}_n(B)$, so we may compare terms of the two sequences $$\text{exp}_n(A)\text{exp}_n(B)=(\sum_{k=0}^n\dfrac{A^k}{k!})(\sum_{k=0}^n\dfrac{B^k}{k!})=\sum_{j=0}^n\sum_{k=0}^n\dfrac{A^jB^k}{j!~k!}=\sum_{j,k ≤ n} \dfrac{A^jB^k}{j!~k!}$$ $$\text{exp}_n(A+B)=\sum_{k=0}^n \dfrac{(A+B)^k}{k!} = \sum_{k=0}^n \dfrac{\sum_{j=0}^k \dfrac{k!}{j!(k-j)!}A^jB^{k-j}}{k!} =$$ $$\sum_{k=0}^n \sum_{j=0}^k \dfrac{A^jB^{k-j}}{j!(k-j)!} = \sum_{j+k ≤ n}\dfrac{A^jB^k}{j!~k!}$$ and then compare their differences (without loss of generality assume $||A|| ≤ ||B||$) $$|\text{exp}_n(A+B)-\text{exp}_n(A)\text{exp}_n(B)| = |\sum_{j,k ≤ n < j+k}\dfrac{A^jB^k}{j!~k!}| ≤ \sum_{j,k ≤ n < j+k} \dfrac{||A||^j||B||^k}{j!~k!} ≤$$ $$\dfrac{n^2||B||^{2n}}{\lfloor n/2 \rfloor !} = \dfrac{||B||^2 \prod_{k=2}^n\dfrac{k^2||B||^{2k}}{(k-1)^2||B||^{2k-2}}}{\lfloor n/2 \rfloor !} ≤ ||B||^2\dfrac{(4||B||^2)^{n-1}}{\lfloor n/2 \rfloor !}$$ When $4||B||^2 < 1$ we clearly have the above vanishing to zero as $n$ approaches infinity, so assume $4||B||^2 ≥ 1$ $$||B||^2\dfrac{(4||B||^2)^{n-1}}{\lfloor n/2 \rfloor !} ≤ ||B||^2\dfrac{(4||B||^2)^n}{\lfloor n/2 \rfloor !}$$ and by lemma 2, the term still tends toward $0$. Thus by lemma 3 we conclude $\text{exp}(A+B)=\text{exp}(A)\text{exp}(B)$.

(44) This is clear from the fact $\text{exp}(λt) = e^{λt}$, the ring isomorphism between $K$ and matrices $KI$, and the preceding exercise.

(45) The first part is clear from the fact that $N^k$ is the matrix with $1$s along the $k^{th}$ superdiagonal (observable most easily by induction from the linear transformation form of $N$), and since $N^r = 0$, we have $\text{exp}(Nt)=\text{exp}_r(Nt)$ is the matrix described above. The second part is clear from the observation in (44) with $M=Nt$ coupled with the first part.$~\square$