## Cyclotomic integers

The object here is to prove that the ring of integers in the $n$-th cyclotomic number field $\Q(\zeta_n)$ is its subring $\Z[\zeta_n]$. Here $\zeta_n$ denotes a primitive $n$-th root of unity in $\C$, e.g. $$\zeta_n=e^{\frac{2i\pi}{n}}.$$ Since $\zeta_n$ is a root of the polynomial $X^n-1$, which has integral coefficients, $\zeta_n$ is integral over $\Z$. This shows that $\Z[\zeta_n]$ is contained in the ring of integers of $\Q(\zeta_n)$. What we need to prove is that any algebraic integer in $\Q(\zeta_n)$ belongs to $\Z[\zeta_n]$. The strategy is to prove it first for $n$ a prime power, and then deduce the general case from this. We need some preliminary statements.

Let $n$ be any natural integer and let, in order to simplify notation, $\zeta$ be a primitive $n$-th root of unity. Let $i$ and $j$ be integers relatively prime to $n$. Then $$\frac{1-\zeta^i}{1-\zeta^j}$$ is a unit of $\Z[\zeta]$. Indeed, since $j$ is relatively prime with $n$, there is an integer $k$ such that $jk\equiv i\pmod n$ by Bezout's Theorem. Since $\zeta^n=1$, one has $$\frac{1-\zeta^i}{1-\zeta^j}= \frac{1-(\zeta^j)^k}{1-\zeta^j}= 1+\zeta^j+(\zeta^{j})^2+\cdots+(\zeta^j)^{k-1}.$$ This shows that the fraction in question belongs to $\Z[\zeta]$. By symmetry, its inverse also belongs to $\Z[\zeta]$. This proves that the fraction in question indeed is a unit of $\Z[\zeta]$.

## A topological explanation for the cone of a morphism of complexes

Let $\mathcal C$ be an additive category, and let $C(\mathcal C)$ be the category of complexes of objects of $\mathcal C$. Let $f\colon P\rightarrow Q$ be a morphism of complexes in $C(\mathcal C)$. Then, one defines a new complex, the cone of $f$, denoted by $\mathrm{cone}(f)$: \[  […]

## Elementary symmetric polynomials

Let $n$ be a natural integer. Let $R$ be a ring and let $R[X_1,\ldots,X_n]$ be the ring of all polynomials in $X_1,\ldots,X_n$ with coefficients in $R$. Let $S_n$ denote the symmetric group of all permutations of the set $\{1,2,\ldots,n\}$. The group $S_n$ acts on the ring  […]

Theorem. Let $K$ be a field of characteristic $p$, with $p$ prime. Let $\phi\colon K\rightarrow K$ be the Frobenius morphism defined by $\phi(x)=x^p$. Then $K$ is perfect if and only if $\phi$ is surjective.Proof. Suppose that $\phi$ is not surjective. Then there is an element $u\in […] Continue reading ## Separable polynomials and separable extensions Definition. Let \(K$ be a field and $\overline K$ an algebraic closure of $K$. Let $P\in K[X]$ be an irreducible polynomial. The polynomial $P$ is called separable if all roots of $P$ in $\overline K$ are simple. A field $K$ is perfect if all irreducible polynomials in $K[X]$ are  […]

We recall here Artin's Theorem giving a necessary and sufficient condition for a field extension to be generated by one element.Definition. Let $L/K$ be a field extension. The field extension is called simple or primitive if there is an element $\theta\in L$ such that $L=K[\theta]$.The word  […]