\ph^r\pd^k$, is Fourier decomposed, integrating over $\th$ forces $P_1 = q_1 + \dots +q_r$ to vanish, $$\int_{S^1} d\th\, e^{iP_1\th} = 2\pi \d_{P_1,0}.
\label{eqn:delta}$$ From eq. , it is clear that a total derivative brings down a factor of $P_1 = q_1 + \dots + q_r$. An operator is therefore a total derivative if and only if the polynomial $w(q_1,\dots,q_r)$ is proportional to $P_1$; hence, the equivalence relation in eq.  is translated to $w_a \sim w_b$ if $w_a = w_b + w_cP_1$ for $w_{a,b,c}\in R_r^{S_r}$. Algebraically, the statement of momentum conservation, $P_1 = 0$, defines an ideal $\avg{P_1}$ of $R_r^{S_r}$. The set of operators containing $r$ powers of $\ph$ fields modulo IBP lie in the quotient ring $R_r^{S_r}/\avg{P_1}$.

In momentum space the EOM translates to $q^2 = 0$, which implies that $P_n = 0$ for $n\ge 2$, as is obvious from eq. . Additionally, the EOM imply $P_1^{r+1}= (q_1 + \dots + q_r)^{r+1} = 0$, since every term in the expansion necessarily involves a $q_i^2$. This embodies the fact that EOM only allow $r$ derivatives to be distributed onto $\phi^r$; application of a further derivative necessarily requires a $\dd^2 \phi$ in the operator. To study the equivalence under the EOM we 