LEC ALG4 1

Existence of bases for vector spaces

Recall that every vector space has a basis. This is NOT true for arbitrary modules!

Example 369.1(Example).

Let $R=k[x,y]$. $I=⟨x,y⟩$ is an $R$-module. $I$ does not have a $R$-basis:

1. $\{x,y\}$ clearly generates $I$, by is not linearly independent: $xy-yx=0$.
2. Any subset of $\{x,y\}$ does not generate $I$; in fact no singleton can generate $I$, since $I$ is not principal.
3. Use the argument in $(1)$ to see that any subset $S\subseteq I$ with $|S|⩾2$ cannot generate $I$.

An $R$-module $M$ is said to be free if it has a basis.