Proof. Let \(F\) denote the free unbased commutative monoid in spaces. The reduced free monoid is the homotopy pushout \[C(X)\simeq F(X)\amalg_{F(*)}*.\] Indeed, mapping this pushout into a commutative monoid \(Y\) gives the homotopy fibre of \(\operatorname{Map}(X,Y)\to\operatorname{Map}(*,Y)=Y\) over the unit. This is exactly the mapping space of based maps from \(X\) to \(Y\), based at its unit. The pushout therefore has the required reduced free universal property, including the basepoint homotopy.

The functor \(\Sigma^\infty_+\) is a strong symmetric monoidal left adjoint from cartesian spaces to spectra. On commutative algebra objects it is again a left adjoint, and hence preserves this pushout. Write \(T=\Sigma^\infty X\). The basepoint section and augmentation give a natural stable splitting \[\Sigma^\infty_+X\simeq S\vee T.\] Under it, the spectrum map induced by the basepoint is the inclusion of the first summand. Since the free commutative algebra functor sends a direct sum to a coproduct of algebras, the pushout after suspension is \[\begin{aligned} \Sigma^\infty_+C(X) &\simeq\mathbb P(S\vee T)\amalg_{\mathbb P(S)}S\\ &\simeq \bigl(\mathbb P(S)\amalg\mathbb P(T)\bigr) \amalg_{\mathbb P(S)}S \simeq\mathbb P(T). \end{aligned}\] Here \(\mathbb P(S)\to S\) sends its generator to the ring unit. The augmentation of the resulting algebra sends \(T\) to zero. Thus the positive-weight summands are precisely the augmentation fibre, or equivalently the reduced suspension spectrum at the unit. In this description the generating spectrum map is the difference between the inclusion of \(X\) and its basepoint section.

We identify the weight-one adjoint using the algebra projection \[\mathbb P(T)\longrightarrow S\oplus T,\] where the target is the split square-zero algebra. For any spectrum \(M\), the fibre over the unit point of \[\mathrm{GL}_1(S\oplus M)\longrightarrow\mathrm{GL}_1S\] identifies with \(\Omega^\infty M\), via \(y\mapsto1+y\), as a grouplike commutative monoid. To see the structure as well as the underlying space, expand a multiproduct using distributivity in \((S\oplus M)^{\wedge t}\). The term with no \(M\) factors gives the base unit. The terms with one \(M\) factor give the sum of the inputs. Every term with two or more \(M\) factors vanishes by the square-zero structure. These formulas respect permutations and iterated products: the structure maps on the fibre are the additive sum-over-fibres maps. They therefore give the stated identification with the additive infinite loop space of \(M\).

The adjunction from the suspension ring now gives an \(E_\infty\) map \(C(X)\to\Omega^\infty T\). On the generating based space it is the usual map \(X\to\Omega^\infty\Sigma^\infty X\). Reduced freeness identifies it with the canonical stabilization map. Equivalently, maps from its group completion into a grouplike commutative monoid \(\Omega^\infty E\) are based maps \(X\to\Omega^\infty E\), or spectrum maps \(\Sigma^\infty X\to E\). This identifies the group completion with \(QX\). For connected \(X\), the reduced free monoid has trivial \(\pi_0\), so it is already grouplike and no group completion is needed. Under \(C(X)\simeq QX\), the adjoint of the weight-one projection is then the identity of \(QX\). By the suspension-spectrum adjunction this says exactly that the projection is the stable counit. ◻

Proof. For \(v\in S^1\), the based map \(x\mapsto v\wedge x\) induces \(C(X)\to C(\Sigma X)\). These maps form a pointed family, since the basepoint of \(S^1\) induces the constant map to the monoid unit. Its adjoint is \[S^1\wedge C(X)\longrightarrow C(\Sigma X).\] Since \(\Sigma X\) is connected, the target is \(Q\Sigma X\). On the generating space \(X\), adjunction identifies this family with the usual stabilization into \(\Omega Q\Sigma X\simeq QX\). Reduced freeness and group completion therefore identify its adjoint with the usual structure map of the infinite loop spaces.

The positive-weight equivalence in Proposition 4 is a natural equivalence between the reduced functors \[X\longmapsto\operatorname{fib}(\Sigma^\infty_+C(X)\to S), \qquad X\longmapsto\bigvee_{r>0}D_r(\Sigma^\infty X).\] Both send the one-point based space to zero. Their action on mapping spaces therefore respects the basepoint given by factorization through that space, including its nullhomotopy. Apply this naturality to the pointed mapping-space family \(v\mapsto(v\wedge-)\), and then take the stable adjunction. On the \(r\)th extended power the same \(v\) is used in every tensor factor. This gives (1) with no mixing of weights. For a sphere, \(D_r(\Sigma^nS)\) is the Thom spectrum of \(n\rho_r\) over \(B\Sigma_r\), where \(\rho_r\) is the permutation representation. The map in question is the Thom map associated to \[n\rho_r\oplus\mathbb R\longrightarrow(n+1)\rho_r,\] using the diagonal line \(\mathbb R\subset\rho_r\).

The commutation rule with upper-index Dyer–Lashof operations is the usual suspension rule for free infinite loop space homology (May 1977, sec. 20, p. 481). The product rules can also be read directly from the adjunction of loop multiplication: the suspension of loop concatenation factors through the pinch map and gives the sum of the two suspensions. Thus a cross product of two positive-dimensional classes has zero image. For a product with a component point, the surviving term is the suspension of the positive-dimensional factor, multiplied by the augmentation of the point class, which is one. ◻

Proof. In terms of the reduced generator, the diagonal is the sum of its two linear inclusions and the term induced by the reduced diagonal \(X\to X\wedge X\). The latter is null for a suspension. Because the source ring is free, a homotopy on the generating spectrum gives the required homotopy of commutative ring maps.

Apply \(D_r\) to the linear map \(T\to T\vee T\). The terms choosing the first summand \(i\) times are indexed by the transitive \(\Sigma_r\)-set \(\Sigma_r/(\Sigma_i\times\Sigma_{r-i})\). Equivariantly, the map into these terms is the diagonal into a finite sum indexed by that set. On homotopy orbits it is the stable transfer to the indicated subgroup. This describes every component, including the two end components. ◻

Proof. The product over a finite binary vector space is additive in \(U\). Indeed, if \(W'=W\oplus\langle a\rangle\) and its root polynomial for \(W\) is \(P_W\), then \[P_{W'}(U)=P_W(U)P_W(U+a) =P_W(U)^2+P_W(a)P_W(U).\] Induction from \(P_{\{0\}}(U)=U\) proves additivity and the displayed powers of \(U\). Homogeneity gives the degrees of the coefficients. They are invariant because an invertible binary linear change of variables permutes the roots.

Put \(B=\mathbb F_2[d_0,\ldots,d_{k-1}]\subset P_k\) and \(L=\mathbb F_2(x_1,\ldots,x_k)\). Each \(x_i\) satisfies the monic root polynomial, so \(P_k\) is integral over \(B\). Therefore \(B\) has transcendence degree \(k\); its \(k\) indicated generators are algebraically independent and \(B\) is integrally closed. The derivative of the root polynomial is the nonzero product \(d_0\) of all nonzero binary linear forms. Its splitting field over \(\operatorname{Frac}B\) is \(L\). Every automorphism of this splitting field acts invertibly and linearly on the root space; conversely every such linear transformation fixes the coefficients. Consequently \(L^{\mathop{\mathrm{GL}}_k(\mathbb F_2)}=\operatorname{Frac}B\). An invariant polynomial lies in this fraction field and is integral over \(B\), hence belongs to \(B\). This proves the assertion. ◻

Proof. We first determine dimensions and the weights in which primitives can occur. The free infinite loop space homology theorem gives, for \(n>0\), \[ H_*QS^n= \mathbb F_2\bigl[Q_{j_1}\cdots Q_{j_k}g_n: 0<j_1\leq\cdots\leq j_k\bigr], \tag{5}\] including the empty word \(g_n\), where \(Q_jz=Q^{j+|z|}z\) and \(|g_n|=n\) (Cohen et al. 1976). The generator \(g_n\) is primitive. Naturality of the operations for the coproduct, together with Cartan and the vanishing of positive operations on the unit, shows that every displayed polynomial generator is primitive.

For \(n=0\), the same polynomial description holds for \(H_*\mathbb P(S)\) on the free degree-zero generator, which we denote here by \(u\). One may see this from the free unbased monoid \(\coprod_rB\Sigma_r\), whose suspension ring is free on a degree-zero spectrum generator. The Dyer–Lashof operations for commutative ring spectra agree with the space operations on such suspension rings (Bruner et al. 1986). Naturality for the linear ring coproduct again makes the polynomial generators primitive. This argument does not identify that coproduct with the actual space diagonal at level zero.

In both calculations, the bottom generator has weight one, a Dyer–Lashof operation doubles weight, and multiplication adds weights. In a polynomial Hopf algebra on primitive generators, the primitives have basis the iterated \(2\)-powers of those generators. For example, this follows by writing the primitive identity as \(f(\mathbf x+\mathbf y)=f(\mathbf x)+f(\mathbf y)\): a monomial involving distinct variables produces a nonzero cross term, and for a single variable all interior binomial coefficients vanish exactly when the exponent is a power of two. Every polynomial involves only finitely many variables, so the argument also applies to the infinite polynomial algebra.

It follows that a primitive basis is given by the words \[Q_{j_1}\cdots Q_{j_k}g_n, \qquad 0\leq j_1\leq\cdots\leq j_k,\] with \(u\) in place of \(g_n\) at \(n=0\). Initial zero indices record iterated squares. A word of length \(k\) has weight \(2^k\) and degree \[2^kn+j_1+2j_2+\cdots+2^{k-1}j_k.\] Thus primitives vanish at the other positive weights. Setting \(a_1=j_1\) and \(a_i=j_i-j_{i-1}\) for \(i>1\) rewrites this degree as \[ 2^kn+\sum_{i=1}^k(2^k-2^{i-1})a_i, \qquad a_i\geq0. \tag{6}\] These are exactly the monomial degrees in \(\Sigma^nd_0^nD_k\). At \(n=0\) we perform this count at fixed weight, allowing degree zero; it remains finite in each degree at that weight.

We next identify the Steenrod module, rather than just its graded dimension. Fix \(r=2^k\) with \(k\geq1\) and let \(V=(\mathbb Z/2)^k\subset\Sigma_r\) be the regular elementary abelian subgroup. The extended power \(D_r(\Sigma^nS)\) is the Thom spectrum of \(n\rho_r\) over \(B\Sigma_r\). Restrict its cohomology to \(BV\). Every proper two-block transfer restricts to zero. Indeed, the double coset formula expresses its restriction as transfers from subgroups \[V\cap g(\Sigma_i\times\Sigma_{r-i})g^{-1},\qquad 0<i<r.\] Each is proper, because \(V\) acts transitively on the particles whereas a two-block subgroup preserves a nonempty proper subset. If \(W<V\), restriction \(H^*BV\to H^*BW\) is surjective. The projection formula then expresses transfer of any class as a class from \(H^*BV\) times the transfer of \(1\), which is the even index \([V:W]\), hence zero. Naturality of the mod-\(2\) Thom class gives the same assertion with the representation twist.

The normalizer of \(V\) acts through \(\mathop{\mathrm{GL}}_k(\mathbb F_2)\), so the restriction image is contained in \(D_k\) times the Thom class. It contains all of that module: the Stiefel–Whitney classes of \(\rho_r\) restrict to the Dickson generators in (2). More explicitly, \(\rho_r|_V\) is the sum of all the real one-dimensional characters, and its total Stiefel–Whitney class is the product of \(1+a\) over all \(a\in V^*\). Thus \(w_{2^k-2^j}(\rho_r)|_{BV}=d_j\), and polynomials in these classes, multiplied by the Thom class, give the claimed image. Restriction therefore gives a surjection from the transfer quotient onto \(D_k\) times the Thom class. The dimension computation (6) makes this surjection an isomorphism in every degree.

Finally, split the regular representation over \(BV\) as \(\mathbb R\oplus\bar\rho_r\). Pulling the Thom class back along the trivial \(n\)-plane inclusion \[\mathbb R^n\longrightarrow n\rho_r\] multiplies it by the Euler class of \(n\bar\rho_r\), namely \(d_0^n\), and leaves the trivial Thom shift \(\Sigma^n\). Multiplication by \(d_0^n\) is injective in the polynomial algebra. All maps used here are stable cohomology maps; hence the resulting identification is the Steenrod module \(\Sigma^nd_0^nD_k\) with its inherited action. In particular this also proves stability of that ideal under \(A\). Weight one is immediate from \(D_1(\Sigma^nS)=\Sigma^nS\).

For compatibility with suspension, the Thom map in Lemma 6 has complement \(\bar\rho_r\). Its pullback sends a Thom coefficient \(f\) to \(d_0f\). The level-\(n\) identification then sends this coefficient to \(d_0^n(d_0f)=d_0^{n+1}f\), which is its image under the level-\((n+1)\) identification. Thus, after the common suspension shift, the map is \[M_k(n+1)=\Sigma^{n+1}d_0^{n+1}D_k \longrightarrow \Sigma M_k(n)=\Sigma^{n+1}d_0^nD_k,\] given by inclusion. This proves (4), with all degree shifts as asserted. ◻

Proof. Under Proposition 4, the particle is the ring element \(t=1+u\). Group completion of \(C(S^0)\) inverts this particle component. For a commutative ring spectrum \(A'\), a map from its suspension ring is a map from the additive \(E_\infty\) particle monoid to the multiplicative \(E_\infty\) space of \(A'\). Such a map extends over group completion precisely when the particle maps to a unit. Equivalently, the resulting ring map inverts \(1+u\). This is the universal property of the localization in (7), including its mapping spaces. The augmentation sends \(u\) to zero and \(t\) to one, so the equivalence is augmented. The underlying module of localization at a degree-zero element is its multiplication telescope. Therefore ordinary homology satisfies \[H_*R=H_*\mathbb P(S\{u\})[(1+u)^{-1}].\]

In \(\mathbb P_{\leq N}(S\{u\})\), the element \(u\) is nilpotent in \(\pi_0\). Thus \(1+u\) is a unit, with inverse \(\sum_{j=0}^N(-u)^j\) in \(\pi_0\), and the truncation extends across localization to give (8). Their compatibility follows from the localization universal property.

Every class of \(H_*R\) can be multiplied by a sufficiently large power of \(t\) to become a class in the free homology algebra \(H_*\mathbb P(S\{u\})\). This algebra is polynomial, as used in Lemma 9, and is a direct sum of weights. Hence such a numerator has only finitely many nonzero weights. If its smallest nonzero component has weight \(r\), multiplying by \((1+u)^a\) leaves that component unchanged: the weight-zero term is one and every other term has positive weight. Conversely, in a finite truncation the inverse of \(1+u\) has weight-zero term one and only positive weights otherwise. Thus a nonzero numerator remains nonzero in every truncation retaining its leading weight. This proves joint detection, independence of the leading component from the chosen numerator, and invariance under powers of \(t\). The inclusion \(H_*Q_0S^0\to H_*QS^0\) is injective, since it is the inclusion of one component, so these conclusions apply to the base component as well.

Before localization, the actual space diagonal on the reduced generator is \[ \Delta u=u\otimes1+1\otimes u+u\otimes u. \tag{9}\] Consequently, a weight-\(r\) class maps only to terms of total weight at least \(r\), and its terms of total weight \(r\) are given by the linear coproduct. This follows by applying extended powers to (9) and then multiplying. It remains true after localization: the diagonal of \(t\) is \(t\otimes t\), whose inverse has constant term one and only positive total weights otherwise. These statements may all be made in finite truncations in the two target factors, so no convergence of an infinite sum is required.

The kernels of the projections through weight \(r-1\) define a decreasing weight filtration. The preceding calculation identifies its associated graded coproduct with the linear coproduct. Since \(z\) is primitive, comparison in its leading total weight gives \[\Delta_{\mathrm{lin}}z_r=z_r\otimes1+1\otimes z_r.\] The weight projections are maps of spectra, so they commute with the stable Steenrod action. Thus \(z_r\) is also \(A\)-annihilated. Lemma 9 makes \(r\) a power of two; weight one has no positive-dimensional homology, so \(r=2^k\) with \(k\geq1\).

For the suspension assertion, choose \(a\geq0\) such that \(t^az\) has a finite numerator in the particle monoid homology. The translation rule in Lemma 6 gives \[\sigma_*(t^az)=\sigma_*z.\] The numerator has the same leading component \(z_r\) and no smaller weights. Apply the weight-preserving family of Lemma 6 before group completion. It follows that the weight-\(r\) component of the suspension is exactly the image of \(z_r\) under (1). By (4), the corresponding functional is the restriction from \(D_k\) to \(d_0D_k\), with one suspension shift.

Finally, all the generators of \(D_k/(d_0)\) have even degrees: \(|d_j|=2^k-2^j\) is even for \(j\geq1\) (and for \(k=1\) the quotient is just \(\mathbb F_2\)). A nonzero odd-degree functional on \(D_k\) therefore cannot vanish on \(d_0D_k\). This proves that the leading component survives suspension when \(d\) is odd. ◻

Proof. The product rule of Lemma 6 kills the decomposables. For the reverse inclusion, use the polynomial generators in (5). Suspension commutes with upper-index operations. When the bottom class changes from \(g_n\) to \(g_{n+1}\), every intermediate word with those same upper indices increases in degree by one. Thus each lower index of a nonempty polynomial generator decreases by one. The original indices satisfy \(0<j_1\leq\cdots\leq j_k\), so the resulting indices satisfy \[0\leq j_1-1\leq\cdots\leq j_k-1.\] These are distinct nonzero primitive words in the target. The empty-word generator \(g_n\) suspends to \(g_{n+1}\), also nonzero. Hence suspension is injective on the span of the polynomial generators modulo decomposables.

At level zero, localization at the particle gives \[H_*QS^0=\mathbb F_2\bigl[t,t^{-1}, Q_{j_1}\cdots Q_{j_k}t: k\geq1,\ 0<j_1\leq\cdots\leq j_k\bigr].\] A word of length \(k\) has component degree \(2^k\). Replacing each such generator by its product with \(t^{-2^k}\) identifies this algebra with the Laurent polynomial algebra on \(t\) over a polynomial algebra concentrated in component zero. Thus \(H_*Q_0S^0\) is polynomial, with generators the translates to component zero of the nonempty positive-excess particle generators \(Q_{j_1}\cdots Q_{j_k}t\) with \(0<j_1\leq\cdots\leq j_k\). A word of length \(k\) is translated by \(t^{-2^k}\). Suspension ignores that translation and commutes with its upper-index operations, giving the same distinct nonzero words on \(g_1\) with all lower indices decreased by one. One may equivalently start with the reduced generator \(u=t-1\): positive operations on the additive monoid unit vanish, so the positive operations on \(u\) and \(t\) agree. This proves injectivity on indecomposables at level zero as well.

The injectivity here is into the full homology of the target. Some of the resulting words have initial lower index zero and are themselves squares; they must not be discarded by projecting the target to indecomposables. Their independence as primitive words is exactly what gives the asserted kernel description. ◻

Proof. Put \(QH=H^+/(H^+)^2\), let \(\pi:H\to QH\) send \(1\) to zero, and define \[\partial=(1\otimes\pi)\Delta:H\longrightarrow H\otimes QH.\] The identity \(\pi(ab)=\epsilon(a)\pi(b)+\epsilon(b)\pi(a)\) shows that \(\partial\) is a derivation. If \(y_i\) are homogeneous polynomial generators, connectedness and grading give \[\partial y_i=1\otimes[y_i] +\text{terms with smaller positive degree in the second factor}.\] For a polynomial \(f\), the derivation rule yields \[\partial f=\sum_i\frac{\partial f}{\partial y_i}\,\partial y_i.\] Only finitely many variables occur in \(f\). If \(\partial f=0\), inspect the coefficient of \([y_i]\) for variables of largest degree occurring in \(f\). There is no contribution from a higher-degree generator, so their ordinary polynomial partial derivatives vanish. Descending through generator degrees shows that every \(\partial f/\partial y_i\) vanishes. Over \(\mathbb F_2\), this means that every exponent in every monomial of \(f\) is even; equivalently, \(f=z^2\) for a unique polynomial \(z\).

If \(f\) is a decomposable primitive, then \(\partial f=1\otimes\pi(f)=0\), so this argument applies. Moreover, \[\bigl(\Delta z-z\otimes1-1\otimes z\bigr)^2=0.\] The tensor product \(H\otimes H\) is again a polynomial algebra, so its Frobenius map is injective. Hence the displayed reduced coproduct is zero and \(z\) is primitive.

Finally, the homology Cartan formula in characteristic two gives \[\mathop{\mathrm{Sq}}_*^{2a}(z^2)=(\mathop{\mathrm{Sq}}_*^a z)^2, \qquad \mathop{\mathrm{Sq}}_*^{2a+1}(z^2)=0.\] If \(z^2\) is \(A\)-annihilated, injectivity of Frobenius forces \(\mathop{\mathrm{Sq}}_*^a z=0\) for every \(a>0\). Since these squares generate the positive Steenrod action, the root is \(A\)-annihilated. The polynomial homology descriptions above verify the algebra hypotheses for the infinite loop spaces under consideration. ◻

Proof. Milnor’s polynomial dual and coproduct description (Milnor 1958, sec. 5, Theorem 3, and Appendix 1) gives \(A_*=\mathbb F_2[\xi_1,\xi_2,\ldots]\). It implies that restriction to \(E_*\) is the quotient by the squares of these generators: their images are the exterior generators dual to the Milnor primitives. Squaring is an injective Hopf algebra homomorphism on \(A_*\); its degreewise dual is the indicated map \(V\). The formulas on the squares follow by pairing with the powers of \(\xi_1\).

For completeness, the kernel can be read directly from the Milnor coproduct. After restricting its first factor to \(E_*\), that coproduct becomes \[\xi_i\longmapsto \bar\xi_i\otimes1+1\otimes\xi_i.\] The other terms have an even positive exponent in the first factor and therefore vanish. A polynomial has trivial coaction under this map exactly when it is a square: the coefficients of the single generators \(\bar\xi_i\) are its ordinary partial derivatives, and their simultaneous vanishing in characteristic two is equivalent to every exponent being even. Conversely, a square has trivial coaction since \(\bar\xi_i^2=0\). Thus the annihilator of \(I_EA\) is exactly the subalgebra of squares, which also annihilates \(\ker V\). Degreewise duality proves \(I_EA=\ker V\). Conjugation preserves \(E\) and the subalgebra of squares in \(A_*\), so it also gives \(AI_E=\ker V\). Finally \(KM=I_EM\), using \(K=I_EA\) and \(AM=M\), and hence the stated quotient module is well defined. ◻

Proof. The \(q_i\) are derivations and \(q_i x_s=x_s^{2^{i+1}}\). The initial formulas on Dickson generators appear in (Wilkerson 1983, Corollary 2.3(b)) and are also recorded, with indexing shifted by one, in (Nguyen Sum 2024, Proposition 2.3); we give the calculation. Apply \(q_i\) to \(P(x_s)=0\). Only the linear power of \(x_s\) contributes a derivative, so \[\sum_{j=0}^{k-1}q_i(d_j)x_s^{2^j}=d_0x_s^{2^{i+1}}.\] This identity holds on the whole binary span of the \(x_s\), by additivity. A linearized polynomial of degree at most \(2^{k-1}\) is uniquely determined by its values on that span: the difference of two such polynomials cannot have \(2^k\) distinct roots unless it is zero. For \(i<k-1\) this gives the partial-derivative formula. For \(i=k-1\), substituting \(x_s^{2^k}=\sum_{j<k}d_jx_s^{2^j}\) gives the Euler formula.

For \(t\geq0\), raising the same root equation to the power \(2^{t+1}\) gives, as an equality of derivations on \(P_k\), \[q_{k+t}=\sum_{j=0}^{k-1}d_j^{2^{t+1}}q_{j+t}.\] Induction on \(t\) expresses every later primitive in the first \(k\) with square coefficients. These coefficients belong to \(D_k\).

Since \(d_0\) is the product of the nonzero binary linear forms and the total square of such a form \(a\) is \(a+a^2\), the total square of \(d_0\) is divisible by \(d_0\). Its quotient is invariant and therefore belongs to \(D_k\). The Cartan formula now proves the stability of every \(d_0^pD_k\).

Each nonzero term in (26) raises the \(d_0\)-exponent by exactly one. Thus no monomial at exponent \(p\) is hit. Above this floor, a monomial with some even \(b_j\) is the \(q_{j-1}\)-image of the monomial obtained by decreasing \(c\) by one and increasing \(b_j\) by one. This preimage is still in \(d_0^pD_k\). If all \(b_j\) are odd, the Euler primitive on the monomial at exponent \(c-1\) has coefficient \[(c-1)+\sum_{j=1}^{k-1}b_j\equiv c+k\pmod2.\] It hits the given monomial exactly when \(c\not\equiv k\pmod2\). The images of the first \(k\) primitives are spanned by these individual monomials. Later primitives add no new images: a square coefficient passes inside any of the derivations. This proves the basis assertion, including the case \(k=1\), when the list of \(b_j\) is empty. ◻

Proof. On \(P_k\) define \(C_x\) to be zero on a monomial unless all its exponents are odd, and in the latter case set \[C_x(x_1^{2a_1+1}\cdots x_k^{2a_k+1}) =x_1^{a_1}\cdots x_k^{a_k}.\] This is the dual Kameko squaring map (Kameko 1990); see (Hung 2005, sec. 2, p. 4069) for the dual formula. It is also the polynomial top-form version of Cartier’s differential-form construction (Cartier 1957); see also (Cartier 1958, II, §6, pp. 200–203). We include the polynomial identities and naturality needed for its restriction to the present ideals. The binomial formula for squares on a one-variable polynomial, together with the Cartan formula, gives \[ C_x\mathop{\mathrm{Sq}}^{2i}=\mathop{\mathrm{Sq}}^iC_x,\qquad C_x\mathop{\mathrm{Sq}}^{2i+1}=0, \qquad C_x(a^2f)=aC_x(f). \tag{29}\] Indeed, on an odd exponent the parity of \(\binom{2a+1}{2i}\) is the parity of \(\binom ai\); an odd square makes that exponent even. An even input exponent can never contribute an odd output exponent, since the corresponding binomial coefficient then vanishes. Applying these facts in all variables proves the identities.

The operator \(C_x\) commutes with \(\mathop{\mathrm{GL}}_k(\mathbb F_2)\). To see this, write a polynomial uniquely as a sum of square-free monomials with square coefficients. The coefficient of \(x_1\cdots x_k\) after a linear change of variables is multiplied by its determinant; no square-free monomial of smaller degree can contribute to that coefficient. Over \(\mathbb F_2\) the determinant is one, and taking the square root of the coefficient commutes with the linear change of variables. Hence \(C_x\) preserves \(D_k\).

We compute this restriction by an elementary form of Cartier naturality. If \(y_1,\ldots,y_k\) are algebraically independent polynomials and \(J=\det(\partial y_j/\partial x_i)\), then \[ C_x\bigl(Jf(y_1,\ldots,y_k)\bigr) =J\,C_y(f)(y_1,\ldots,y_k). \tag{30}\] Here \(C_y\) is the same odd-exponent extractor in formal variables \(y_j\). For a proof, the polynomial differential-form complex in one variable has cohomology generated by \(1\) and \([x\,dx]\) over \(\mathbb F_2[x^2]\). Tensoring these complexes gives the corresponding assertion in \(k\) variables. Thus there is an isomorphism from the algebra of forms, regarded without differential, to its cohomology, given by \[f\longmapsto f^2,\qquad dx_i\longmapsto[x_i\,dx_i].\] This isomorphism is natural under polynomial substitution. On differentials the required fact is that \(f\mapsto[f\,df]\) is additive and satisfies the derivation rule with coefficients squared: its additivity follows because the cross term is \(d(fg)\), and its product rule follows by expanding \((fg)d(fg)\). In top degree the inverse isomorphism sends \([h\,dx_1\cdots dx_k]\) to \(C_x(h)\,dx_1\cdots dx_k\). Pulling back a top form in the \(y\) variables proves (30).

For Dickson coordinates \(y_j=d_{j-1}\) the Jacobian is \[ J=d_0^{k-1}. \tag{31}\] In fact, differentiating \(P(x_s)=0\) with respect to each \(x_i\) gives the matrix identity whose two factors have entries \(x_s^{2^j}\) and \(\partial d_j/\partial x_i\), and whose product is \(d_0\) times the identity matrix. The determinant of the first factor is \(d_0\): it is nonzero, is divisible by every nonzero binary linear form, and has degree \(2^k-1\). Taking determinants gives (31).

For \(c\geq k-1\), apply (30) to \(f=d_0^{c-k+1}\prod_{j\geq1}d_j^{b_j}\). It is zero unless \(c\equiv k\pmod2\) and every \(b_j\) is odd; otherwise it is exactly (28). The same assertion holds for smaller nonnegative \(c\). Multiply first by a sufficiently large even power \(d_0^{2N}\), use \(C_x(d_0^{2N}f)=d_0^NC_x(f)\), and cancel \(d_0^N\) in the polynomial domain. The output exponents in the nonzero cases are nonnegative.

The description of the hits in Lemma 22 now shows that \(C_x\) descends to coinvariants and is a bijection on the stated above-floor monomials. Squares do not decrease the \(d_0\)-exponent: this follows from the divisibility of the total square of \(d_0\) used above and the Cartan formula. Consequently the above-floor span is a submodule. Equation (29) gives the claimed action on its image. The least possible target exponent is the one displayed in the statement, and a direct degree calculation in (28) gives the change \(d\mapsto(d-k)/2\). ◻

Proof. The monomial lies above the exponent floor \(e\) in \(M_k(e)\). Since the functional factors through exterior coinvariants, Lemma 22 forces \(c\equiv k\pmod2\) and all its other exponents to be odd. Its total degree there is \(m-1\), so (24) gives \(n\equiv k\pmod2\).

Restrict the induced functional to the above-floor submodule and use Lemma 23. The first allowed exponent in this submodule is \(c=n\), so its image has first exponent \[e'=\frac{n+k-2}{2}.\] Give that image the grading shift \(e'\), and put \[ m'=\frac m2,\qquad n'=\frac{n+k}{2},\qquad e'=n'-1,\qquad c'=\frac{c+k-2}{2}. \tag{33}\] The image is exactly \(M_k(e')\), and its paired monomial has degree \(m'-1\): if the old internal degree is \(d\), then \(e'+(d-k)/2=(m-2)/2\). The new functional is again annihilated, because the Cartier map intertwines the halved action. If \(m'\) is even, its paired monomial cannot be at the floor \(c'=e'\); a monomial at that floor has even total degree, whereas \(m'-1\) is odd. It therefore lies in the next ideal and satisfies the same hypothesis, allowing the step to repeat.

Iterate while the paired parameter \(m\) is even, for \(t=\nu_2(m)\) steps. With \(b=2^t\), the resulting parameters obey \[n_t=\frac{n+(b-1)k}{b},\qquad c_t-n_t+2=\frac{c-n+2}{b},\qquad c_t\geq n_t-1.\] The integrality of \(n_t\) gives the congruence in (32), and the last two relations give its inequality. All shifts \(e_t=n_t-1\) are nonnegative.

If \(m=b\), the final paired monomial has degree zero in \(M_k(e_t)\). The smallest degree in that module is \(2^ke_t\), so \(e_t=0\). Thus \[n+(m-1)k=m.\] Since \(n,k\geq1\), this equality forces \(n=k=1\). ◻

Proof. Put \(M=M_k(n)\) and \(B=A/K\). The exact sequence of right \(A\)-modules \(K\to I\to I_B\to0\), followed by tensoring with \(M\), shows that the kernel of (34) is the image of \(K\otimes_A M\). Here we have used \[I_B\otimes_A M =I_B\otimes_B(B\otimes_A M) =I_B\otimes_BH_0(E,M).\] Multiplication gives a surjection \(I_E\otimes_EA\to K=I_EA\); tensoring again gives a surjection \(I_E\otimes_EM\to K\otimes_AM\). Thus every element in this kernel has a preimage in \(I_E\otimes_EM\). The action maps from both tensor products to \(M\) commute with this map. A preimage of a cycle is therefore a cycle itself.

We prove that every even-total-degree cycle in \(I_E\otimes_EM\) is zero. Write a homogeneous such cycle as \[z=\sum_i q_i\otimes z_i.\] Each \(z_i\) has odd total degree in \(M\). By (24), its monomials have \(d_0\)-exponent \(c\geq n+1\) and parity opposite to \(n\), hence opposite to \(k\). Lemma 22 gives \(z_i\in I_EM\).

This last property allows us to replace later primitives by the first \(k\) inside the tensor product. If \(q_i=\sum_{j<k}a_j^2q_j\) on \(M\), then \[ q_i\otimes z_i=\sum_{j<k}q_j\otimes a_j^2z_i \qquad(z_i\in I_EM). \tag{35}\] It suffices to check this for \(z_i=q_sy\). Balancing, commutativity of the exterior generators, and their vanishing on squares give \[\begin{aligned} q_i\otimes q_sy &=q_s\otimes q_iy =\sum_{j<k}q_s\otimes a_j^2q_jy\\ &=\sum_{j<k}q_s\otimes q_j(a_j^2y) =\sum_{j<k}q_j\otimes a_j^2q_sy. \end{aligned}\] Thus \(z\) has a cycle representative using only \(q_j\), \(j<k\), whose coefficients still have exponents at least \(n+1\).

Let \(E_{[k]}=\Lambda(q_0,\ldots,q_{k-1})\) and \(L=D_k[d_0^{-1}]\). We first show that \(L\) is projective over \(E_{[k]}\). For \(T\subseteq\{0,\ldots,k-1\}\) write \(q_T=\prod_{i\in T}q_i\), and put \[v_0=d_0^{-k}\prod_{j=1}^{k-1}d_j.\] The first \(k-1\) primitives take \(v_0\) to \(d_0^{-1}\), and the last, Euler primitive takes \(d_0^{-1}\) to \(1\). Hence \(q_{\{0,\ldots,k-1\}}v_0=1\). The action map from the free module \(E_{[k]}\otimes_{\mathbb F_2}L\) to \(L\) has the following section: \[ s(x)=\sum_{T\subseteq\{0,\ldots,k-1\}} q_T\otimes v_0\bigl(q_{T^c}x\bigr). \tag{36}\] It is \(E_{[k]}\)-linear: multiplying its first factor by \(q_i\) and reindexing the subsets gives \(s(q_ix)\), since terms with repeated \(q_i\) vanish. To compute the composite with the action map, expand each \(q_T(v_0q_{T^c}x)\) by the Cartan formula. For a fixed subset \(U\), the term \((q_Uv_0)(q_{U^c}x)\) occurs once for every \(T\) with \(U\subseteq T\subseteq\{0,\ldots,k-1\}\). Its coefficient is \(2^{k-|U|}\), so only the full subset remains. The composite is therefore \(x\). The formula preserves degrees, since \(v_0\) has the negative of the degree of the product of all the \(q_i\). This proves the asserted projectivity.

Consider now the tensor product of the usual periodic free resolutions for the one-generator exterior algebras. After tensoring with an \(E_{[k]}\)-module, its first differential is \[(u_0,\ldots,u_{k-1})\longmapsto\sum_{i<k}q_iu_i.\] The degree-two terms have one generator for each repeated pair \(i,i\) and each distinct pair \(i,j\). Writing \(e_i\) for the degree-one generators, their images are \[e_i\otimes q_i y,\qquad e_i\otimes q_jy+e_j\otimes q_iy\quad(i<j).\] The cokernel of this differential is \(I_{E_{[k]}}\otimes_{E_{[k]}}M\): the displayed relations become \(q_i^2\otimes y=0\) and \((q_iq_j+q_jq_i)\otimes y=0\) under balancing. Projectivity of \(L\) implies that every degree-one cycle in this complex with coefficients in \(L\) has a degree-two filling.

Apply this to the cycle representative obtained above, ignoring its harmless shift by \(n\). Choose a filling with Laurent-polynomial coefficients. In each of the first two differentials, every nonzero term raises the \(d_0\)-exponent of its coefficient by exactly one, by (26). Delete from the filling all monomials with exponent less than \(n\). Their differential has exponents at most \(n\), whereas the retained part has exponents at least \(n+1\). Since the given cycle has only exponents at least \(n+1\), the retained part is still a filling. Its coefficients belong to \(d_0^nD_k\). All Laurent polynomials and sums used here are finite, so this deletion requires no completion.

We have proved that the representative is a boundary already over \(E_{[k]}\) with coefficients in \(M\). Equivalently, it is zero in \(I_{E_{[k]}}\otimes_{E_{[k]}}M\), and hence also in \(I_E\otimes_EM\). Its image in \(I\otimes_AM\) is zero, which proves the lemma. ◻

Proof. The Adem relation \(\mathop{\mathrm{Sq}}^m=\mathop{\mathrm{Sq}}^1\mathop{\mathrm{Sq}}^{m-1}\) gives \(\mathop{\mathrm{Sq}}^m\otimes v=\mathop{\mathrm{Sq}}^1\otimes\mathop{\mathrm{Sq}}^{m-1}v\). For \(n\geq2\) the second operation exceeds the internal polynomial degree \(m-n\), so this is zero. For \(n=1\) it is the internal square \(v^2\). By linearity we may take \(v=d_0^c\prod_{j\geq1}d_j^{b_j}\), where \(c\geq1\). For \(k\geq2\), formula (26) gives \[v^2=\mathop{\mathrm{Sq}}^1\left( d_0^{2c-1}d_1^{2b_1+1}\prod_{j=2}^{k-1}d_j^{2b_j} \right).\] The preimage belongs to \(d_0D_k\). For \(k=1\) use instead \(\mathop{\mathrm{Sq}}^1(d_0^{2c-1})=d_0^{2c}\). Balancing once more and using \((\mathop{\mathrm{Sq}}^1)^2=0\) proves the assertion. ◻

Proof. Apply the total square simultaneously to the variables \(x_s\) and \(U\) in (23). Since \(\mathop{\mathrm{Sq}}(U+a)=(U+a)(U+a+1)\) and \(P\) is additive, \[\sum_{j=0}^k\mathop{\mathrm{Sq}}(d_j)(U+U^2)^{2^j} =P(U)P(U+1)=P(U)^2+P(1)P(U).\] Comparison of the coefficients of \(U^{2^j}\) gives \(\mathop{\mathrm{Sq}}(d_0)=P(1)d_0\) and, for \(1\leq j<k\), \[\mathop{\mathrm{Sq}}(d_j)+\mathop{\mathrm{Sq}}(d_{j-1})=d_{j-1}^2+P(1)d_j.\] Summing these identities yields \[\mathop{\mathrm{Sq}}(d_j)=P(1)\sum_{i=0}^jd_i+\sum_{i=0}^{j-1}d_i^2,\] which is (37). The term of degree \(2|d_j|-1\) in that formula is exactly \(d_0d_{j+1}\). In the Cartan expansion of an operation one below the top on a product, exactly one factor receives its operation one below the top and every other factor receives its top operation. Equal choices cancel in pairs, leaving precisely (38). ◻

Proof. The tensor is a cycle, because \(\mathop{\mathrm{Sq}}^m\) exceeds the internal degree \(m-n\) of \(v\). We induct on \(\nu_2(m)\), using Lemma 26 throughout; its parity assumption follows from (32). Here we halve coinvariants of the current ideal, with floor \(n\). This differs from the floor \(n-1\) used for the functional in Lemma 25.

Suppose first that \(b\geq4\). The image of \(v\) in exterior coinvariants is either zero, in which case the injectivity lemma gives the result immediately, or is an above-floor survivor. Indeed \(c\geq n+b-2\geq n+2\). In this case the first allowed exponent in the above-floor part of \(M_k(n)\) is \(n+2\), and its Cartier image has first exponent \[n'=\frac{n+k}{2}.\] With this shift the image is \(M_k(n')\), with total degrees halved. The image of our cycle in (34) consequently comes from \(\mathop{\mathrm{Sq}}^{m/2}\otimes v'\) in \(I\otimes_AM_k(n')\), where \(v'=C(v)\). Put \(c'=(c+k-2)/2\) and \(b'=b/2\). Then \[n'\equiv k\pmod{b'},\qquad c'-n'+2=\frac{c-n+2}{2}\geq b'.\] Thus the same bounds hold for \(m/2,n',c'\). Since \(m/2\) is still not a power of two, induction makes this latter tensor zero, and injectivity gives the required vanishing of the original cycle.

It remains to treat \(b=2\). Write \(m=2L\), with \(L\geq3\) odd. All coinvariants of \(M_k(n)\) have even total degree: this is clear at the floor from (24), and above it from \(c\equiv k\equiv n\pmod2\). Regrade them by half and call the resulting \(A\)-module \(C_n\). The projected tensor is \[\mathop{\mathrm{Sq}}^L\otimes[v] =\mathop{\mathrm{Sq}}^1\otimes\mathop{\mathrm{Sq}}^{L-1}[v].\] The second factor is the class of the original operation \(\mathop{\mathrm{Sq}}^{m-2}v\). It suffices to show that this class is zero or an image of \(\mathop{\mathrm{Sq}}^1\) in \(C_n\), since balancing then introduces \((\mathop{\mathrm{Sq}}^1)^2=0\).

If \(n\geq3\), the operation already exceeds the internal degree of \(v\). If \(n=2\), then \(k\) is even and at least two, and the operation is the internal square \(v^2\). This lies above the floor and has all nonzero-index exponents even, so it is zero in coinvariants by Lemma 22.

Suppose \(n=1\), so \(k\) is odd. For \(k=1\) the coinvariant basis, including the floor, consists of all odd monomials in \(d_0\). The map \(C_x\) identifies \(C_1\) with \(\Sigma D_1\), and the image of \(v=d_0^{2L-1}\) is \(d_0^{L-1}\). Its image under \(\mathop{\mathrm{Sq}}^{L-1}\) is \[d_0^{2L-2}=\mathop{\mathrm{Sq}}^1(d_0^{2L-3}),\] as required.

Finally, suppose \(n=1\) and \(k\geq3\) is odd. The internal degree of \(v\) is \(m-1\), so Lemma 28 applies to \(\mathop{\mathrm{Sq}}^{m-2}v\). In each summand of (38), at most one exponent of \(d_j\) with \(j\geq1\) is odd. Since there are at least two such indices, every term above the floor dies in exterior coinvariants. The only possible floor term comes from the index \(j=0\) when \(v=d_0w\), where \(w\) involves only \(d_j\), \(j\geq1\). It is \(d_0d_1w^2\).

This remaining term is an image for the halved first square. Formula (37) gives \(\mathop{\mathrm{Sq}}^2d_2=d_1\) and \(\mathop{\mathrm{Sq}}^1d_0=\mathop{\mathrm{Sq}}^2d_0=0\) for \(k\geq3\). Together with (26) and the Cartan formula, it therefore gives the explicit identity \[ \mathop{\mathrm{Sq}}^2(d_0d_2w^2) =d_0d_1w^2+ d_0^3d_2\left(\frac{\partial w}{\partial d_1}\right)^2. \tag{39}\] The extra term is an exterior hit in \(d_0D_k\), with preimage \[q_0\left( d_0^2d_1d_2\left(\frac{\partial w}{\partial d_1}\right)^2 \right) =d_0^3d_2\left(\frac{\partial w}{\partial d_1}\right)^2.\] Hence the class of \(d_0d_1w^2\) is the original \(\mathop{\mathrm{Sq}}^2\)-image of \([d_0d_2w^2]\), or equivalently a \(\mathop{\mathrm{Sq}}^1\)-image after halving. This completes the final case, and the injectivity lemma finishes the proof. ◻

Proof of Lemma 20. The odd assertion is Lemma 27. For even \(m\), the functional hypothesis gives (32) by Lemma 25. If \(m\) is not a power of two, Lemma 29 contradicts (25). If \(m\) is a power of two, the last assertion of Lemma 25 gives \(n=k=1\). ◻