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The Dedekind factorization theorem says that, when a rational prime $p$ does not divide the index of an order $\mathbb{Z}[\theta ]\subseteq {\mathcal{O}}_K$, a factorizationof the minimal polynomial modulo $p$ gives

Let $K=\mathbb{Q}(\sqrt{d})$ with square-free $d$, and let $p$ be odd. The resulting splitting of rational primes in a quadratic field is:

Finally suppose $d>0$ and
$\alpha =A/C+(B/C)\sqrt{d}$ has norm $-1$, with integers
$A,B,C$ chosen coprime. ThenIf an odd $p\equiv 3(\mathrm{mod}4)$ ramified, then $p\mid d$. Reduction modulo $p$ gives
$A^2\equiv -C^2(\mathrm{mod}p)$. Since $-1$ is not a square modulo such a prime, $p\mid A,C$. Dividing the equation's divisibility shows $p\mid B$ as well (use that square-free $d$ has $p$-adic valuation one), contradicting coprimality. Hence no prime $p\equiv 3(\mathrm{mod}4)$ ramifies.

The binomial theorem givesThe central binomial coefficient is the largest of these $2n+1$ nonnegative binomial coefficients, so

For the upper bound, fix a prime number $p\le 2n$. By Legendre formula,Each summand is either zero or one. Hence the P-adic valuation satisfiesIf $p\le \sqrt{2n}$, this impliesThere are at most $\sqrt{2n}$ such prime numbers. If instead $p>\sqrt{2n}$, then $p^2>2n$, so the P-adic valuation is at most one. Multiplying the prime-power contributions therefore yields

The strong form of Hensel lemma says the following. Let $v$ be a discrete valuation on a complete field $K$. If $f\in {\mathcal{O}}_K[X]$ andthen $f$ has a root $x\in {\mathcal{O}}_K$ satisfying $v(x-x_0)>v(f^{\prime }(x_0))$.

To prove it, apply Newton iteration over a valued field:The initial inequality says that the first correction has valuation greater than $v(f^{\prime }(x_0))$. Taylor expansion then shows inductively that $v(f^{\prime }(x_n))=v(f^{\prime }(x_0))$, while the valuations of the corrections tend to infinity. Hence $(x_n)$ is a Cauchy sequence. Completeness gives a limit $x$, and continuity gives $f(x)=0$.

Now decompose the multiplicative group asThe P-adic valuation givesand cubing is the identity on $\{\pm 1\}$. Put $U_2=1+9{\mathbb{Z}}_3$. Expansion gives $U_1^3\subseteq U_2$. Conversely, for $u=1+9a\in U_2$, choose $b\equiv a(\mathrm{mod}3)$ and put $x_0=1+3b$. For $f(X)=X^3-u$,so the strong form of Hensel lemma produces a cube root in $U_1$. Thus $U_1^3=U_2$. Finally,is an isomorphism. Combining the valuation and principal-unit factors proves the cube-class group of the 3-adic numbers identity

Let $c_1,\dots ,c_n$ be the columns of the rational partition regular matrix $A$, and clear denominators so that they are integer vectors. We use the P-adic columns lemma. For a large prime number $p$, color each positive integer by a sufficiently long initial block of the unit part of its P-adic valuation, together with its valuation modulo the block length. Partition regularity supplies a monochromatic $x=(x_1,\dots ,x_n)$ withGroup the indices according to the successive $p$-adic orders of the $x_i$. At the lowest order, division by the common power of $p$ and reduction modulo the chosen large power showsComparing the next nonzero blocks of base-$p$ digits shows successively thatFor completeness, these congruences may be made exact by taking the digit block longer than every determinant and coordinate formed from the fixed columns: a nonzero such integer cannot be divisible by the resulting power of $p$. There are only finitely many ordered partitions of $[n]$, so passing through arbitrarily long blocks leaves one partition satisfying all the displayed identities. This is precisely the columns property.

For a prime $p$ and a positive integer $N$, the P-adic valuation is