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,xin 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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