why does $\varphi'(N)=0$ in this proof?

Question

Fulton's book on page 105 defines $N$:

Afterwards Fulton writes this solution for this lemma:

I didn't understand why $\varphi'(N)=0$

Thanks

Answer 1

It is enough to check that $\varphi'$ vanishes on the generating set for $N$. For this, we compute directly \begin{eqnarray} \varphi'([x+y] - [x] - [y]) &=& D(x+y) - D(x) - D(y) \\ &=& D(x) + D(y) - D(x) - D(y) \\ &=& 0. \end{eqnarray} Similarly, \begin{eqnarray} \varphi'([\lambda x] - \lambda[x]) &=& D(\lambda x) - \lambda D(x) \\ &=& \lambda D(x) - \lambda D(x) &=& 0. \end{eqnarray}

Finally,

\begin{eqnarray} \varphi'([xy] - x[y] - y[x]) &=& D(xy)- xD(y) - yD(x) \\ &=& (D(x)y + xD(y)) - xD(y) - yD(x) \\ &=& 0. \end{eqnarray} The takeaway from these computation is that the generators of $N$ are defined precisely to be killed by the map $\varphi'$. Compare their definitions with that of the derivation $D$.