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inline typename Stream::char_type
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{ return t(s); }
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The invention concerns an electric submersible pump which consists of a pump vessel which is connected to an electric motor and whose central shaft is connected to a pump head provided with impeller, flange and impeller shaft. The pump vessel has a central pump shaft. The motor comprises a steel housing and a steel stator which surrounds the steel housing and on which an electric coil is provided.
A known electric submersible pump of this kind is described in EP 0 922 344 A1. Here, the magnetic coil is fastened on the steel stator.
A disadvantage of the known submersible pump is that the steel stator is unable to be fastened on the magnetic coil firmly. This is due to tolerances. Since there are both different installation heights and dimensional deviations between the steel stator and the magnetic coil, an alignment of the steel stator and the magnetic coil is not possible. The result is that the magnetic coil is detached at the steel stator. the last seven years there are
7,500 to 10,000 additional vehicles on the roads per week, as
compared to the same period in 2011. The number of observed collisions
between cars and pedestrians have increased from 2,000 in 2011 to
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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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