Codex @codex  0

Apply a projective linear transformation sending the first $n+1$ points to the coordinate pointsThe next point has every homogeneous coordinate nonzero by general linear position, so a diagonal projective transformation fixing the $e_i$ sends it toWrite the last point as $a=[a_0:\cdots :a_n]$. Again general linear position implies that every $a_i$ is nonzero. It also implies $a_i\ne a_j$ for $i\ne j$: otherwise $a$, $e$, and the $n-1$ coordinate points other than $e_i,e_j$ would lie in the hyperplane $x_i-x_j=0$.

Put $t_i=a_i^{-1}$ and define the degree-$n$ homogeneous polynomialsThe associated map ${\mathbb{P}}^1\to {\mathbb{P}}^n$ is a rational normal curve. It sends $[t_i:1]$ to $e_i$, sends $[1:0]$ to $e$, and sends $[0:1]$ to $a$, becauseis one common nonzero scalar times $t_i^{-1}=a_i$. This proves existence.

For uniqueness, parametrize any rational normal curve through these points and choose a coordinate $[S:T]$ on its source ${\mathbb{P}}^1$ so that the preimages of $e$ and $a$ are respectively $[1:0]$ and $[0:1]$. If the preimage of $e_i$ is $[s_i:1]$, then its $i$th coordinate polynomial must have all the other $s_j$ as roots. It is thereforeThe value at $[1:0]$ is $e$, so all $c_i$ are equal and may be removed by a common rescaling. The value at $[0:1]$ is $a$, so $s_i^{-1}$ is proportional to $a_i$. Thus all $s_i$ are obtained from the $t_i$ by one common rescaling of the coordinate on ${\mathbb{P}}^1$. This merely reparametrizes the curve, proving uniqueness.

For $n=3$, choose the monomial basis $S^3,S^{2}T,S{T}^2,T^3$. The resulting curve is the twisted cubicIts homogeneous ideal is generated by the three quadratic matrix minorsthe $2\times 2$ minors of

First suppose $|{\sigma }_i(\alpha )|=1$ for every complex embedding. The coefficients of the monic polynomialare rational algebraic integers and hence integers. They are uniformly bounded in $m$, because each is an elementary symmetric sum of numbers of modulus one. Only finitely many such integer polynomials can occur, so only finitely many algebraic integers ${\alpha }^m$ occur. Two powers coincide; since $\alpha \ne 0$, this gives ${\alpha }^{r-s}=1$. Thus $\alpha$ is a root of unity. This is Kronecker theorem on algebraic integers in the unit disk.

The second printed assertion is literally false for $\alpha =0$. For the intended statement with $\alpha \ne 0$, use the Minkowski embedding: algebraic integers of $K$ form a lattice, so only finitely many have all conjugates of modulus at most $2$. A nonzero algebraic integer with every conjugate of modulus at most one hasso every conjugate has modulus one and the first part applies. Consequently every non-root in that finite box has ${\max }_i|{\sigma }_i(\alpha )|>1$. Choose $c>1$ no larger than the least of these finitely many maxima, taking $c=2$ if there are no non-roots. Then every nonzero $\alpha \in {\mathcal{O}}_K$ satisfying $|{\sigma }_i(\alpha )|<c$ for all $i$ is a root of unity.

Dirichlet unit theorem states that for signature $(r_1,r_2)$,and the logarithmic embedding of number field units maps the free part to a full lattice in the hyperplane where the weighted coordinate sum is zero.

For a real quadratic field, the roots of unity are $\{\pm 1\}$ and the logarithms of positive units form a nonzero discrete subgroup $a\mathbb{Z}\subset \mathbb{R}$. Its smallest positive element is $\log ϵ$ for a unit $ϵ>1$. Given any positive unit $u>1$, choose $n$ so that $1\le u{ϵ}^{-n}<ϵ$; minimality forces the quotient to be $1$. Applying signs and inverses shows

Since $11\equiv 3(\mathrm{mod}4)$, the ring of integers of a quadratic field is $\mathbb{Z}[\sqrt{11}]$. A unit $a+b\sqrt{11}$ satisfies the Pell equation $a^2-11b^2=\pm 1$. The elementhas norm $1$. To prove it is the smallest unit above one without invoking the continued-fraction algorithm, suppose $1<a+b\sqrt{11}<10+3\sqrt{11}$. Replacing by its inverse or negative if needed makes $b>0$, and the conjugate relation implies $b\sqrt{11}<(u+u^{-1})/2<10$, so $b\le 3$. For $b=1,2,3$, the numbers $11b^2\pm 1$ are respectivelyOnly the last list contains a square, namely $100=10^2$, which yields the endpoint itself. Hence no smaller unit exists and

Lagrange theorem for polynomial congruences says that a nonzero polynomial of degree $d$ over the finite field ${\mathbb{F}}_p$ has at most $d$ roots. To prove it, use mathematical induction on $d$. A constant nonzero polynomial has no roots. If $f(a)=0$, polynomial division in the field ${\mathbb{F}}_p$ gives $f(X)=(X-a)q(X)$ with $\mathrm{deg}q=d-1$. Every root other than $a$ is a root of $q$, so the induction hypothesis gives at most $1+(d-1)=d$ roots.

Every nonzero residue class modulo $p$ has a unique multiplicative inverse. In the product $(p-1)!$, pair each class with its inverse. The unpaired classes satisfy $x=x^{-1}$, hence $x^2-1=0$. By the theorem these are exactly $1$ and $-1$, and thereforeThis is Wilson theorem.

Now write $p-1=kq$. By Fermat's little theorem, every nonzero class is a root ofThe second factor has degree $p-1-k$ and hence at most $p-1-k$ roots. Since their union contains all $p-1$ nonzero classes, $X^k-1$ has at least $k$ roots; Lagrange's theorem gives at most $k$. Thus $x^k\equiv 1(\mathrm{mod}p)$ has exactly $k$ solutions.