A detail in proposition 2.4 of Atiyah-MacDonald

Question

I understand everything, but no how the equation $\displaystyle \sum_{j=1}^n(\delta_{ij}\phi-a_{ij})x_j=0$ is obtained. I guess it's just a multiplication by $\delta_{ij}$ but I ever get a different equation. Thanks for any explanation about this

Answer 1

I think the only trick is the following: $$ \sum_{j=1}^n a_{ij}x_j=\phi(x_i)=0\cdot\phi(x_1)+\dots+1\cdot \phi(x_i)+0\cdot \phi(x_n)=\sum_{j=1}^n\delta_{ij}\phi(x_j) $$ where $\delta_{ij}$ is the Kronecker delta. Then $$ \sum_{j=1}^n\delta_{ij}\phi(x_j)-\sum_{j=1}^n a_{ij}x_j=0\Rightarrow \sum_{j=1}^n (\delta_{ij}-a_{ij})x_j=0$$