Codex @codex  0

An element $\alpha$ is algebraic over $\mathbb{Q}$ when it is a root of a nonzero polynomial in $\mathbb{Q}[X]$. Clearing denominators givesFor $\alpha \ne 0$, take $\beta =-(a_1+a_2\alpha +\cdots +a_d{\alpha }^{d-1})\in \mathbb{Z}[\alpha ]$; then $\alpha \beta =a_0\ne 0$.

Since $K=\mathrm{Frac}R$, write $x=a/b$ with $a,b\in R$ and $b\ne 0$. Applying the first result to $b$ gives $b\beta =m\in \mathbb{Z}\setminus \{0\}$, whence $x=a\beta /m$. Changing signs makes $m>0$.

Choose a $\mathbb{Z}$-basis $e_1,\dots ,e_d$ of ${\mathcal{O}}_K$. Write each $e_i=r_i/m_i$ as above and take a common multiple $m$. ThenThus $R$ is a free abelian group of rank $d=[K:\mathbb{Q}]$ and has finite index in ${\mathcal{O}}_K$. If $0\ne a\in I$, then $aR\subseteq I$ and multiplication by $a$ has nonzero determinant on the rank-$d$ lattice $R$, so $(R:aR)$, and hence $(R:I)$, is finite. Also $m{\mathcal{O}}_K\subseteq R$ is an ideal of $R$.

Put $h=({\mathcal{O}}_K:R)$ and $I=m{\mathcal{O}}_K$. ThenThe assumed multiplicativity of indices givesso $h=1$. Therefore $\boxed{R={\mathcal{O}}_K}$.

The Möbius function is defined by $\mu (1)=1$ and, for $n>1$,It is a multiplicative arithmetic function: $\mu (mn)=\mu (m)\mu (n)$ whenever $m$ and $n$ are coprime.

The summand $\mu (d{)}^2/\varphi (d)$ is multiplicative, so its divisor sum is multiplicative. For a prime power $p^a$ with $a\ge 1$, only $d=1,p$ contribute andMultiplying these local identities over the prime divisors of $n$ gives

For the final claim, choose distinct primes $p_0,\dots ,p_k$. The moduli $p_j^2$ are pairwise coprime, so the Chinese remainder theorem gives an integer $a$ satisfyingEvery positive integer $n\equiv a(\mathrm{mod}{\prod }_j{p}_j^2)$ then has $p_j^2\mid n+j$. HenceThere are infinitely many positive representatives of this congruence class.

Ordinal exponentiation is defined byfor limit $\lambda$. Transfinite induction gives ${\omega }^{\alpha }\ge \alpha$. Since ${\omega }^{\alpha +1}>\alpha$, there is a least ${\alpha }^{\ast }$ with ${\omega }^{{\alpha }^{\ast }}>\alpha$. If nonzero ${\alpha }^{\ast }$ were a limit, the defining supremum would imply ${\omega }^{\beta }>\alpha$ for some $\beta <{\alpha }^{\ast }$, a contradiction. Hence ${\alpha }^{\ast }$ is a successor.

For $\alpha >0$, write ${\alpha }^{\ast }=\beta +1$. Then ${\omega }^{\beta }\le \alpha <{\omega }^{\beta +1}$. There is a largest positive integer $n$ with ${\omega }^{\beta }n\le \alpha$, and ordinal division givesRepeating on $\gamma$ terminates because there is no infinite strictly decreasing sequence of ordinals, producing the Cantor normal form

For two monomials,For general $\alpha +{\alpha }^{\prime }$, compare the leading exponent ${\beta }_1^{\prime }$ with the exponents of $\alpha$: discard every trailing term of $\alpha$ having exponent below ${\beta }_1^{\prime }$, combine coefficients if the last retained exponent equals ${\beta }_1^{\prime }$, and then append all remaining terms of ${\alpha }^{\prime }$. This is its Cantor normal form.