Proof with Galois theory that $K(sqrt alpha) = K(sqrt beta)$ if and only if it exists $gamma$ with $alpha =...












1












$begingroup$


Let E/K be a field extension, $1+1neq0$ in K and $alpha$, $beta in K^x={a in K |, exists , b: ab = ba =1}$.



Proof with Galois theory: $K(sqrt alpha) = K(sqrt beta) Leftrightarrow exists , gamma in K^x : alpha = gamma^2beta$




Got so far that $sqrt alpha, sqrt beta notin K Rightarrow [K(sqrt alpha):K]= 2 =[K(sqrt beta):K] overset{char(K) neq 2}Rightarrow K(sqrt alpha), K(sqrt beta)$ are galois.






abstract-algebra galois-theory galois-extensions







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    1












    $begingroup$


    Let E/K be a field extension, $1+1neq0$ in K and $alpha$, $beta in K^x={a in K |, exists , b: ab = ba =1}$.



    Proof with Galois theory: $K(sqrt alpha) = K(sqrt beta) Leftrightarrow exists , gamma in K^x : alpha = gamma^2beta$




    Got so far that $sqrt alpha, sqrt beta notin K Rightarrow [K(sqrt alpha):K]= 2 =[K(sqrt beta):K] overset{char(K) neq 2}Rightarrow K(sqrt alpha), K(sqrt beta)$ are galois.






    abstract-algebra galois-theory galois-extensions







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Let E/K be a field extension, $1+1neq0$ in K and $alpha$, $beta in K^x={a in K |, exists , b: ab = ba =1}$.



      Proof with Galois theory: $K(sqrt alpha) = K(sqrt beta) Leftrightarrow exists , gamma in K^x : alpha = gamma^2beta$




      Got so far that $sqrt alpha, sqrt beta notin K Rightarrow [K(sqrt alpha):K]= 2 =[K(sqrt beta):K] overset{char(K) neq 2}Rightarrow K(sqrt alpha), K(sqrt beta)$ are galois.






      abstract-algebra galois-theory galois-extensions







      share|cite|improve this question











      $endgroup$




      Let E/K be a field extension, $1+1neq0$ in K and $alpha$, $beta in K^x={a in K |, exists , b: ab = ba =1}$.



      Proof with Galois theory: $K(sqrt alpha) = K(sqrt beta) Leftrightarrow exists , gamma in K^x : alpha = gamma^2beta$




      Got so far that $sqrt alpha, sqrt beta notin K Rightarrow [K(sqrt alpha):K]= 2 =[K(sqrt beta):K] overset{char(K) neq 2}Rightarrow K(sqrt alpha), K(sqrt beta)$ are galois.






      abstract-algebra galois-theory galois-extensions




      abstract-algebra galois-theory galois-extensions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 13 '18 at 10:40







      Zorro_C

















      asked Dec 13 '18 at 9:57









      Zorro_CZorro_C

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          1 Answer
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          1












          $begingroup$

          $implies$: Assuming $K(sqrtalpha) = K(sqrtbeta)$.



          If $sqrtalpha = xin K$, then $K(sqrtalpha) = K$, so we have $sqrtbetain K$. Set $gamma = frac{sqrtalpha}{sqrtbeta}$ and we're done.



          Otherwise, we have $sqrtalphain K(sqrtbeta)$. In other words, there are $x, yin K$ such that
          $$
          sqrtalpha = x + ysqrtbeta\
          alpha = x^2 + beta y^2 + 2xysqrtbeta
          $$
          From this we can deduce that $2xy = 0$, which because $1+1neq 0$ gives us $xy = 0$, meaning either $x = 0$ or $y = 0$. However, $sqrtalphanotin K$ implies $yneq 0$, so we must have $x = 0$. Set $gamma = y$, and we're done.



          $Longleftarrow,$: Assuming $alpha = gamma^2beta$.



          We have $sqrtalpha = gammasqrtbeta in K(sqrtbeta)$ and $sqrtbeta = sqrtalpha/gammain K(sqrtalpha)$, meaning the two fields must be the same.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            The solution is absolutely right and I've solved the problem already that way. But I wanted to come up with a solution in which Galois theory would be applied.
            $endgroup$
            – Zorro_C
            Dec 13 '18 at 10:24










          • $begingroup$
            @Zorro_C If you've done this already, that's the kind of thing you would want to put in the question post so that people don't spend time and effort duplicating your work.
            $endgroup$
            – Arthur
            Dec 13 '18 at 10:28












          • $begingroup$
            Sorry, but I have mentioned in every line, also in the title, that I like to get a solution with application of Galois theory
            $endgroup$
            – Zorro_C
            Dec 13 '18 at 10:39










          • $begingroup$
            What do you mean by "application of Galois theory".? If you admit that Kummer theory (see any text book) is contained in Galois theory, then you can apply it directly. For a summary of Kummer theory, see e.g. math.stackexchange.com/a/1609061/300700
            $endgroup$
            – nguyen quang do
            Dec 13 '18 at 17:33












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          active

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          1












          $begingroup$

          $implies$: Assuming $K(sqrtalpha) = K(sqrtbeta)$.



          If $sqrtalpha = xin K$, then $K(sqrtalpha) = K$, so we have $sqrtbetain K$. Set $gamma = frac{sqrtalpha}{sqrtbeta}$ and we're done.



          Otherwise, we have $sqrtalphain K(sqrtbeta)$. In other words, there are $x, yin K$ such that
          $$
          sqrtalpha = x + ysqrtbeta\
          alpha = x^2 + beta y^2 + 2xysqrtbeta
          $$
          From this we can deduce that $2xy = 0$, which because $1+1neq 0$ gives us $xy = 0$, meaning either $x = 0$ or $y = 0$. However, $sqrtalphanotin K$ implies $yneq 0$, so we must have $x = 0$. Set $gamma = y$, and we're done.



          $Longleftarrow,$: Assuming $alpha = gamma^2beta$.



          We have $sqrtalpha = gammasqrtbeta in K(sqrtbeta)$ and $sqrtbeta = sqrtalpha/gammain K(sqrtalpha)$, meaning the two fields must be the same.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            The solution is absolutely right and I've solved the problem already that way. But I wanted to come up with a solution in which Galois theory would be applied.
            $endgroup$
            – Zorro_C
            Dec 13 '18 at 10:24










          • $begingroup$
            @Zorro_C If you've done this already, that's the kind of thing you would want to put in the question post so that people don't spend time and effort duplicating your work.
            $endgroup$
            – Arthur
            Dec 13 '18 at 10:28












          • $begingroup$
            Sorry, but I have mentioned in every line, also in the title, that I like to get a solution with application of Galois theory
            $endgroup$
            – Zorro_C
            Dec 13 '18 at 10:39










          • $begingroup$
            What do you mean by "application of Galois theory".? If you admit that Kummer theory (see any text book) is contained in Galois theory, then you can apply it directly. For a summary of Kummer theory, see e.g. math.stackexchange.com/a/1609061/300700
            $endgroup$
            – nguyen quang do
            Dec 13 '18 at 17:33
















          1












          $begingroup$

          $implies$: Assuming $K(sqrtalpha) = K(sqrtbeta)$.



          If $sqrtalpha = xin K$, then $K(sqrtalpha) = K$, so we have $sqrtbetain K$. Set $gamma = frac{sqrtalpha}{sqrtbeta}$ and we're done.



          Otherwise, we have $sqrtalphain K(sqrtbeta)$. In other words, there are $x, yin K$ such that
          $$
          sqrtalpha = x + ysqrtbeta\
          alpha = x^2 + beta y^2 + 2xysqrtbeta
          $$
          From this we can deduce that $2xy = 0$, which because $1+1neq 0$ gives us $xy = 0$, meaning either $x = 0$ or $y = 0$. However, $sqrtalphanotin K$ implies $yneq 0$, so we must have $x = 0$. Set $gamma = y$, and we're done.



          $Longleftarrow,$: Assuming $alpha = gamma^2beta$.



          We have $sqrtalpha = gammasqrtbeta in K(sqrtbeta)$ and $sqrtbeta = sqrtalpha/gammain K(sqrtalpha)$, meaning the two fields must be the same.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            The solution is absolutely right and I've solved the problem already that way. But I wanted to come up with a solution in which Galois theory would be applied.
            $endgroup$
            – Zorro_C
            Dec 13 '18 at 10:24










          • $begingroup$
            @Zorro_C If you've done this already, that's the kind of thing you would want to put in the question post so that people don't spend time and effort duplicating your work.
            $endgroup$
            – Arthur
            Dec 13 '18 at 10:28












          • $begingroup$
            Sorry, but I have mentioned in every line, also in the title, that I like to get a solution with application of Galois theory
            $endgroup$
            – Zorro_C
            Dec 13 '18 at 10:39










          • $begingroup$
            What do you mean by "application of Galois theory".? If you admit that Kummer theory (see any text book) is contained in Galois theory, then you can apply it directly. For a summary of Kummer theory, see e.g. math.stackexchange.com/a/1609061/300700
            $endgroup$
            – nguyen quang do
            Dec 13 '18 at 17:33














          1












          1








          1





          $begingroup$

          $implies$: Assuming $K(sqrtalpha) = K(sqrtbeta)$.



          If $sqrtalpha = xin K$, then $K(sqrtalpha) = K$, so we have $sqrtbetain K$. Set $gamma = frac{sqrtalpha}{sqrtbeta}$ and we're done.



          Otherwise, we have $sqrtalphain K(sqrtbeta)$. In other words, there are $x, yin K$ such that
          $$
          sqrtalpha = x + ysqrtbeta\
          alpha = x^2 + beta y^2 + 2xysqrtbeta
          $$
          From this we can deduce that $2xy = 0$, which because $1+1neq 0$ gives us $xy = 0$, meaning either $x = 0$ or $y = 0$. However, $sqrtalphanotin K$ implies $yneq 0$, so we must have $x = 0$. Set $gamma = y$, and we're done.



          $Longleftarrow,$: Assuming $alpha = gamma^2beta$.



          We have $sqrtalpha = gammasqrtbeta in K(sqrtbeta)$ and $sqrtbeta = sqrtalpha/gammain K(sqrtalpha)$, meaning the two fields must be the same.






          share|cite|improve this answer









          $endgroup$



          $implies$: Assuming $K(sqrtalpha) = K(sqrtbeta)$.



          If $sqrtalpha = xin K$, then $K(sqrtalpha) = K$, so we have $sqrtbetain K$. Set $gamma = frac{sqrtalpha}{sqrtbeta}$ and we're done.



          Otherwise, we have $sqrtalphain K(sqrtbeta)$. In other words, there are $x, yin K$ such that
          $$
          sqrtalpha = x + ysqrtbeta\
          alpha = x^2 + beta y^2 + 2xysqrtbeta
          $$
          From this we can deduce that $2xy = 0$, which because $1+1neq 0$ gives us $xy = 0$, meaning either $x = 0$ or $y = 0$. However, $sqrtalphanotin K$ implies $yneq 0$, so we must have $x = 0$. Set $gamma = y$, and we're done.



          $Longleftarrow,$: Assuming $alpha = gamma^2beta$.



          We have $sqrtalpha = gammasqrtbeta in K(sqrtbeta)$ and $sqrtbeta = sqrtalpha/gammain K(sqrtalpha)$, meaning the two fields must be the same.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 13 '18 at 10:15









          ArthurArthur

          122k7122211




          122k7122211








          • 1




            $begingroup$
            The solution is absolutely right and I've solved the problem already that way. But I wanted to come up with a solution in which Galois theory would be applied.
            $endgroup$
            – Zorro_C
            Dec 13 '18 at 10:24










          • $begingroup$
            @Zorro_C If you've done this already, that's the kind of thing you would want to put in the question post so that people don't spend time and effort duplicating your work.
            $endgroup$
            – Arthur
            Dec 13 '18 at 10:28












          • $begingroup$
            Sorry, but I have mentioned in every line, also in the title, that I like to get a solution with application of Galois theory
            $endgroup$
            – Zorro_C
            Dec 13 '18 at 10:39










          • $begingroup$
            What do you mean by "application of Galois theory".? If you admit that Kummer theory (see any text book) is contained in Galois theory, then you can apply it directly. For a summary of Kummer theory, see e.g. math.stackexchange.com/a/1609061/300700
            $endgroup$
            – nguyen quang do
            Dec 13 '18 at 17:33














          • 1




            $begingroup$
            The solution is absolutely right and I've solved the problem already that way. But I wanted to come up with a solution in which Galois theory would be applied.
            $endgroup$
            – Zorro_C
            Dec 13 '18 at 10:24










          • $begingroup$
            @Zorro_C If you've done this already, that's the kind of thing you would want to put in the question post so that people don't spend time and effort duplicating your work.
            $endgroup$
            – Arthur
            Dec 13 '18 at 10:28












          • $begingroup$
            Sorry, but I have mentioned in every line, also in the title, that I like to get a solution with application of Galois theory
            $endgroup$
            – Zorro_C
            Dec 13 '18 at 10:39










          • $begingroup$
            What do you mean by "application of Galois theory".? If you admit that Kummer theory (see any text book) is contained in Galois theory, then you can apply it directly. For a summary of Kummer theory, see e.g. math.stackexchange.com/a/1609061/300700
            $endgroup$
            – nguyen quang do
            Dec 13 '18 at 17:33








          1




          1




          $begingroup$
          The solution is absolutely right and I've solved the problem already that way. But I wanted to come up with a solution in which Galois theory would be applied.
          $endgroup$
          – Zorro_C
          Dec 13 '18 at 10:24




          $begingroup$
          The solution is absolutely right and I've solved the problem already that way. But I wanted to come up with a solution in which Galois theory would be applied.
          $endgroup$
          – Zorro_C
          Dec 13 '18 at 10:24












          $begingroup$
          @Zorro_C If you've done this already, that's the kind of thing you would want to put in the question post so that people don't spend time and effort duplicating your work.
          $endgroup$
          – Arthur
          Dec 13 '18 at 10:28






          $begingroup$
          @Zorro_C If you've done this already, that's the kind of thing you would want to put in the question post so that people don't spend time and effort duplicating your work.
          $endgroup$
          – Arthur
          Dec 13 '18 at 10:28














          $begingroup$
          Sorry, but I have mentioned in every line, also in the title, that I like to get a solution with application of Galois theory
          $endgroup$
          – Zorro_C
          Dec 13 '18 at 10:39




          $begingroup$
          Sorry, but I have mentioned in every line, also in the title, that I like to get a solution with application of Galois theory
          $endgroup$
          – Zorro_C
          Dec 13 '18 at 10:39












          $begingroup$
          What do you mean by "application of Galois theory".? If you admit that Kummer theory (see any text book) is contained in Galois theory, then you can apply it directly. For a summary of Kummer theory, see e.g. math.stackexchange.com/a/1609061/300700
          $endgroup$
          – nguyen quang do
          Dec 13 '18 at 17:33




          $begingroup$
          What do you mean by "application of Galois theory".? If you admit that Kummer theory (see any text book) is contained in Galois theory, then you can apply it directly. For a summary of Kummer theory, see e.g. math.stackexchange.com/a/1609061/300700
          $endgroup$
          – nguyen quang do
          Dec 13 '18 at 17:33


















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