The exterior resolution needed for odd primary bordism
We give an explicit free resolution over an exterior algebra, induce it on the correct module side, and compute the possible Ext degrees. The Thom-module calculation supplies the actual module to which it will be applied.
A free resolution over an exterior algebra
Let \(k\) be a field of odd characteristic. Let
\[ E=\Lambda_k(Q_0,Q_1,\ldots),\qquad |Q_i|=d_i, \]where every \(d_i\) is a positive odd integer. Thus \(Q_i^2=0\) and \(Q_iQ_j=-Q_jQ_i\) for \(i\ne j\). Let \(k\) be the trivial graded left \(E\)-module, by the augmentation sending every \(Q_i\) to zero. Infinite exterior algebras and direct sums below are algebraic: each element has finite support.
For every finitely supported sequence \(\alpha=(\alpha_0,\alpha_1,\ldots)\) of nonnegative integers, introduce a symbol \(g_\alpha\) of homological degree \(\|\alpha\|=\sum_i\alpha_i\) and internal degree \(\sum_i\alpha_i d_i\). Define the free graded left \(E\)-modules
\[ P_s=\bigoplus_{\|\alpha\|=s}E g_\alpha, \qquad d(g_\alpha)=\sum_{i:\alpha_i>0}Q_i g_{\alpha-e_i}. \tag{1} \]Extend \(d\) by left \(E\)-linearity. The map has homological degree \(-1\) and internal degree zero. There is one summand per distinct index \(i\), not \(\alpha_i\) summands. In particular, \(g_{n e_i}\) is a divided-power symbol for purposes of this resolution, not an ordinary monomial to which a polynomial derivative is applied. No multiplication on the symbols is required.
The augmentation \(P_0\to k\) sends \(a g_0\) to the augmentation of \(a\). Equation (1) defines a complex: terms with the same index in \(d^2\) contain \(Q_i^2=0\), and the two terms with distinct indices \(i,j\) have coefficient \(Q_iQ_j+Q_jQ_i=0\). This argument also covers repeated entries of \(\alpha\).
Here is an explicit contraction of its positive-weight part. Write \(Q_I=Q_{i_1}\cdots Q_{i_m}\), where \(I=\{i_1<\cdots<i_m\}\) is a finite set. These elements form the exterior basis. For the basis element \(Q_I g_\alpha\), put
\[ \beta=\alpha+\mathbf1_I. \]The differential preserves \(\beta\). For fixed \(\beta\ne0\), let \(S=\{i:\beta_i>0\}\) and \(j=\min S\). The basis elements of this weight space are exactly
\[ Q_I g_{\beta-\mathbf1_I},\qquad I\subseteq S. \]Right multiplication by \(Q_i\), followed by increasing-order rewriting, gives
\[ d\bigl(Q_I g_{\beta-\mathbf1_I}\bigr) =\sum_{i\in S\setminus I} (-1)^{\#\{u\in I:u>i\}} Q_{I\cup\{i\}}g_{\beta-\mathbf1_{I\cup\{i\}}}. \tag{2} \]Define the \(k\)-linear map of homological degree \(+1\)
\[ h\bigl(Q_I g_{\beta-\mathbf1_I}\bigr)= \begin{cases} (-1)^{\#\{u\in I:u>j\}} Q_{I\setminus\{j\}}g_{\beta-\mathbf1_{I\setminus\{j\}}},&j\in I,\\ 0,&j\notin I. \end{cases} \tag{3} \]The internal degree is unchanged. The map \(h\) is not asserted to be \(E\)-linear; a \(k\)-linear contraction suffices for exactness of a complex of \(E\)-modules.
We check \(dh+hd=1\), including the signs. If \(j\notin I\), then \(h\) first is zero, and after \(d\) only its \(i=j\) term survives under \(h\). The two signs are both \((-1)^{\#\{u\in I:u>j\}}\), so their product is (1). If \(j\in I\), the \(i=j\) term of \(dh\) gives the original element with coefficient (1). For each \(i\ne j\), adding \(i\) and removing \(j\) in the two orders gives opposite signs: the sign exponents differ by
\[ \mathbf1_{i>j}+\mathbf1_{j>i}=1\pmod2. \]Thus all off-diagonal terms cancel. The choice \(j\in S\) is fixed on the entire weight subcomplex; neither operation changes \(\beta\). This also proves the identity when some \(\beta_i>1\).
Weight zero consists only of \(k g_0\) in degree zero and maps isomorphically to the augmented \(k\). Give this two-term complex the inverse augmentation as its contraction. Consequently
\[ \cdots\longrightarrow P_2\longrightarrow P_1\longrightarrow P_0 \longrightarrow k\longrightarrow0 \tag{4} \]is exact. Every chain is a finite sum of weight components, so the proof remains valid for infinitely many generators. Each \(P_s\) is free over \(E\); (4) is a free graded \(E\)-resolution of \(k\).
The absence of multiplicities in (1) is necessary. In characteristic \(p\), the alternative rule \(d(g_{n e_0})=nQ_0g_{(n-1)e_0}\) gives \(d(g_{p e_0})=0\). In the one-generator complex, no boundary has a nonzero \(g_{p e_0}\) coefficient, since every differential has coefficient divisible by \(Q_0\). It therefore introduces a false positive-degree homology class.
Induction to a larger graded algebra
Let \(A\) be an augmented graded \(k\)-algebra containing \(E\), and suppose \(A\) is free as a graded right \(E\)-module. This side matters because \(P_s\) is a left \(E\)-module. A direct sum of shifted copies of \(E\) preserves exact sequences when tensored over \(E\): tensoring each copy is a grading shift of the original sequence, and exactness of a direct sum is checked component by component. Thus
\[ A\otimes_E P_\bullet\longrightarrow A\otimes_E k \tag{5} \]is an exact complex of free graded left \(A\)-modules. Its differential on the free generators is again (1), now with \(Q_i\in A\). The target is the quotient left module \(A/AE_+\). Nothing here identifies that left ideal with a two-sided ideal without further hypotheses.
Suppose a graded left \(A\)-module \(M\) is actually given with an isomorphism
\[ M\cong\bigoplus_{\lambda}\Sigma^{m_\lambda}(A\otimes_E k). \tag{6} \]Taking the corresponding direct sum of shifts of (5) is a free resolution of \(M\). Applying graded \(A\)-linear maps into the trivial module \(k\) gives zero cochain differential: each differential entry is one of the \(Q_i\), and every \(Q_i\) acts by zero on \(k\). Under the convention that a free generator of internal degree \(t\) contributes to \(\operatorname{Ext}^{s,t}_A(M,k)\), the possible bidegrees are exactly those of the resolution generators:
\[ s=\sum_i\alpha_i, \qquad t=m_\lambda+\sum_i\alpha_i d_i. \tag{7} \]If the index sets are infinite, Hom of a direct sum is a product; this does not introduce a class in an absent bidegree. No finite-dimensionality assumption is needed for the vanishing conclusion below.
The odd prime degree calculation
Let \(p\) be odd, set \(d_i=2p^i-1\), and assume every \(m_\lambda\) is divisible by four. The index \(i=0\) is included: \(d_0=1\), and this generator accounts for the degree-one Bockstein. From (7),
\[ t-s=m_\lambda+\sum_i2\alpha_i(p^i-1)\equiv0\pmod4. \tag{8} \]Hence the hypotheses (5)–(6) imply
\[ \operatorname{Ext}^{s,t}_A(M,k)=0 \quad\text{whenever }t-s\not\equiv0\pmod4. \tag{9} \]In particular the entire stem \(t-s=7\) vanishes. This is an algebraic conclusion about the stated module \(M\), not yet a calculation of a bordism group.
Application to the actual Thom module
The full operation-algebra computation, Section 9, and the Thom-module calculation, Sections 2–5, supply the hypotheses of (5)–(6) for \(M=H^*(MSO;\mathbb F_p)\). Consequently (9) applies. Finite Adams-tower detection gives the required homotopy divisibility, and the bounding theorem eliminates the remaining finite primary groups.
René Thom, Travaux de Milnor sur le cobordisme, p. 173, describes this exterior-generator resolution. Equations (2)–(3) specify its contraction and signs.