# Extending a derivative to its exterior algebra: Rotman, incorrect?

This is page 770, Lemma 9.165(ii)

If $\varphi:M \rightarrow M$ is a $k$_map, there exists a unique derivation $d_\varphi: \wedge(M) \rightarrow \wedge (M)$ which is graded with $d|_\varphi M=\varphi$ for all $p \ge 0$.

From i, there is a unique map $D_\varphi:T(M) \rightarrow T(M)$ on the tensor algebras. Given by $$D_\varphi^p(M) ( m_1 \otimes \cdots \otimes m_p) = \sum_{i=1}^p m_1 \otimes \cdots \otimes \varphi(m_i) \otimes \cdots \otimes m_p$$ So it suffices to show $D_\varphi(J) \subseteq J$, where $J$ is ideal generated by $\{v \otimes v \, : \, v \in M \}$.

I don't understand the proof given by Rotman, particularly the inductive step in page 771.

We note that any element in $J$ is of the form $\sum a_i \otimes v_i \otimes v_i \otimes b_i$, where $a_i,b_i$ homogenous and $v_i \in M$.

But the inductive step is trivial ? Since we must have something of the form $$D_\varphi(a_i) \otimes ( v_i \otimes v_i) \otimes b_i + a_i \otimes D_\varphi(v_i \otimes v_i) \otimes b_i+ a_i \otimes (v_i \otimes v_i) \otimes D_\varphi (b_i)$$
each of which is in $J$ since $J$ is an ideal.