I have a question about the use of projections in Lang's proof of the Euclidean division algorithm for power series (Algebra - Serge Lang, Chapter IV, section 9, Theorem 9.1). Specifically, there is a line which states that the problem of solving for a quotient $q$ is equivalent to solving for $Z$ below: $$ \tau(g) = \tau\bigg(Z \frac{\alpha(f)}{\tau(f)}\bigg) + Z = \bigg(I + \tau \circ \frac{\alpha(f)}{\tau(f)}\bigg)Z$$ where $f$ and $g$ are power series, $\alpha$ is projection onto the beginning of the series (terms of degree at most $n-1$), and $\tau$ indicates projection onto the tail end of the series (terms of degree at least $n$).
Why does the rightmost equality hold? It appears to me that, writing $Y = \frac{\alpha(f)}{\tau(f)}$ with corresponding coefficients $(y_i)$, $$\tau\bigg(Z \frac{\alpha(f)}{\tau(f)}\bigg) = \tau \big(Z\alpha(Y) + X^nZ \tau(Y)\big) \\ = \tau\big(Z\alpha(Y)\big) + \tau(Y)Z \\ = \tau\big(Z\alpha(Y)\big) + \bigg(\tau \circ \frac{\alpha(f)}{\tau(f)}\bigg)Z$$ and also $$\tau\big(Z\alpha(Y)\big) = y_{n-1}(z_1 + z_2X + z_3X^2 + ...) \\ + y_{n-2}(z_2 + z_3X + z_4X^2 + ...) \\ . \\ . \\ . \\ + y_1(z_{n-1} + z_nX + z_{n+1}X^2 + ...) \\ + y_0(z_n + z_{n+1}X + z_{n+2}X^2 + ...)$$ but it is not obvious to me from this that $\tau(Z\alpha(Y)) = BZ$ for some power series $B$. Could somebody please explain why such a $B$ exists and what form it would take? (I'm assuming we would then take $I = B + 1$ and the rest of the proof makes sense)