Subring Of The Ring Of Endomorphisms Of A Vector Space
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Subring Of The Ring Of Endomorphisms Of A Vector Space
Let \( \displaystyle \mathbb{K} \) be a field, let \( \displaystyle V \) be a vector space over \( \displaystyle \mathbb{K} \) and let \( \displaystyle End_{\mathbb{K}}(V) \) the ring of endomorphisms of (the abelian group) \( \displaystyle (V,+) \). Let \( \displaystyle \mathcal{S} \) be a non empty set of linear maps \( \displaystyle f : V \longrightarrow V \) onto \( \displaystyle V \) such that \[ \displaystyle W \text{ subspace of } V \wedge f(W) \subseteq W \implies W = \{ \vec{0} \} \vee W = V \] Show that the set \[ \displaystyle \mathcal{D} = \left\{ g \in End_{\mathbb{K}}(V) \big| g \circ f = f \circ g , \forall f \in \mathcal{S} \right\} \]is a subring of \( \displaystyle End_{\mathbb{K}}(V) \) and, additionally, it is a division ring.
Re: Subring Of The Ring Of Endomorphisms Of A Vector Space
We see that for each \(\displaystyle{g,h \in D}\) and \(\displaystyle{f\in S}\) ,
\(\displaystyle{\left (g-h \right )\circ f=g\circ f-h\circ f=f\circ g-f\circ h=f\circ \left ( g-h \right )}\)
\(\displaystyle{\left ( g\circ h \right )\circ f=g\circ \left ( h\circ f \right )=\left ( g\circ f \right )\circ h=f\circ \left ( g\circ h \right )}\)
so \(\displaystyle{D}\) is a subring of \(\displaystyle{End_{K}V}\)
For the second part we see that, for each \(\displaystyle{g\in D}\) and \(\displaystyle{f\in S}\) :
If \(\displaystyle{x\in kerg}\) then \(\displaystyle{g\left ( f\left ( x \right ) \right )=f\left ( g\left ( x \right ) \right )=f\left ( 0 \right )=0\Rightarrow f\left ( x \right )\in kerg}\)
so \(\displaystyle{f\left ( kerg \right )\subseteq kerg}\) and since \(\displaystyle{kerg}\) is a subspace of \(\displaystyle{V}\) we have that
\(\displaystyle{kerg=V\vee kerg=\left \{ 0 \right \}\Rightarrow g=0_{End_{K}V}\vee g}\) is \(\displaystyle{1-1}\)
\(\displaystyle{\Rightarrow D=\left \{ 0_{End_{K}V} \right \} \vee g}\) is \(\displaystyle{1-1}\)
but since \(\displaystyle{Id_{V} \in D , D}\) cannot contain only the zero element of \(\displaystyle{End_{K}V}\) so \(\displaystyle{g}\) is \(\displaystyle{1-1}\).
Also , if \(\displaystyle{x\in img}\) then \(\displaystyle{x=g\left ( y \right )}\) for some \(\displaystyle{y\in V}\) and \(\displaystyle{f\left ( x \right )=f\left ( g\left ( y \right ) \right )=g\left ( f\left ( y \right ) \right )\in img}\)
so \(\displaystyle{f\left ( img \right )\subseteq img}\) and since \(\displaystyle{img}\) is a subspace of \(\displaystyle{V}\) we have that
\(\displaystyle{img=\left \{ 0 \right \}\vee img=V\Rightarrow g=\left \{ 0_{End_{K}V} \right \}\vee g}\) is onto
\(\displaystyle{\Rightarrow D=\left \{ 0_{End_{K}V} \right \}\vee g }\) is onto,so \(\displaystyle{g}\) must be onto.
Since \(\displaystyle{g}\) is \(\displaystyle{1-1}\) and onto there exist \(\displaystyle{g^{-1}}\) such that \(\displaystyle{g\circ g^{^{-1}}=Id_{V}=g^{-1}\circ g}\)
wich is linear and \(\displaystyle{g^{-1}\circ f\circ g=g^{-1}\circ g\circ f=f\Rightarrow\left ( g^{-1}\circ f\circ g \right )\circ g^{-1}=f\circ g^{-1}\Rightarrow g^{-1}\circ f=f\circ g^{-1}}\)
we take that \(\displaystyle{g\in D}\) so finally \(\displaystyle{D}\) is a division ring.
\(\displaystyle{\left (g-h \right )\circ f=g\circ f-h\circ f=f\circ g-f\circ h=f\circ \left ( g-h \right )}\)
\(\displaystyle{\left ( g\circ h \right )\circ f=g\circ \left ( h\circ f \right )=\left ( g\circ f \right )\circ h=f\circ \left ( g\circ h \right )}\)
so \(\displaystyle{D}\) is a subring of \(\displaystyle{End_{K}V}\)
For the second part we see that, for each \(\displaystyle{g\in D}\) and \(\displaystyle{f\in S}\) :
If \(\displaystyle{x\in kerg}\) then \(\displaystyle{g\left ( f\left ( x \right ) \right )=f\left ( g\left ( x \right ) \right )=f\left ( 0 \right )=0\Rightarrow f\left ( x \right )\in kerg}\)
so \(\displaystyle{f\left ( kerg \right )\subseteq kerg}\) and since \(\displaystyle{kerg}\) is a subspace of \(\displaystyle{V}\) we have that
\(\displaystyle{kerg=V\vee kerg=\left \{ 0 \right \}\Rightarrow g=0_{End_{K}V}\vee g}\) is \(\displaystyle{1-1}\)
\(\displaystyle{\Rightarrow D=\left \{ 0_{End_{K}V} \right \} \vee g}\) is \(\displaystyle{1-1}\)
but since \(\displaystyle{Id_{V} \in D , D}\) cannot contain only the zero element of \(\displaystyle{End_{K}V}\) so \(\displaystyle{g}\) is \(\displaystyle{1-1}\).
Also , if \(\displaystyle{x\in img}\) then \(\displaystyle{x=g\left ( y \right )}\) for some \(\displaystyle{y\in V}\) and \(\displaystyle{f\left ( x \right )=f\left ( g\left ( y \right ) \right )=g\left ( f\left ( y \right ) \right )\in img}\)
so \(\displaystyle{f\left ( img \right )\subseteq img}\) and since \(\displaystyle{img}\) is a subspace of \(\displaystyle{V}\) we have that
\(\displaystyle{img=\left \{ 0 \right \}\vee img=V\Rightarrow g=\left \{ 0_{End_{K}V} \right \}\vee g}\) is onto
\(\displaystyle{\Rightarrow D=\left \{ 0_{End_{K}V} \right \}\vee g }\) is onto,so \(\displaystyle{g}\) must be onto.
Since \(\displaystyle{g}\) is \(\displaystyle{1-1}\) and onto there exist \(\displaystyle{g^{-1}}\) such that \(\displaystyle{g\circ g^{^{-1}}=Id_{V}=g^{-1}\circ g}\)
wich is linear and \(\displaystyle{g^{-1}\circ f\circ g=g^{-1}\circ g\circ f=f\Rightarrow\left ( g^{-1}\circ f\circ g \right )\circ g^{-1}=f\circ g^{-1}\Rightarrow g^{-1}\circ f=f\circ g^{-1}}\)
we take that \(\displaystyle{g\in D}\) so finally \(\displaystyle{D}\) is a division ring.
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Re: Subring Of The Ring Of Endomorphisms Of A Vector Space
Thank you, Alex, for your effort!
I would like solely to mention that, since you want to prove that \( \mathcal{D} \) is a division ring, you could have considered an arbitrary \( \displaystyle g \in \mathcal{D} \smallsetminus \{ 0 \} \) and thus avoid considering so many cases.
I would like solely to mention that, since you want to prove that \( \mathcal{D} \) is a division ring, you could have considered an arbitrary \( \displaystyle g \in \mathcal{D} \smallsetminus \{ 0 \} \) and thus avoid considering so many cases.
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