Ivan Shishkin, Birch Grove

pp-adic valuation

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Let nn be a nonzero integer and pp a prime integer. We define vp ⁣:ZNv_p\colon \Z^*\longrightarrow \N the pp-adic valuation of nn.

If pp is a not a prime factor of nn, one defines vp(n):=0v_p(n):=0.

Now say that pp is a prime factor of nn, then define vp(n):=max{iN/pin}v_p(n):=\max\{i\in\N^*/ p^i\mid n\} to be the maximal power of pp that divides nn.

Remarks
\bullet One can generalize this definition to non trivial rationnal numbers by setting vp(ab)=vp(a)vp(b)v_p(\frac{a}{b})=v_p(a)-v_p(b). Note that this definition does not depend on (a,b,c,d)(Z)4(a,b,c,d)\in(\Z^*)^4 such that ab=cd\frac{a}{b}=\frac{c}{d}.
\bullet With this definition, the fundamental theorem of arithmetic can be written as
nN,n=pprimepvp(n)\forall n\in\N^*, n=\prod_{p\,\,\text{prime}}p^{v_p(n)}and this equality can be generalized to all non vanishing rationnal numbers.

Examples
\bullet The 22-adic valuation of 88 is 33 since 8=238=2^3.
\bullet The 33-adic valuation of 2424 is 11 since 24=23324=2^3\cdot 3.
\bullet The 55-adic valuation of 1025\displaystyle\frac{10}{25} is v5(10)v5(25)=v5(52)v5(52)=12=1v_5(10)-v_5(25)=v_5(5\cdot2)-v_5(5^2)=1-2=-1.

Practice this concept with exercises

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  • Let us assume that 2\sqrt{2} is a rationnal number. Set a,ba,b two coprime integers such that 2=ab\sqrt{2}=\frac{a}{b}.

    Show that this statement is absurd.

    Open exerciseDifficulty 18/100 · 1 solution · 1 hint
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