Differential Graded Algebras and Homological Foundations: Examples and Counterexamples

By [F1] and [F2], $\mathbb{Z}$ with its standard operations is a commutative ring with unit, so it is an admissible base for [F6].

Applying [F3] first to $\mathbb{Z}$ and then to $\mathbb{Z}[t]$ shows that $S=(\mathbb{Z}[t])[ϵ]$ is a commutative ring. By [F4], $({ϵ}^2)=S{ϵ}^2$ is a two-sided ideal; [F5] therefore gives the quotient ring $A$ with its associative multiplication and unit. For cosets, $[s][t]=[st]=[ts]=[t][s]$, so $A$ is commutative. The constant classes define a unital ring homomorphism $\mathbb{Z}\to A$; its induced scalar action is central, and associativity and distributivity make multiplication $\mathbb{Z}$-bilinear.

Every polynomial in $ϵ$ with coefficients in $\mathbb{Z}[t]$ has a unique form $f_0(t)+f_1(t)ϵ+{ϵ}^{2}h(t,ϵ)$. By [F4], the last term is in $({ϵ}^2)$, and a polynomial of degree at most one in $ϵ$ lies in $({ϵ}^2)$ only when both its coefficients vanish. Thus every quotient class has a unique representative $f+gϵ$, giving the stated direct sum of $\mathbb{Z}$-modules. The quotient multiplication is $(f+gϵ)(h+kϵ)=fh+(fk+gh)ϵ$; this preserves the grading and gives ${ϵ}^2=0$ and $tϵ=ϵt$.

The unique expression in step 3.1 makes the formula for $d$ well defined. Coefficientwise differentiation is additive and commutes with multiplication by each integer, so $d$ is $\mathbb{Z}$-linear; it sends $A^0$ to $A^1$ and $A^1$ to $A^2=0$, hence has degree $+1$.

For every $f+gϵ$, $d^2(f+gϵ)=d(f^{\mathrm{\prime }}ϵ)=0$, because $d$ vanishes on $A^1$. Thus [F7] holds.

If $f={\sum }_i{a}_i{t}^i$ and $h={\sum }_j{b}_j{t}^j$ have degree zero, the finite coefficient sums give $(fh{)}^{\mathrm{\prime }}={\sum }_{i+j\ge 1}(i+j)a_i{b}_j{t}^{i+j-1}=f^{\mathrm{\prime }}h+f{h}^{\mathrm{\prime }}$. Hence $d(fh)=d(f)h+fd(h)$.

If $a=f$ and $b=gϵ$ have degrees zero and one, both $d(ab)$ and $d(a)b+ad(b)$ vanish because $d(gϵ)=0$ and $(f^{\mathrm{\prime }}ϵ)(gϵ)=0$. If $a=gϵ$ and $b=f$, then both sides of $d(ab)=d(a)b-ad(b)$ vanish because $d(gϵ)=0$ and $(gϵ)(f^{\mathrm{\prime }}ϵ)=0$. If both have degree one, their product and both terms on the right vanish. These are all remaining homogeneous degree pairs, so [F6] holds.

step 1.1, step 2.1, step 3.1 establish the graded unital associative $\mathbb{Z}$-algebra structure, and step 4.1, step 5.1, step 5.2, step 5.3 establish the degree-one, square-zero differential and signed Leibniz rule. Thus all conditions of [F6] hold, so $A$ is a cohomological DGA.

By [F1] and [F2], $\mathbb{Q}$ is a commutative ring with unit, and [F3] is the unital base-algebra convention used in the construction of $B$. The words form a $\mathbb{Q}$-basis, so the displayed formula defines a unique linear map. Every nonzero generator image has degree one higher than its input. For homogeneous words $s,t$, the summands replacing letters of $s$ give $\delta (s)t$, while those replacing letters of $t$ acquire the extra sign $(-1{)}^{\mathrm{\mid }s\mathrm{\mid }}$; hence $\delta (st)=\delta (s)t+(-1{)}^{\mathrm{\mid }s\mathrm{\mid }}s\delta (t)$. Bilinearity extends the identity to all homogeneous elements.

Applying the generator rules twice gives ${\delta }^2(x)=\delta (u)=v$. The word $v$ is a nonzero basis element of the free algebra, so ${\delta }^2(x)\mathrm{\ne }0$.

By [F4], square-zero is required for a cohomological DGA differential. Thus $\delta$ is a degree-one graded derivation that is not square-zero, refuting the statement.

Associativity and the unit of $A$ make left multiplication a unital left $A$-action on $M=A$, and degree preservation gives $A^p{M}^q\subseteq M^{p+q}$; zero acts by zero and the unit acts as the identity.

The differential $d_M=d_A$ has degree $+1$ and squares to zero by [F4]. For homogeneous $a,m\in A$, the required identity $d_M(am)=d_A(a)m+(-1{)}^{\mathrm{\mid }a\mathrm{\mid }}a{d}_M(m)$ is the DGA Leibniz identity [F2].

The action and differential therefore satisfy every condition in [F3], so $A$ is a left DG module over itself.