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This ternary quadratic form has a nonzero zero over ${\mathbb{Q}}_p$ exactly when $p\ne 7$. Away from $2,5,7,13$, the finite-field square-set intersection proof supplies a nonsingular residue zero, and Hensel lemma lifts it. At 5 use $(0,1,1)$ modulo 5; at 13 use $(3,1,0)$. At 2 hold $x=2,z=1$ and apply the strong form of Hensel lemma to $7Y^2+33$ at $Y=1$: the function and derivative have valuations 3 and 1. At 7, the equation modulo 7 forces $x=z=0$ because 5 is not a square; reduction after dividing by 7 then forces $y=0$. Any normalized integral nonzero solution would therefore have all coordinates divisible by 7, a contradiction.

Starting from $a_0$, Newton iteration sets $a_{n+1}=a_n-f(a_n)/f^{\prime }(a_n)$. Under the strong form of Hensel lemma inequality, the correction valuations grow at least geometrically, so $(a_n)$ is a Cauchy sequence converging to a root.

For a local field $F$ containing the Nth roots of unity, fix the Local Artin map with arithmetic Frobenius convention. The Hilbert norm residue symbol can be defined byIt is independent of the chosen root, continuous and bimultiplicative, factors through $F^{\times }/(F^{\times }{)}^N$ in each variable, and gives a nondegenerate pairing on that finite group. It satisfies $(a,b{)}_{F,N}=(b,a{)}_{F,N}^{-1}$ and $(a,1-a{)}_{F,N}=1$ for $a\ne 1$. Its norm interpretation isFor $N=2$ the quadratic Hilbert symbol is symmetric and sign-valued. It is also characterized by the existence of a nonzero solution of $z^2=a{x}^2+b{y}^2$. Moreover $(a,-a{)}_F=1$, and therefore $(a,a{)}_F=(a,-1{)}_F$. We now compute both the square-class groups and the pairing explicitly.

For odd $p$, write $a=p^{\alpha }u$ with $u\in {\mathbb{Z}}_p^{\times }$. An element is a square exactly when $\alpha$ is even and the residue of $u$ is a square in ${\mathbb{F}}_p^{\times }$. Necessity follows by reduction; sufficiency follows from the simple-root form of Hensel lemma applied to $X^2-u$, since $2x$ is then a unit. If $\eta$ is any unit with nonsquare residue, this givesThe nonsquare unit defines an unramified quadratic extension. Its field norms include every unit and have even valuation, as follows from the local norm calculation below. Thus $(\eta ,u{)}_p=1$ for every unit, but $(\eta ,p{)}_p=-1$. Finally $(p,p{)}_p=(p,-1{)}_p=(-1/p)$, by symmetry and the preceding unit calculation. Bimultiplicativity now proves explicit quadratic Hilbert symbols over the p-adic numbers:Here the factors are Legendre symbols of the nonzero residues. The formula works for arbitrary integer valuations $\alpha ,\beta$, since only their parity matters.

At $p=2$, the square of every odd integer is one modulo eight. Conversely, for an odd $u\equiv 1(\mathrm{mod}8)$, apply the strong form of Hensel lemma to $X^2-u$ at $1$: the error has valuation at least three, while the derivative has valuation one. Thus the unit square classes of the p-adic integers are precisely their four residues modulo eight. Combining this with parity of the valuation givesFor an odd unit putInspection of the residues $1,3,5,7$ gives $[u]=[-1{]}^{ϵ(u)}[5{]}^{\omega (u)}$. It remains to compute six basis pairings. Explicit field norms giveindeed $2=1^2+1^2$, $5=1^2+2^2$, $2=2^2-2\cdot 1^2$, and $5=5^2-5\cdot 2^2$.

In contrast $(-1,-1{)}_2=-1$. To justify the obstruction even for rational 2-adic coordinates, write $x=2^t{x}_0$, $y=2^t{y}_0$ with $x_0,y_0$ integral and at least one odd. Then $v_2(x_0^2+y_0^2)$ is zero if exactly one is odd, and one if both are odd. A sum of squares equal to the unit $-1$ consequently forces $t=0$ and exactly one coordinate odd. Its residue modulo four would then be one, whereas $-1$ is three.

Similarly $(2,5{)}_2=-1$. If $x^2-2y^2$ is a unit, the valuations of the two terms have opposite parity, so cannot cancel. It follows that $x$ is an odd integral element and $y$ is integral. Modulo eight its norm is then $1$ or $7$, never $5$. These signs, symmetry, and bimultiplicativity giveThe two boxed formulas determine the quadratic Hilbert symbol on all nonzero elements, including every rational prime $p$.

Write $\mathcal{O}$ for the valuation ring of $K$, $\pi$ for a uniformizer, $v(\pi )=1$, and $k={\mathbb{F}}_q$ for the residue field. A useful strong form of Hensel lemma is the following: if $f\in \mathcal{O}[X]$ and $a_0\in \mathcal{O}$ satisfythen there is a root $a\in \mathcal{O}$ with $v(a-a_0)>v(f^{\prime }(a_0))$, unique among roots in that ball. If $f(a_0)\ne 0$, its distance satisfies $v(a-a_0)=v(f(a_0))-v(f^{\prime }(a_0))$.

Here is a proof by successive correction. Put $s=v(f^{\prime }(a_0))$, $d_j=v(f(a_j))$, andStop if a root has already been reached. Otherwise the correction has valuation $d_j-s>s$. Integral Taylor expansions show that $v(f^{\prime }(a_{j+1}))=s$ andThus $d_j-2s$ grows at least geometrically, the corrections tend to zero, and completeness gives a limit $a$ with $f(a)=0$. The first correction has strictly smaller valuation than every later nonzero correction, proving the asserted distance. If $x,y$ are in the ball $v(x-a_0),v(y-a_0)>s$, thenThe second factor cannot vanish. Two roots in the ball must therefore coincide. This proves Hensel lemma; in particular a simple root modulo $\pi$ has a unique lift with that residue.

We use this to establish existence and uniqueness of unramified local extensions. For each integer $d\ge 1$, choose a monic irreducible polynomial $\bar{f}\in k[X]$ of degree $d$ and lift it to a monic $f\in \mathcal{O}[X]$. Reduction and Gauss lemma show that $f$ remains irreducible over $K$. LetThe ring $A$ is finite free and complete over $\mathcal{O}$, and $A/\pi A\cong {\mathbb{F}}_{q^d}$. An element outside $\pi A$ is invertible: lift an inverse from the residue field and invert the remaining $1+\pi z$ by a convergent geometric series. Every nonzero element of $A$ is ${\pi }^t$ times such a unit, since ${\bigcap }_t{\pi }^{t}A=0$ in a finite free module. Thus $A$ is a discrete valuation ring, its fraction field is $L$, andThis constructs an unramified extension of each degree.

Conversely, let $M/K$ be an unramified extension of degree $d$. Its residue field is the unique degree-$d$ finite field extension ${\mathbb{F}}_{q^d}/{\mathbb{F}}_q$. Choose a residue root of $\bar{f}$ in it. The root is simple, so Hensel lemma lifts it to a root of $f$ in $M$. This embeds $L$ into $M$, and equality of degrees makes it an isomorphism. All $d$ residue conjugates lift in $L$ itself; hence $L/K$ is Galois, and reduction identifies its Galois group with the cyclic Galois group of the residue field extension. Its distinguished Frobenius automorphism reduces to $x\mapsto x^q$.

There is exactly one unramified extension $K_d$ of each degree inside a fixed algebraic closure, andFor the middle equality, every nonzero element of ${\mathbb{F}}_{q^d}$ lifts uniquely to a $(q^d-1)$st root of unity, because that exponent is prime to $p$. A primitive such residue generates the finite field extension, so its lift generates $K_d$. Taking the union gives the maximal unramified extension, with Galois group $\hat{\mathbb{Z}}$.

Use the strong form of Hensel lemma: for $f\in {\mathbb{Z}}_p[T]$ and $a\in {\mathbb{Z}}_p$, ifthere is a unique polynomial root $b$ in the ball $v_p(b-a)>v_p(f^{\prime }(a))$. In particular, a simple polynomial root modulo $p$ lifts uniquely in its residue class. Newton iteration gives $v_p(b-a)=v_p(f(a))-v_p(f^{\prime }(a))$ when $f(a)\ne 0$.

For odd $p\notin \{5,7,13\}$, the setseach have $(p+1)/2$ elements, so they intersect. This gives a zero modulo $p$ with $z=1$. The derivative with respect to $z$ is $26$, nonzero modulo $p$, so Hensel's lemma lifts $z$ while holding $x,y$ fixed. This is the elementary proof of isotropy of nondegenerate ternary quadratic forms over finite fields followed by a simple-root lift.

At $p=5$, the residue vector $(0,1,1)$ is a zero, and the derivative $14y$ is nonzero. At $p=13$, use $(3,1,0)$, for which the value is 52 and the derivative $10x=30$ is nonzero modulo 13. Again a single-variable Hensel lift suffices.

At $p=2$, hold $x=2,z=1$ and use $f(Y)=7Y^2+33$. At $Y=1$, $f(1)=40$ has valuation 3, while $f^{\prime }(1)=14$ has valuation 1. Since $3>2$, the strong Hensel inequality gives a polynomial root $y\in {\mathbb{Z}}_2$, indeed $y\equiv 1(\mathrm{mod}4)$. This yields a nonzero solution.

At $p=7$, suppose a nonzero solution exists and scale it so that all coordinates are in ${\mathbb{Z}}_7$ and at least one is a unit. Reduction modulo 7 gives $z^2=5x^2$. Since 5 is not a square modulo 7, both $x,z$ are divisible by 7. In the original equation their terms are then divisible by 49, so $7y^2$ is divisible by 49 as well, forcing $y$ divisible by 7. This contradicts the normalization. Thus local isotropy of the five seven thirteen form gives

A useful version is the strong form of Hensel lemma. Let $K$ be complete for a Non-Archimedean absolute value, let $f\in {\mathcal{O}}_K[X]$, and suppose $a\in {\mathcal{O}}_K$ satisfiesThen there is a unique root $\alpha$ in the ball $|\alpha -a|<|f^{\prime }(a)|$, and $|\alpha -a|\le |f(a)/f^{\prime }(a)|$.

Put $h=|f^{\prime }(a)|$ and $\eta =|f(a)|/h^2<1$. Starting with $x_0=a$, apply Newton iteration over a valued field $x_{n+1}=x_n-f(x_n)/f^{\prime }(x_n)$. Taylor expansion over the valuation ring givesfor integral $x,u$. Inductively $|f^{\prime }(x_n)|=h$ and $|f(x_n)|/h^2\le {\eta }^{2^n}$. Indeed the Newton increment has size at most $h{\eta }^{2^n}<h$, so the derivative size is unchanged, and the new function value is bounded by the square of the increment. The increments tend to zero, and the ultrametric inequality makes the sequence Cauchy. Completeness supplies a root with the stated distance bound.

If $b,c$ are distinct points of the ball, their divided difference is $f^{\prime }(a)$ plus a term of size at most $\max (|b-a|,|c-a|)<h$, so $(f(b)-f(c))/(b-c)$ is nonzero. Two roots in that ball are impossible. This proves the stated Hensel lemma. In particular a simple root modulo the maximal ideal lifts uniquely to a root in its residue class.

For $p=2$, the even residue class contains exactly one root, by the simple-root form of Hensel lemma at $a=0$: $f(0)=20$ and $f^{\prime }(0)=-5$ is a unit. For the odd classes use the strong form of Hensel lemma:In both cases $v_2(f(a))>2v_2(f^{\prime }(a))=2$. Each gives exactly one root in $a+4{\mathbb{Z}}_2$. These two balls partition the odd integers, so there are no further roots.

For $p=3$, reduction gives $\overline{f}=X^3+X+2$, whose sole root is $2$. The derivative $3X^2-5$ is always a unit modulo $3$, so that root lifts uniquely. For $p=5$, the only possible residue is zero. But if $x\in 5{\mathbb{Z}}_5$, then $x^3-5x\in 25{\mathbb{Z}}_5$ while $20$ has valuation one, so $f(x)$ cannot vanish. The exact root counts are

For each finite place $v\in S$, choose $d_v\in K_v^{\times }$ such that $K_v(\sqrt{d_v})/K_v$ is the unramified quadratic extension. Such a choice exists even at residue characteristic two. At odd residue characteristic use a unit with nonsquare residue. At residue characteristic two, choose $\overline{c}$ for which $X^2-X-\overline{c}$ is irreducible over the finite residue field, lift $c$ to ${\mathcal{O}}_v$, and set $d_v=1+4c$. The lifted polynomial has unit derivative at every residue point and gives the unramified degree-two extension; its root is $(1+\sqrt{d_v})/2$.

The square classes are open in $K_v^{\times }$. Indeed the strong form of Hensel lemma applied to $X^2-u$ at $1$ shows that $u$ is a square whenever $v(u-1)>2v(2)$. Thus any sufficiently close approximation $a$ to $d_v$ has $a/d_v$ square. By weak approximation for number fields, choose $a\in K$ satisfying this at every finite $v\in S$. Then the local extensions $K_v(\sqrt{a})$ are all unramified of degree two. At least one such condition ensures $a$ is not a square in $K$. If $S$ contains no finite place, impose the same condition at one auxiliary finite place.

Set $L=K(\sqrt{a})$. At every specified finite prime there is one prime of $L$ above it, with ramification index one and residue degree two. Hence $\boxed{\mathfrak{p}{\mathcal{O}}_L\text{ is prime for every finite }\mathfrak{p}\in S}$. The dyadic case is covered by the unramified construction, rather than by a nonexistent nonsquare residue unit in characteristic two.

We count extensions up to $K$-isomorphism. An extension of degree $n$ has a maximal unramified extension $E/K$ inside it, with $[E:K]=f\mid n$, and $L/E$ is totally ramified of degree $e=n/f$. The unramified extension of each degree is unique, so there are only finitely many possibilities for $E$.

Fix $E$ and $e$. The coefficient space of monic degree-$e$ Eisenstein polynomials is compact: its nonconstant nonleading coefficients lie in the compact maximal ideal, and its constant coefficient lies in the compact set ${\mathfrak{m}}_E\setminus {\mathfrak{m}}_E^2$. Each such polynomial $F$ has a root $\alpha$ generating a degree-$e$ extension. For sufficiently close coefficients of a polynomial $G$, the numbers $G(\alpha )$ and $G^{\prime }(\alpha )-F^{\prime }(\alpha )$ are arbitrarily small, while $F^{\prime }(\alpha )\ne 0$ because the characteristic is zero. The strong form of Hensel lemma, applied in $E(\alpha )$, supplies a root $\beta$ of $G$ as close to $\alpha$ as desired. Choose the neighborhood so that the Krasner's lemma inequality holds. Then $E(\alpha )\subseteq E(\beta )$, and both degrees are $e$, so the extensions are equal. Compactness gives a finite cover by such neighborhoods, proving finiteness of degree-$n$ extensions. If extensions are realized as subfields of a fixed algebraic closure rather than counted up to isomorphism, each has only finitely many embeddings, so that count is finite too.

For odd $p$, write $K^{\times }={\pi }^{\mathbb{Z}}{\mathcal{O}}_K^{\times }$. The square map is bijective on $1+{\mathfrak{m}}_K$ by Hensel lemma, and the finite residue field has cyclic multiplicative group of even order. ThusThe quadratic extensions from square classes correspond to its three nontrivial classes. Representatives are $u,\pi ,u\pi$, where $u$ is a unit with nonsquare residue. There are $\boxed{3}$ quadratic extensions: $K(\sqrt{u})$ is unramified, while $K(\sqrt{\pi })$ and $K(\sqrt{u\pi })$ are totally ramified. For completeness, two nonsquare parameters give the same extension only if their ratio is a square: writing $\sqrt{b}=x+y\sqrt{a}$ and squaring forces $x=0$, hence $b/a=y^2$.

A strong lifting lemma. Let $R$ be a complete discrete valuation ring, with fraction field $F$ and normalized discrete valuation $v$. A useful strong form of Hensel lemma says that for $f\in R[T]$ and $a\in R$ satisfyingthere is a root $b\in R$ withIf $f(a)=0$, take $b=a$. Otherwise the hypothesis ensures that $f^{\prime }(a)\ne 0$. The root is unique among $b$ satisfying $v(b-a)>v(f^{\prime }(a))$. In particular, a simple root of the reduction of $f$ lifts uniquely with its specified residue.

Here is a proof by Newton iteration over a valued field. Put $a_0=a$ andWrite $A_j=v(f(a_j))$ and $B=v(f^{\prime }(a_0))$. Taylor expansion over $R$ shows, inductively, that the increment has valuation $A_j-B>B$, that $v(f^{\prime }(a_{j+1}))=B$, and thatIndeed, the constant and linear terms of $f(a_{j+1})$ cancel, leaving terms of valuation at least twice that of the increment. The change in $f^{\prime }$ has valuation greater than $B$, so its valuation stays $B$. Thus all iterates lie in $R$ and form a Cauchy sequence. Completeness and continuity give a root $b$. The later increments have strictly larger valuations than the first, proving the displayed distance formula. For uniqueness, if $b,c$ are in the stated ball, Taylor expansion factorsThe second factor is nonzero, so two roots must coincide. This proves Hensel's lemma in the form needed here.

Odd residue characteristic. For odd $p$, a p-adic unit $u$ is a square exactly when its reduction is a square in ${\mathbb{F}}_p^{\times }$. Necessity follows by reduction. For sufficiency, a nonzero residue root of $T^2-u$ has derivative $2T$ invertible modulo $p$, so the Hensel lemma lifts it. The multiplicative group of a finite field is cyclic, and its subgroup of squares has index two. ThereforeRepresentatives are $1$ and any unit whose reduction is a quadratic nonresidue.

Residue characteristic two. An odd integer has square congruent to one modulo eight, so a square 2-adic unit must lie in $1+8{\mathbb{Z}}_2$. Conversely, if $u\equiv 1(\mathrm{mod}8)$, take $f(T)=T^2-u$ and $a=1$. Then $v_2(f(1))\ge 3>2v_2(f^{\prime }(1))=2$, and the strong form of Hensel lemma produces a square root. HenceThe four unit square classes of the p-adic integers have representatives $1,3,5,7$.

A number-field completion is finite. Let $F$ be the given number field and $L=F_{\mathfrak{p}}$ its completion. Normalize its absolute value on a field so that the restriction to $\mathbb{Q}$ is $|\cdot {|}_p$: if $v_{\mathfrak{p}}(p)=e$, use $|x|=p^{-v_{\mathfrak{p}}(x)/e}$. Changing the original absolute value by this positive power does not change its completion. The closure of $\mathbb{Q}$ in $L$ is an isometric copy of ${\mathbb{Q}}_p$.

Take a $\mathbb{Q}$-basis of $F$ and let $V$ be its ${\mathbb{Q}}_p$-span in $L$. Choose a maximal linearly independent subset of that finite spanning set. The compact-sphere proof of finite-dimensional non-Archimedean norm equivalence applies to this basis too, so $V$ is complete and therefore closed in $L$. Since $F\subset V$ is dense in $L$, it follows that $V=L$. Hence

Every finite extension is obtained this way. Let $E/{\mathbb{Q}}_p$ be finite. Use its extending Non-Archimedean absolute value, whose completeness and uniqueness were proved above. In characteristic zero the primitive element theorem gives $E={\mathbb{Q}}_p(\alpha )$; scaling $\alpha$ by a power of $p$, we may arrange that it is integral. Its monic minimal polynomial $f$ belongs to ${\mathbb{Z}}_p[T]$. If the degree is one, $E={\mathbb{Q}}_p$ is already the completion of $\mathbb{Q}$.

Otherwise choosewhere ${\alpha }^{\prime }$ runs over the conjugates in an algebraic closure. Approximate the coefficients of $f$ by integers, obtaining a monic $g\in \mathbb{Z}[T]$ of the same degree. Since $f^{\prime }(\alpha )\ne 0$, sufficiently accurate approximation gives $|g(\alpha )|<|g^{\prime }(\alpha ){|}^2$ and makes $|g(\alpha )/g^{\prime }(\alpha )|<\delta$. The strong form of Hensel lemma in $E$ produces a root $\beta \in E$ of $g$ with $|\beta -\alpha |<\delta$.

For completeness, the required Krasner's lemma follows from the unique extension of an absolute value to a finite extension. In a finite Galois extension containing $\alpha ,\beta$, every automorphism $\sigma$ fixing $\beta$ is an isometry, andIf $\sigma \alpha \ne \alpha$, this contradicts the choice of $\delta$. Thus every automorphism fixing $\beta$ fixes $\alpha$, and Galois correspondence gives ${\mathbb{Q}}_p(\alpha )\subseteq {\mathbb{Q}}_p(\beta )$. Since $\beta \in E$, equality follows. In particular $g$ is irreducible over ${\mathbb{Q}}_p$, and therefore over $\mathbb{Q}$.

Put $F=\mathbb{Q}(\beta )\subset E$. It is a number field, and the closure of $F$ in $E$ contains ${\mathbb{Q}}_p(\beta )=E$, so $F$ is dense. LetEvery element of the ring of integers of a number field has absolute value at most one: otherwise its highest-degree term would strictly dominate all other terms of a monic integer equation. Thus reduction into the residue field of $E$ makes $\mathfrak{p}$ a prime ideal, and $\mathfrak{p}\cap \mathbb{Z}=p\mathbb{Z}$. The restriction of the valuation to $F$ is the valuation of this prime, up to normalization. Indeed, the localized discrete valuation ring $({\mathcal{O}}_F{)}_{\mathfrak{p}}$ maps into ${\mathcal{O}}_E$, its units have absolute value one, and its uniformizer has absolute value less than one. Density and completeness giveThis proves the local realization as a completion of a number field.

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

Assume $|a{|}_2>1$, equivalently $v_2(a)<0$, and writeFor $f(Y)=Y^2-(1+4/a^4)$ at $Y=1$,The strong form of Hensel lemma therefore gives $y\in {\mathbb{Q}}_2$ with $y^2=1+4/a^4$. Taking $X=a^{2}y$ yields $X^2=a^4+4$. Hence every $a$ with $|a{|}_2>1$, and in particular every sufficiently large $a$, works.

At finitely many finite places of a number field, choose square-class parameters for the local unramified quadratic extensions. Local square classes are open, including at dyadic places by the strong form of Hensel lemma. Weak approximation for number fields gives a global parameter in all the prescribed local classes. Its quadratic extension is unramified of residue degree two at those places, so their extended prime ideals remain prime. If no finite place is prescribed, an auxiliary local nonsquare condition ensures the global extension is nontrivial.

For odd $p$, a p-adic unit is a square exactly when its residue is a square, by the Hensel lemma. Its unit square-class group is $C_2$. For $p=2$, a unit is a square exactly when it is congruent to one modulo eight, by the strong form of Hensel lemma. The four square classes are represented by $1,3,5,7$, giving $C_2\times C_2$.