Generating set of a free module is its basis

Question

Let $F \cong R^d$ be a free R-module, where R is Noetherian. Let $Y$ be a generating set for $F$ with $|Y| \leqslant d$. Show that $Y$ is a basis for $F$ and $|Y|=d$.

Thought that this must be like that, bug got the outcome "prove it!".

Answer 1

Since $R$ is Noetherian, you know that $F$ is a Noetherian $R$-module, since $d$ is finite.

Fix a basis $\{x_1,\ldots,x_d\}$ of $F$, and suppose that $Y=\{y_1,\ldots,y_\ell\}$ with $\ell\leqslant d$. Define the map $f:F\to F$ by

$$f:\sum_{j=1}^{d}a_j x_j\mapsto \sum_{j=1}^{\ell}a_j y_j$$

Note that $f$ is necessarily surjective since $Y$ is a generating set. But, it's a common fact that every endomorphism of a Noetherian $R$-module is an isomorphism, and thus $f$ must be an isomorphism. This automatically implies both results.

EDIT: I assumed in the above that $d$ was finite. I assume that is what you meant.