Proofs without words 2

A visual proof that $$\sin 2x = 2\sin x \cos x, \quad  \cos 2x = \cos^2 x−\sin^2x$$  The argument only requires similarity of triangles.

Credits: Fermat's Library on Twitter (@fermatslibrary)

Wedderburn's little theorem

Theorem. Every finite division ring $K$ is a field.

This classical result was first proven by Joseph Wedderburn in 1905. It is actually a rather unexpected fact that the finiteness of elements in $K$ imposes such a strong condition as the commutativity of multiplication. The finiteness assumption is clearly an essential one, because there exist infinite, non commutative division rings, such as the ring $\mathbb{H} $ of real quaternions.

The proof of Wedderburn's theorem usually given in modern textbooks (as the ones cited in the references below) is due to Ernst Witt (1931) and goes as follows.

Proof. Let $Z:=Z(K)$ be the center of $K$.  By assumption, $Z$ is a finite field and then it has $q=p^s$ elements, where $p$ is the characteristic and $s= [Z : \mathbb{F}_p]$ is the dimension of $Z$ as a $\mathbb{F}_p$-vector space.  We want to show that $Z=K$.

Since $K$ is a finite-dimensional vector space over $Z = \mathbb{F}_q$, it has $q^t$ elements for some $t \geq 1$. For all $x \in K$, let us denote by $C_x$ the centralizer of $x$ in $K$, namely $$C_x :=\{y \in K \, | \, xy=yx \}.$$ Then $C_x$ is a $Z$-vector space, and we call $q^{d(x)}$  its dimension.  Then $d(x)$ divides $t$ and $d(x)=t$ if and only if $x \in Z$.

Setting $G:=K^*$ for the multiplicative group of $K$, we see that $G$ is a finite group whose center is $Z^*$; moreover the centralizer in $G$ of an element $x$ is $C_x^*$. Then the equation of conjugate classes for $G$ yields
\begin{equation} \label{eq:classes}
q^t-1 = |G| = |Z^*| + \sum \frac{|G|}{|{C_x}^*|} = q-1 + \sum \frac{q^t-1}{q^{d(x)}-1},
\end{equation} where the sum is extended to a complete system of representatives for the conjugacy classes of $G \setminus Z^*= K \setminus Z$.

We want now to show that the only possibility is $d(x)=t=1$, namely $K=Z$. In order to do this, we exploit the theory of cyclotomic polynomials. In fact, for any positive divisor $d$ of $t$ such that $1 < d < t$, we have $$X^t -1 = \prod_{k | t} \Phi_k(X) =\Phi_t(X)(X^d-1) \left(\prod_{k<t, \, k |t, \, k \nmid d } \Phi_k(X) \right).$$ This implies that $$\frac{X^t-1}{X^d-1}$$ is a polynomial with integer coefficients, which is divided by $\Phi_t(X)$. In particular, setting $X=q$ and $d=d(x)$, we deduce that $\Phi_t(q)$ divides $(q^t-1)/(q^{d(x)}-1)$, hence by  using \eqref{eq:classes} we infer that $\Phi_t(q)$ divides $q-1$.

We now obtain the desired contradiction: indeed, if $\zeta$ is a primitive $t^{\mathrm{th}}$ root of unity, since $t >1$ and $q \in \mathbb{N}$  we have  $|q - \zeta| > q-1,$ so  $$\Phi_t(q) = \prod_{(k, \, t)=1} |q - \zeta^k| > q-1, $$ which is absurd.   $\; \Box$

Wedderburn's theorem is a purely algebraic result; however, it has unexpected consequences in projective geometry. In fact, it is known that a projective plane $\mathbb{P}$ is Desarguesian if and only if it is of the form $\mathbb{P}^2(K)$ for some division ring  $K$, and that in that case it is Pappian if and only if $K$ is a field. So we obtain the following 

Corollary. A finite projective plane $\mathbb{P}^2(K)$ is Desarguesian if and only if it is Pappian.

It is remarkable that this is the only known proof of the implication $$\textrm{Desarguesian} \Longrightarrow \textrm{Pappian}$$ for a finite projective plane $\mathbb{P}^2(K).$

References

• E. Artin: Algèbre Géométrique, Gauthier-Villars, p. 37.
• S. Gabelli: Teoria delle equazioni e teoria di Galois, Springer, p. 171.
• I. N. Herstein: Algebra, Editori Riuniti, p. 397.
• https://en.wikipedia.org/wiki/Wedderburn%27s_little_theorem

Blindness

"Should you just be an algebraist or a geometer?" is like saying "Would you rather be deaf or blind?"
Sir Micheal Atiyah 

Source
M. Atiyah: Mathematics in the 20th century, Bull. London Math. Soc. 34 (2002) 1–15.
https://doi.org/10.1112/S0024609301008566

Proofs without words 1

Every odd number is the difference of two consecutive squares.

Pick an odd number $n$ (in red), "bend it" in the middle and fill the square (with the blue part). In the picture, we see that for $n=13$ we obtain $13=7^2-6^2$.

Credits: MSE question 263101

When length does not matter

L'articolo di 5 righe con il quale, nel 1966, Lander e Parkin smentirono la congettura di Eulero nel caso $n=5$. 

La congettura è ancora aperta per $n  \geq  6$.

 

L'irrazionalità di $e$

Esponiamo la celebre dimostrazione in poche righe dell'irrazionalità di $e$, dovuta a Joseph Fourier. L'idea è quella di ragionare per assurdo, spezzando in modo opportuno il noto sviluppo in serie e maggiorando una delle quantità risultanti per mezzo di una serie geometrica. 

Teorema. Il numero di Nepero $e$ è irrazionale. 

Dimostrazione.  Supponiamo per assurdo che $e$ sia razionale, cioè che si possa scrivere $e=m/n$ con $m, \, n \in \mathbb{N}$. Allora si ha 

\begin{equation*}
\frac{m}{n}=\sum_{k=0}^{+ \infty} \frac{1}{k!} \end{equation*} da cui, dopo avere moltiplicato entrambi i termini per $n!$, si ottiene \begin{equation*}\begin{split} & m(n-1)! -  \sum_{k=0}^n \frac{n!}{k!} = \sum_{k=n+1}^{+ \infty} \frac{n!}{k!} \\ = &  \frac{1}{n+1} + \frac{1}{(n+1)(n+2)}+ \frac{1}{(n+1)(n+2)(n+3)}+ \ldots \\ <  &  \frac{1}{n+1} + \frac{1}{(n+1)^2}+ \frac{1}{(n+1)^3}+ \ldots \\ = & \frac{1}{n} \end{split}\end{equation*} Il primo termine a sinistra è un intero positivo, mentre l'ultimo a destra è compreso strettamente fra $0$ e $1$, contraddizione. $\square$

Walks

A line is a dot that went for a walk.
Paul Klee

Source.
The Diaries of Paul Klee (1964)