Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 179955 by Shrinava last updated on 04/Nov/22

Answered by aleks041103 last updated on 04/Nov/22

(K,+)≇(K/{O_2 }, ∙)⇒ ∄s:K→K/{O_2 }:s(x+y)=s(x)∙s(y)∧s−bijective denote ((a,b),((−b),a) ) :=(a,b) then (a,b)+(c,d)=(a+c,b+d) (a,b)∙(c,d)= ((a,b),((−b),a) ) ∙ ((c,d),((−d),c) ) = = (((ac−bd),(ad+bc)),((−(ad+bc)),(ac−bd)) ) = (ac−bd,ad+bc) But for complex numbers we have (a+bi)+(c+di)=(a+c)+(b+d)i and (a+bi)(c+di)=(ac−bd)+i(ad+bc) ⇒obviously(if this is not obvious, I′ll clarify − just ask) (K,+)≅(C,+) and (K/{O_2 },∙)≅(C/{0},∙) Therefore to prove that (K,+)≇(K/{O_2 }, ∙) is equivalent to proving that (C,+)≇(C/{0}, ∙). suppose (C,+)≅(C/{0}, ∙). then ∃s:C→C/{0}:s(z+w)=s(z)s(w) and ∃s^(−1) :C/{0}→C:s(s^(−1) (z))=z and s^(−1) (s(z))=z but if s(z+w)=s(z)s(w)⇒s(z)=v^z , for some v∈C/{0} ⇒v=re^(2πif) , r∈R^+ and f∈[0,1) ⇒s(a+bi)=r^a e^(−2πbf) (r^(ib) e^(2πifa) )=e^(a ln(r)−2πbf) e^(i(2πfa+b ln(r))) now we need to find (a,b) for given s(a+bi)=e^u e^(i(p+2kπ)) , where u∈R^+ and p∈[0,2π). ⇒ { ((aln(r)−2πbf=u)),((bln(r)+2πaf=p+2kπ)) :} i.e. (((ln(r)),(−2πf)),((2πf),(ln(r))) ) ((a),(b) ) = ((u),((p+2kπ)) ) but det( (((ln(r)),(−2πf)),((2πf),(ln(r))) ))=(ln(r))^2 +4π^2 f^2 >0 ⇒we can always invert the matrix and find (a,b) for ∀k∈Z. ⇒if s:C→C/{0} : s(z+w)=s(z)s(w) then for ∀w∈C/{0}, ∃z_1 ≠z_2 (z_(1,2) ∈C):s(z_1 )=s(z_2 ) ⇒s is not bijective ⇒ every homomorphism from (C,+) to (C/{0},∙) is not bijective. ⇒(K,+)≅(C,+) ≇(C/{0},∙)≅(K/{O_2 },∙) ⇒(K,+)≇(K/{O_2 },∙) Q.E.D

$$\left({K},+\right)\ncong\left({K}/\left\{{O}_{\mathrm{2}} \right\},\:\centerdot\right)\Rightarrow \\ $$$$\nexists{s}:{K}\rightarrow{K}/\left\{{O}_{\mathrm{2}} \right\}:{s}\left({x}+{y}\right)={s}\left({x}\right)\centerdot{s}\left({y}\right)\wedge{s}−{bijective} \\ $$$${denote}\:\begin{pmatrix}{{a}}&{{b}}\\{−{b}}&{{a}}\end{pmatrix}\::=\left({a},{b}\right) \\ $$$${then}\:\left({a},{b}\right)+\left({c},{d}\right)=\left({a}+{c},{b}+{d}\right) \\ $$$$\left({a},{b}\right)\centerdot\left({c},{d}\right)=\begin{pmatrix}{{a}}&{{b}}\\{−{b}}&{{a}}\end{pmatrix}\:\centerdot\begin{pmatrix}{{c}}&{{d}}\\{−{d}}&{{c}}\end{pmatrix}\:= \\ $$$$=\begin{pmatrix}{{ac}−{bd}}&{{ad}+{bc}}\\{−\left({ad}+{bc}\right)}&{{ac}−{bd}}\end{pmatrix}\:=\:\left({ac}−{bd},{ad}+{bc}\right) \\ $$$${But}\:{for}\:{complex}\:{numbers}\:{we}\:{have} \\ $$$$\left({a}+{bi}\right)+\left({c}+{di}\right)=\left({a}+{c}\right)+\left({b}+{d}\right){i} \\ $$$${and} \\ $$$$\left({a}+{bi}\right)\left({c}+{di}\right)=\left({ac}−{bd}\right)+{i}\left({ad}+{bc}\right) \\ $$$$\Rightarrow{obviously}\left({if}\:{this}\:{is}\:{not}\:{obvious},\:{I}'{ll}\:{clarify}\:−\:{just}\:{ask}\right) \\ $$$$\left({K},+\right)\cong\left(\mathbb{C},+\right)\:{and}\:\left({K}/\left\{{O}_{\mathrm{2}} \right\},\centerdot\right)\cong\left(\mathbb{C}/\left\{\mathrm{0}\right\},\centerdot\right) \\ $$$${Therefore}\:{to}\:{prove}\:{that}\:\left({K},+\right)\ncong\left({K}/\left\{{O}_{\mathrm{2}} \right\},\:\centerdot\right) \\ $$$${is}\:{equivalent}\:{to}\:{proving}\:{that} \\ $$$$\left(\mathbb{C},+\right)\ncong\left(\mathbb{C}/\left\{\mathrm{0}\right\},\:\centerdot\right). \\ $$$${suppose}\:\left(\mathbb{C},+\right)\cong\left(\mathbb{C}/\left\{\mathrm{0}\right\},\:\centerdot\right). \\ $$$${then}\:\exists{s}:\mathbb{C}\rightarrow\mathbb{C}/\left\{\mathrm{0}\right\}:{s}\left({z}+{w}\right)={s}\left({z}\right){s}\left({w}\right) \\ $$$${and}\:\exists{s}^{−\mathrm{1}} :\mathbb{C}/\left\{\mathrm{0}\right\}\rightarrow\mathbb{C}:{s}\left({s}^{−\mathrm{1}} \left({z}\right)\right)={z}\:{and}\:{s}^{−\mathrm{1}} \left({s}\left({z}\right)\right)={z} \\ $$$${but}\:{if}\:{s}\left({z}+{w}\right)={s}\left({z}\right){s}\left({w}\right)\Rightarrow{s}\left({z}\right)={v}^{{z}} ,\:{for}\:{some}\:{v}\in\mathbb{C}/\left\{\mathrm{0}\right\} \\ $$$$\Rightarrow{v}={re}^{\mathrm{2}\pi{if}} ,\:{r}\in\mathbb{R}^{+} \:{and}\:{f}\in\left[\mathrm{0},\mathrm{1}\right) \\ $$$$\Rightarrow{s}\left({a}+{bi}\right)={r}^{{a}} {e}^{−\mathrm{2}\pi{bf}} \left({r}^{{ib}} {e}^{\mathrm{2}\pi{ifa}} \right)={e}^{{a}\:{ln}\left({r}\right)−\mathrm{2}\pi{bf}} {e}^{{i}\left(\mathrm{2}\pi{fa}+{b}\:{ln}\left({r}\right)\right)} \\ $$$${now}\:{we}\:{need}\:{to}\:{find}\:\left({a},{b}\right)\:{for}\:{given}\:{s}\left({a}+{bi}\right)={e}^{{u}} {e}^{{i}\left({p}+\mathrm{2}{k}\pi\right)} , \\ $$$${where}\:{u}\in\mathbb{R}^{+} \:{and}\:{p}\in\left[\mathrm{0},\mathrm{2}\pi\right). \\ $$$$\Rightarrow \\ $$$$\begin{cases}{{aln}\left({r}\right)−\mathrm{2}\pi{bf}={u}}\\{{bln}\left({r}\right)+\mathrm{2}\pi{af}={p}+\mathrm{2}{k}\pi}\end{cases} \\ $$$${i}.{e}. \\ $$$$\begin{pmatrix}{{ln}\left({r}\right)}&{−\mathrm{2}\pi{f}}\\{\mathrm{2}\pi{f}}&{{ln}\left({r}\right)}\end{pmatrix}\:\begin{pmatrix}{{a}}\\{{b}}\end{pmatrix}\:=\begin{pmatrix}{{u}}\\{{p}+\mathrm{2}{k}\pi}\end{pmatrix} \\ $$$${but}\:{det}\left(\begin{pmatrix}{{ln}\left({r}\right)}&{−\mathrm{2}\pi{f}}\\{\mathrm{2}\pi{f}}&{{ln}\left({r}\right)}\end{pmatrix}\right)=\left({ln}\left({r}\right)\right)^{\mathrm{2}} +\mathrm{4}\pi^{\mathrm{2}} {f}^{\mathrm{2}} >\mathrm{0} \\ $$$$\Rightarrow{we}\:{can}\:{always}\:{invert}\:{the}\:{matrix}\:{and}\:{find} \\ $$$$\left({a},{b}\right)\:{for}\:\forall{k}\in\mathbb{Z}. \\ $$$$\Rightarrow{if}\:{s}:\mathbb{C}\rightarrow\mathbb{C}/\left\{\mathrm{0}\right\}\::\:{s}\left({z}+{w}\right)={s}\left({z}\right){s}\left({w}\right) \\ $$$${then}\:{for}\:\forall{w}\in\mathbb{C}/\left\{\mathrm{0}\right\},\:\exists{z}_{\mathrm{1}} \neq{z}_{\mathrm{2}} \left({z}_{\mathrm{1},\mathrm{2}} \in\mathbb{C}\right):{s}\left({z}_{\mathrm{1}} \right)={s}\left({z}_{\mathrm{2}} \right) \\ $$$$\Rightarrow{s}\:{is}\:{not}\:{bijective} \\ $$$$\Rightarrow\:{every}\:{homomorphism}\:{from}\:\left(\mathbb{C},+\right)\:{to}\: \\ $$$$\left(\mathbb{C}/\left\{\mathrm{0}\right\},\centerdot\right)\:{is}\:{not}\:{bijective}. \\ $$$$\Rightarrow\left({K},+\right)\cong\left(\mathbb{C},+\right)\:\ncong\left(\mathbb{C}/\left\{\mathrm{0}\right\},\centerdot\right)\cong\left({K}/\left\{{O}_{\mathrm{2}} \right\},\centerdot\right) \\ $$$$\Rightarrow\left({K},+\right)\ncong\left({K}/\left\{{O}_{\mathrm{2}} \right\},\centerdot\right) \\ $$$${Q}.{E}.{D} \\ $$

Commented by Shrinava last updated on 05/Nov/22

cool dear professor thank you so much

$$\mathrm{cool}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com