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A nontrivial additive valuation has rank one when its ordered value group is Archimedean and therefore embeds order-preservingly in $\mathbb{R}$. Exponentiating a real-valued representative gives a Non-Archimedean absolute value; changing the embedding rescales the valuation positively. Higher-rank ordered value groups cannot be represented by real-valued absolute values on a field.

For integral $m\ge 2$, suppose $E:y^2=x(x+1)(x+m^2)$ has good reduction at five or seven. At five, $m^2\equiv 4$; at seven, $m^2\equiv 2$ or $4$. Each resulting reduced elliptic curve has eight points. These restrictions follow from $\Delta =16m^4(m^2-1{)}^2$ and $c_4=16(m^4-m^2+1)$: if an odd prime divides $m(m^2-1)$, the j-invariant of an elliptic curve has negative valuation. The torsion-freeness of the formal group over Qp for odd p bounds the whole rational torsion by eight. The point $(m,m(m+1))$ doubles to $(0,0)$, and $(-1,0)$ is an independent 2-torsion point. These generate the stated group of order eight.

For a degree-$d$ polynomial, reflecting the usual coefficient-index axis replaces $(j,v(a_j))$ by $(d-j,v(a_j))$. The reflected lower polygon has slopes equal to root valuations, rather than their negatives. Horizontal lengths, and therefore multiplicities, are unchanged. This convention makes positive lower ramification groups breaks appear as positive slopes of a reduced ramification polygon.

For odd $p$, parity of valuation and the square class of the unit residue give a group $C_2\times C_2$, represented by $1,u,p,pu$ for a unit $u$ with nonsquare residue. For $p=2$, valuation parity and unit residue modulo eight give $C_2^3$, with generators $-1,5,2$. The unit criteria follow from the Hensel lemma, using its strong derivative version for $p=2$.

The valuation contributes one factor $C_2$. The unit decomposition of the 2-adic Gaussian field contributes ${\mu }_4/{\mu }_4^2\cong C_2$ and the additive quotient ${\mathbb{Z}}_2[i]/2{\mathbb{Z}}_2[i]\cong C_2^2$. Thus there are sixteen square classes and fifteen quadratic extensions over this field.

For a number field $K$ and a finite set $S$ of finite primes, an S-unit is an element of $K^{\times }$ whose discrete valuation is zero at every prime outside $S$. Its group is the unit group of the ring obtained by allowing denominators at $S$. The Dirichlet unit theorem and the valuation homomorphism to ${\mathbb{Z}}^S$ show that this group is a finitely generated abelian group.

A nonidentity rational point in the formal kernel of a minimal Weierstrass equation at two distinct primes has infinite order of a group element. Indeed, the prime-to-residue-characteristic multiplication on a formal group implies that a finite order in the kernel at $p$ must be a power of $p$. Membership in the kernel at a different prime $q$ also forces a power of $q$. The only common possibility is order one, contradicting nonidentity. For an integral model, negative valuations of the affine coordinates provide a convenient kernel-membership test.

Suppose the ring of integers of a number field is a principal ideal domain, $a,b$ are integral, and $b$ is a unit. On $y^2=x(x^2+ax+b)$ every nonzero $x$-coordinate has even valuation at every finite prime: positive valuation gives $2v(y)=v(x)$, and negative valuation gives $2v(y)=3v(x)$. Thus the two-torsion square-class homomorphism takes values among unit square classes. For the Gaussian integers, their unit group modulo squares has two elements, represented by $1,i$. This can reduce a two-isogeny descent to a very small calculation even though the field contains no ordering.

Suppose an irreducible monic integral cubic has discriminant $-p$ for a prime $p>3$, and let $L$ be its splitting field. Its Galois group is $S_3$ and its quadratic subfield is $\mathbb{Q}(\sqrt{-p})$. The extension over that subfield is cyclic cubic and everywhere unramified. Away from $p$, separable reduction and Hensel lemma put the roots in an unramified local extension. At $p$, discriminant valuation one gives a double residue root and a separate simple root; lift the simple root, and the remaining quadratic's discriminant has square class $-p$. Its splitting field is already the quadratic subfield's completion. The imaginary quadratic base has no real places.

For an unramified extension of local fields $L/K$, an integral element whose residue generates $k_L/k_K$ generates $L/K$: its degree is at least the residue degree $f=[L:K]$. The reduced powers $1,\bar{\alpha },\dots ,{\bar{\alpha }}^{f-1}$ are a basis of the residue field extension. If $x=\sum c_j{\alpha }^j$, factor out a coefficient of least valuation. Reduction of the remaining sum is nonzero by that residue basis, so $v_L(x)={\min }_j{v}_K(c_j)$. Consequently $x$ is integral exactly when all $c_j$ are integral.

Two additive valuations are equivalent when they have the same valuation ring. Their ordered value-group images then identify compatibly with the valuations; for nontrivial real-valued rank-one valuations this is multiplication by a positive constant.

For a polynomial over a discretely valued field and a rational weight $t$, this coefficient-minimum function is multiplicative. Scale the variable by an element of valuation $t$ in a valued extension and reduce the initial terms; nonzero residue polynomials have nonzero product. For a monic polynomial factored into roots, $w_t(f)={\sum }_r\min (t,v(r))$. Its supporting lines determine the lower Newton polygon, giving the Newton polygon root valuation theorem and its multiplicities.