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An associator ring is a ring in which equality of the principal ideals generated by two elements forces them to be associate elements: $(a)=(b)$ implies $a=ub$ for a unit in a ring $u$. The reverse implication always holds, since $a=ub$ and $b=u^{-1}a$ give both ideal inclusions. The property separates equality of generated ideals from equality of generators up to an invertible factor.

Every integral domain is an associator ring. If $a\ne 0$ and $(a)=(b)$, write $a=rb$ and $b=sa$. Then $(1-rs)a=0$, so cancellation gives $rs=1$ and $a=rb$ with $r$ a unit. The zero case is immediate. The converse fails: $\mathbb{Z}/4\mathbb{Z}$ has generator classes $\{0\}$, $\{2\}$ and $\{1,3\}$ for its three ideals, and $1,3$ differ by a unit, although $2$ is a nonzero zero divisor.

Finite direct product of rings preserve this property. Equality of principal ideals is coordinatewise, and a tuple of units is a unit. Every quotient ring of a principal ideal domain is also an associator ring: first, modulo a prime power $p^k$, every nonzero class is a power $p^e$, $e<k$, times a unit, and distinct powers generate distinct ideals. Next factor a nonzero nonunit modulus into prime powers and apply the Chinese remainder theorem. The zero modulus gives the original integral domain, while a unit modulus gives the zero ring.

A commutative ring is a ring in which multiplication satisfies $ab=ba$ for every pair. The integers and the polynomial ring $\mathbb{Z}[X]$ are examples. So is $\mathbb{Z}/n\mathbb{Z}$, the quotient ring whose elements are residue classes in modular arithmetic. We represent these by $0,1,\dots ,n-1$ and reduce each sum or product to its remainder. All following calculations modulo $6$ use this convention.

In $\mathbb{Z}/6\mathbb{Z}$, $2\cdot 3=0$, although neither factor is zero. Thus $2$ and $3$ are zero divisors: nonzero elements whose product with another nonzero element can vanish. Also $2\cdot 1=2\cdot 4=2$. Removing the common factor $2$ would falsely assert $1=4$. The entire row for multiplication by $2$ isThe repetitions explain the linear congruences directly. The equation $2x=2$ has exactly the two solutions $x=1,4$, whereas $2x=1$ has none. A common factor can erase information, and a nonzero coefficient does not guarantee that an equation has a solution.

A unit in a ring is an element $u$ with a multiplicative inverse, meaning $uv=vu=1$ for some $v$. In $\mathbb{Z}/6\mathbb{Z}$, the units are exactly $1$ and $5$, since $5^2=1$. The other nonzero elements, $2,3,4$, are zero divisors; for $4$, use $4\cdot 3=0$. In a nonzero commutative ring, a unit cannot be a zero divisor: multiplying $ub=0$ by its multiplicative inverse forces $b=0$. But a nonzero element need not be a unit or a zero divisor: the integer $2$ is neither in $\mathbb{Z}$.

By contrast, in $\mathbb{Z}/5\mathbb{Z}$, every row for a nonzero multiplier is a permutation of $0,1,2,3,4$. Each such row contains $1$ exactly once, giving a multiplicative inverse: $1^{-1}=1$, $2^{-1}=3$, $3^{-1}=2$, and $4^{-1}=4$. Hence this commutative ring is a field. For example, $2x=1$ now has the unique solution $x=3$. Reading the tables reveals which algebraic steps the arithmetic permits.

variation of parameters with the fundamental solutions $x,x^2$ givesThe kernel is continuous across $x=t$ and its first derivative jumps by $t^{-2}$, which gives unit in a ring source after multiplication by the leading coefficient. It satisfies both endpoint conditions. The homogeneous solution $Ax+B{x}^2$ satisfies them only when $A=B=0$, proving uniqueness.

If $a=ub$ with $u$ a unit in a ring, then $a$ lies in the principal ideal generated by $b$. This gives $(a)\subseteq (b)$. Since $u$ has a multiplicative inverse, $b=u^{-1}a$, giving the reverse inclusion. ThereforeThus associate elements always generate the same principal ideal; the associator ring condition asks whether equality of these ideals also forces the generators to differ by a unit.

For $z=a+b\sqrt{d}$, let $\bar{z}=a-b\sqrt{d}$ denote the algebraic conjugate, not necessarily ordinary complex conjugation when $d>0$. Conjugation respects multiplication, and $z\bar{z}=a^2-d{b}^2$. ThereforeIf $z$ is a unit in a ring, choose its inverse $w$ in the same ring. Then $N(z)N(w)=N(1)=1$, and both norms are nonnegative integers, forcing $N(z)=1$. Conversely, if $N(z)=1$, then $z\bar{z}=\pm 1$, so $z^{-1}=\pm \bar{z}$ belongs to the ring. Thus $z$ is a unit in a ring exactly when $N(z)=1$.

For negative $d$, the equation becomes $a^2+|d|b^2=1$. If $d=-1$, its integer solutions give the four units $\pm 1,\pm i$. If $d<-1$, the term $|d|b^2$ forces $b=0$, leaving $a=\pm 1$. HenceFor $d=2$, the element $u=1+\sqrt{2}$ has field norm $|1-2|=1$. Its powers $u^n$, $n\ge 0$, are all units by multiplicative function and are distinct, since their positive real values strictly increase with $n$. Thus $\mathbb{Z}[\sqrt{2}]$ has infinitely many units. This argument concerns the stated integer-coefficient ring; it does not replace it by a larger ring of algebraic integers.

For $y=\alpha x+\beta x^2$, the derivatives are $y^{\prime }=\alpha +2\beta x$ and $y^{\prime \prime }=2\beta$. Substituting gives $2\beta x^2-2x(\alpha +2\beta x)+2(\alpha x+\beta x^2)=0$. The independent solutions $x,x^2$ have Wronskian $x^2\ne 0$ on $[1,2]$, so they form a fundamental system of solutions.

Divide the inhomogeneous linear differential equation by $x^2$ and use variation of parameters. Choosing integrals from one gives the particular solutionThus $z=Ax+B{x}^2+z_p$. The first boundary condition gives $A=-B$. The integral endpoint terms cancel when differentiating $z_p$, so the second givesConsequentlyCollect the coefficient of $f(t)$ on either side of $t=x$. A suitable Green function isThe two pieces agree at $x=t$, $G(1,t)=0$, and $G_x(2,t)=0$ for $t<2$. For an interior source point its derivative jump isThus multiplication by the leading coefficient $t^2$ gives unit in a ring source strength. Away from $x=t$, both pieces solve the homogeneous linear differential equation. With continuous $f$, the integral representation has the required second derivative and solves the equation pointwise in $(1,2)$.

For uniqueness, the difference of two solutions is $Ax+B{x}^2$. Its boundary conditions give $A+B=0$ and $A+4B=0$, forcing $A=B=0$. Hence the solution is unique. This is the mixed Green function for an Euler operator with exponents one and two.

A unit in a ring is an element $u$ with a multiplicative inverse in $R$, so $uv=1$ for some $v\in R$.

An irreducible element is a nonzero nonunit $a$ such that every factorization $a=bc$ has at least one of $b,c$ a unit in a ring.

A prime element is a nonzero nonunit $p$ such that $p\mid bc$ implies $p\mid b$ or $p\mid c$. Equivalently its principal ideal $(p)$ is a nonzero prime ideal. These are definitions in the given commutative ring; the equivalence between irreducibility and primality requires additional hypotheses on the ring.

The expansion isThe notation has the units digit before the comma and successive positive powers afterwards, as defined in the source. To check it, the geometric series converges in the 3-adic norm and its sum is $1+(3+2\cdot 3^2)/(1-3^2)=1-21/8=-13/8$.

For the general repeating block, $|p^2{|}_p<1$ givesBoth numerator and denominator are integers, and the denominator is a $p$-adic unit.

For $p=3$, a root would have valuation zero because five is a unit. It would therefore reduce to a unit cube modulo nine. The only cube residue classes modulo nine are $0,1,-1$, none equal to five. Thus no root exists in ${\mathbb{Q}}_3$.

For $p=5$, the valuation equation would be $3v_5(x)=1$, impossible because the valuation of a nonzero element of ${\mathbb{Q}}_5$ is an integer. Thus no root exists in ${\mathbb{Q}}_5$.

For $p=13$, $7^3\equiv 5(\mathrm{mod}13)$ and the derivative $3\cdot 7^2\equiv 4\not\equiv 0(\mathrm{mod}13)$. Hensel's lemma lifts this simple root to a root in ${\mathbb{Z}}_{13}\subseteq {\mathbb{Q}}_{13}$. Thus a root exists in ${\mathbb{Q}}_{13}$.

The curve has good reduction at five and seven: its discriminant is $-432D^2$, and the hypotheses make $D$ a unit at those primes. Over ${\mathbb{F}}_5$ the cubing map is a bijection, soThe character sum vanishes after replacing $x^3+D$ by a variable ranging over the whole field. Over ${\mathbb{F}}_7$, where $D=2$, the possible cubes are zero once, one three times, and minus one three times. The corresponding right sides are two, three and one. Two and one are nonzero squares while three is not, so the count is $1+2+0+6=9$.

Rational torsion injects into the reduction group at a good odd prime, a standard reduction theorem. Thus its order divides $\gcd (6,9)=3$. If a point of order three existed, its coordinates would be integral by the Nagell–Lutz theorem. Since $2P=-P$, the duplication formula givesThus either $x=0$ and $y^2=D$, or $x^3=-4D$ and $y^2=-3D$.

At the prime in the hypothesis, $D$ is a nonsquare unit. Since $p\equiv 1(\mathrm{mod}3)$, there is a nontrivial cube root of unity $\zeta \in {\mathbb{F}}_p$, and $(2\zeta +1{)}^2=-3$. Hence $-3$ is a nonzero square, so both $D$ and $-3D$ are nonsquares. Neither ordinate equation is possible even modulo $p$. There is no rational point of order three, proving

For rank-one two-isogeny descent with parameters three and minus three, use the standard two-isogeny rank formula. For $C:y^2=x(x^2+ax+b)$ and its dual curve $D:y^2=x(x^2-2ax+a^2-4b)$, let ${\alpha }_C,{\alpha }_D$ be the abscissa square-class maps, with their usual values at $(0,0)$ and the identity. If $r_C$ is the Mordell–Weil rank, thenFor the present curve $a=3,b=-3$, its possible classes are $1,-1,3,-3$, and the identity and $(0,0)$ supply $1,-3$. Exclude the class three as follows. A point in that class has $x=3(U/V{)}^2$, with coprime integers $U,V$, and substitution in the curve gives the necessary quartic equationwith integer $N$. Modulo three this forces $3\mid V,N$. Coprimality then makes $U$ a unit modulo three, and modulo nine the right side is $3U^4\equiv 3$, while $N^2\equiv 0$, a contradiction. The class minus one is excluded as well, since multiplying it by the known class minus three would give three. Therefore the first image has size two.

The dual curve is $D:y^2=x(x^2-6x+21)$. Its quadratic factor is $(x-3{)}^2+12>0$ on the real line, so every nonzero rational abscissa is positive. By (i) its image is contained in $\{1,3,7,21\}$. The point $(0,0)$ supplies 21 and the point $(3,6)$ supplies three. As the square-class map is a homomorphism, their product supplies seven. All four occur. Hence the descent formula gives $2^{r_C+2}=2\cdot 4=8$, so

For a point with $u\ne 0$, consider the valuation $v_{\ell }$ at any rational prime. If $v_{\ell }(u)<0$, the term $u^3$ has strictly lower valuation than $a_1{u}^2$ and $b_{1}u$, since the coefficients are integral. Thus $2v_{\ell }(v)=3v_{\ell }(u)$, forcing $v_{\ell }(u)$ even. If $v_{\ell }(u)>0$ and $\ell \nmid b_1$, the factor $u^2+a_{1}u+b_1$ is a unit, so $2v_{\ell }(v)=v_{\ell }(u)$, again even. At valuation zero there is no odd exponent to remove.

Therefore every odd valuation in the square class of $u$ lies at a prime dividing $b_1$, and all negative valuations are even. Removing rational squares gives a signed squarefree integer $r$ with $r\mid b_1$. For $(0,0)$ the assigned class is that of $b_1$, whose squarefree representative also divides $b_1$; the identity has representative one. HenceThe brackets stress that these integers represent square classes, which is the target of $q$.

The three suggested values areThere is an obstruction at $p=3$: every integer residue satisfies $x^3-x\equiv 0(\mathrm{mod}3)$, whereas $-1\equiv 2$ is not a square modulo three. Reduction of a solution in the p-adic integers would give an impossible solution modulo three.

At $p=2$, take $x=-2$. The odd 2-adic unit in a ring $-7$ is congruent to one modulo eight, so it is a square in ${\mathbb{Z}}_2$. We use the standard 2-adic square criterion that an odd unit in a ring is a square exactly when it is one modulo eight. Its sufficiency follows by successively adjusting an odd square root modulo $2^m$ by $2^{m-1}$ to obtain the next required square congruence. Thus $y^2=-7$ has a 2-adic integer solution.

For $p\ge 5$ other than $23$, the discriminant of $y^2=x^3-x-1$ isso this model has good reduction of an elliptic curve. Hasse's theorem on elliptic curves givesSince there is just one point at infinity, there is an affine point. Smoothness means that at least one partial derivative of $y^2-x^3+x+1$ is a unit in a ring modulo $p$, so Hensel's lemma lifts the affine point to ${\mathbb{Z}}_p^2$. The last exceptional odd prime is also soluble: at $p=23$, $h(-2)=-7\equiv 16$ has square roots $\pm 4$, which are simple roots of $Y^2+7$ modulo $23$, and lift by Hensel's lemma.

For small-prime checks the suggested values also give $x=0$ at $p=5$ and $x=3$ at $p=7$, where respectively $-1\equiv 2^2$ and $23\equiv 3^2$. ThereforeThe same prime list holds if one only asks for a solution modulo $p$, but the lifting above establishes the asserted p-adic integer solutions. This is local solubility of the cubic x cubed minus x minus one.

In the elliptic curve method, addition is attempted modulo the composite number. A denominator that is not a unit in a ring supplies a factor through its greatest common divisor with that number.

First double $P$ using the elliptic-curve addition formula. Its tangent slope isThese calculations are valid modulo $221$ since the only initial denominator is two, which is invertible. Thus $2P=(14,-53)$, equivalently $(14,168)$ modulo $221$.

To form $3P=P+2P$, the chord slope would requireThe denominator hasIt cannot be inverted, and thirteen is the requested proper factor. The failure has a geometric explanation: modulo thirteen, $2P=(1,-1)=-P$, so $3P$ is the point at infinity. Modulo seventeen that chord denominator remains nonzero. The differing behavior at the two primes is exactly what the elliptic curve method detects.

Let ${\omega }_G=h_G(T)dT$. The pullback through the formal group homomorphism isThe identity $f(F(X,T))=G(f(X),f(T))$ and the chain rule show that this pullback is invariant under $F$: translation by $T$ before $f$ becomes translation by $f(T)$ afterward. By part (i) it is a scalar multiple of ${\omega }_F$. Its coefficient at zero is $h_G(0)f^{\prime }(0)=f^{\prime }(0)$, so

Apply this to the multiplication-by-$p$ endomorphism $q(T)=[p](T)$. Since the linear coefficient of $F$ is $X+Y$, repeated addition gives $q^{\prime }(0)=p$. If ${\omega }_F=h(T)dT$ with $h(0)=1$, the pullback formula givesbecause $h(q(T))$ is a unit in a ring. Write $q(T)={\sum }_{m\ge 1}{a}_m{T}^m$. Then $m{a}_m\in pR$ for every $m$. If $p\nmid m$, choose integers $u,v$ with $um+vp=1$. It follows thatChoose $b_m\in R$ with $a_m=p{b}_m$ for those $m$ not divisible by $p$, and defineCoefficient comparison now provesNo cancellation of $p$ was used, so this prime-multiplication coefficient splitting for formal groups remains valid when $p$ is a zero divisor in $R$.

Put $U(T)=F_X(0,T)$. The formal group identity $F(X,0)=X$ gives $U(0)=1$, so $U$ is a unit in a ring in $R[[T]]$, even if $R$ has zero divisors.

A differential $h(T)dT$ is invariant under formal translation ifas an identity of formal power series. Setting $X=0$ gives $h(T)U(T)=h(0)$. Consequently every invariant differential of a formal group law, if it exists, must have the form $h(0)U(T{)}^{-1}dT$. In particular there can be only one with constant coefficient one.

Existence follows directly from associativity. Differentiatewith respect to $Z$ at zero, obtainingEvery displayed $U$ is a unit in a ring, soThis is precisely invariance of $U(T{)}^{-1}dT$. ThusNormalization means its constant coefficient is one. This proves both existence and uniqueness over an arbitrary commutative ring, without division by any integer.

A unique factorization domain is an integral domain in which every nonzero nonunit is a finite product of irreducible elements, uniquely up to reordering and multiplication of factors by a unit in a ring. A principal ideal domain is an integral domain in which every ideal is generated by one element.

Gauss lemma for polynomials says that over a unique factorization domain $D$, the product of two primitive polynomials is primitive, and the polynomial content satisfies $c(fg)=c(f)c(g)$ up to a unit. In particular a primitive polynomial of positive degree is irreducible over $D$ exactly when it is irreducible over the fraction field of $D$. Consequently a polynomial ring over a unique factorization domain is itself a unique factorization domain.

The Eisenstein criterion says that if $f(X)=a_n{X}^n+\cdots +a_0\in D[X]$ has $n\ge 1$ and there is a prime element $p$ with $p\nmid a_n$, $p\mid a_j$ for every $j<n$, and $p^2\nmid a_0$, then $f$ is irreducible over the fraction field. If $f$ is primitive, it is also irreducible in $D[X]$. Over $\mathbb{Z}$ one may take $p$ to be an ordinary prime number. The primitivity qualification prevents a nonunit constant content from being overlooked.

For $Q=\phi (P)$, the local ring of a nonsingular curve at $Q$ is a discrete valuation ring. Choose a uniformizer $t$ there. The ramification index of a morphism of curves isEquivalently, ${\phi }^{\ast }t=u{z}^{e_P}$ for a local uniformizer $z$ at $P$ and a unit in a ring $u$. A different choice $t^{\prime }=vt$ changes its pullback by a unit in a ring, so the index is well-defined. This local definition applies also to inseparable morphisms; conclusions involving separability or tame ramification require those extra hypotheses.

Any unit in a ring of $R$ must also be a unit of $\mathbb{Q}[X]$, hence a nonzero constant $q\in \mathbb{Q}$. Membership of both $q$ and $q^{-1}$ in $R$ requires $q,q^{-1}\in \mathbb{Z}$, so

Let $J$ be an ideal of $R^{\prime }=S^{-1}R$ and contract it toThis is an ideal of $R$, so $I=(d)$ because $R$ is a principal ideal domain. Every $x^n/1$ is a unit in a ring $R^{\prime }$. Consequently, if $r/x^n\in J$, then $r/1=(x^n/1)(r/x^n)\in J$, so $r\in (d)$ and $r/x^n\in (d/1)$. Conversely $d/1\in J$, hence every multiple of it belongs to $J$. Thusso every ideal of $R^{\prime }$ is principal. The localization remains an integral domain; hence $R^{\prime }$ is a principal ideal domain.