# Module question

1. Feb 27, 2009

### sid_galt

Let $$A, M$$ be a commutative ring and a finitely generated A-module respectively. Let $$\phi$$ be an A-module endomorphism of M such that $$\phi (M)\subseteq \alpha\ M$$ where $$\alpha$$ is an ideal of A. Let $$x_1,\dots,x_n$$ be the generators of M. Then we know that $$\displaystyle{\phi(x_i)=\sum_{j=1}^{n} a_{ij}x_j\ (1\leq i\leq n;\ a_{ij}\in \alpha)}$$.

Then the book I have (commutative algebra by atiyah goes on to say) - That means
$$\sum_{j=1}^{n} (\delta_{ij}\phi - a_{ij})x_j=0,\ \delta_{ij}$$ being the kronecker delta function This is the part I can't understand - how can you separate $$\phi$$ form it's argument $$x_j$$. How can $$\phi(x) = \phi\cdot x$$?

2. Mar 3, 2009

### arnaldur

The equation you quote from the book, is correct but the term within parenthesis is a mapping which is applied to xj instead of multiplying it. The aij part is actually aij times the identity map.

Last edited: Mar 4, 2009