In Weibel's K-book, there is a lemma saying that for a local ring $R$, with maximal ideal $\mathfrak{m}$, every finitely generated projective $R$-module is free. The proof uses a simple statement that "every element $u\in R$ is invertible in $R/\mathfrak{m}$ is a unit of $R$".
This statement itself is very easy to prove:
Since $R/\mathfrak{m}$ is the residue field, then $u+\mathfrak{m}$ is invertible if and only if it's non-zero, i.e. $u\not\in\mathfrak{m}$. Suppose $u$ is not a unit, then by Atiyah-MacDonald's Corollary 1.5, it's contained in $\mathfrak{m}$, a contradiction.
However, I'm confused by Weibel's proof of this statement: Weibel's proof. More specifically, how is the following true?
Indeed, by multiplying by a representative for the inverse of $\bar{u}\in R/\mathfrak{m}$ we may assume that $u\in 1+\mathfrak{m}$.
Here's my understanding: To "assume that $u\in 1+\mathfrak{m}$", I guess (could be wrong) that we need to have another representative $u'$ for $\bar{u}$, with $u'\in 1+\mathfrak{m}$, so that we may replace $u$ with $u'$ whenever $u\not\in 1+\mathfrak{m}$. Weibel seems to suggest that the choice of $u'$ is $uv$, where $v$ is the representative for the inverse of $\bar{u}$. If this is the case, then $uv$ need to be a representative for $\bar{u}$, but I can't see why this is true. Is this requirement necessary at all?
I really appreciate any help and clarification!
Weibel is saying the following: to prove the claim, it is enough to show that it is true for $u\in 1 + \mathfrak{m}.$
The key (implicit) claim is that $u$ is invertible in $R$ if $uv$ is invertible in $R.$ This is true very generally: if $uv$ is invertible with inverse $x,$ then the equation $uvx = 1$ implies that $vx$ is the inverse of $u$ in $R.$ So, we can reduce from the case of an arbitrary $u$ which becomes a unit in $R/\mathfrak{m}$ to only those $u\in 1 + \mathfrak{m}.$
In particular, $uv$ is not a different representative for $\overline{u} = u + \mathfrak{m}$ in general; it is a representative for $\overline{1} = 1 + \mathfrak{m}$ which may be distinct from $u + \mathfrak{m}.$ But by the above, this is OK, because $u$ divides $uv.$