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How can I tell if $1$ and $x$ are inverses in a field $F$ if $\det(1+x)=\det(1-x)$?

In $\mathbb R$, if $1$ and $x$ are inverses then $\det(1+x)=\det(1-x)$. How can I show that the same holds for an arbitrary field?

A:

If $E$ is any field, then $\{x \in E : \exists z \in E : z+z = x\}$ is a subgroup of $E$ with $1$ as its identity. Now apply the First Isomorphism Theorem to see that
$$\{x \in E : \exists z \in E : z+z = x\} \cong \frac{E}{\langle 1-x \rangle}$$
If $1$ and $x$ are inverses, then $1-x$ generates the trivial ideal in the right hand group, meaning that the right hand group is an infinite set (i.e. $|E|=\infty$) so it’s not isomorphic to any group. Therefore, $1$ and $x$ cannot be inverses in $E$.

A:

Suppose $1,x\in F$ and $1-x$ is invertible (thus $F(x)$ is a field). Let $f(t)=t(1-x)+1$. Then $f$ is a polynomial over $F$, and if $f$ is a polynomial over $F$ then $f’$ is a polynomial over $F$. So
$$f'(1)=\frac{f(1)-f(0)}{1-0}=1-1=0$$
by Lagrange’s Theorem.
Therefore, $1-x$ and $1-1=0$ are inverses.

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