Completions and p-adic fields

For nonzero rational x, write $x=p^ku/v$ with integers u,v not divisible by p and define $v_p(x)=k$. Set $v_p(0)=\infty$ and $|x|_p=p^{-v_p(x)}$, with $|0|_p=0$. The valuation is additive under multiplication and satisfies $v_p(x+y)\ge\min(v_p(x),v_p(y))$. Hence the absolute value is nonarchimedean: the size of a sum is at most the larger input size. If the two input valuations differ, cancellation cannot raise the smaller one, so equality holds.

Cauchy sequences for the resulting distance can fail to converge in Q, just as real Cauchy sequences can. Their completion is Q_p, and the field operations extend continuously. The p-adic integers Z_p consist of elements of nonnegative valuation. Its units have valuation zero, its maximal ideal is pZ_p, and its residue field is F_p. This local ring is different from the rational localization Z_(p), which is not complete, even though both have the same rational valuation and residue field.

Every p-adic integer has a unique expansion $\sum_{j\ge0}a_jp^j$ with digits from zero through p−1. Truncating after N digits determines its residue modulo $p^N$. The inverse-limit description uses compatible residues at every precision, not independent choices of residues at unrelated moduli. A p-adic field element can have finitely many negative powers as well. Infinite positive powers converge because their valuations grow without bound, rather than because their real values become small.

For a convergent geometric series, the usual finite-sum identity still determines the limit when interpreted in this metric. The same formal series may diverge over the reals. Negative rational numbers can therefore have endless nonnegative p-adic digit expansions, and that does not make them positive in an ordered-field sense; the p-adic field has no ordering compatible with this digit intuition. Congruence stability explains the approximation process more reliably than comparing the real sizes of the successive integers.

For a finite extension L of Q_p, one often normalizes its discrete valuation $v_L$ by $v_L(\pi_L)=1$ for a uniformizer. Then $v_L(p)=e$, the ramification index. The valuation extending the original $v_p$ is instead $v_L/e$, which can have fractional values. Confusing these conventions changes the stated value of a root of p and can make the identity degree equals ramification times residue degree appear inconsistent. State which valuation is in use before comparing base and extension elements.

The residue field of L has p to the f elements, where f is its residue degree, and the extension degree is ef. The extended absolute value with the base p-adic normalization is p raised to minus v_L divided by e. A residue field of p squared elements is not generally the ring Z/p²Z: the latter has a nonzero nilpotent class p, while the former is a field. Distinguish extension degree, residue-field size, and finite-precision residue rings throughout local computations.