If $alpha ^{2}$ is algebraic over $F$, then show that $alpha$ is algebraic over $F$.












1















If $alpha ^{2}$ is algebraic over $F$, then show that $alpha$ is algebraic over $F$.




I have only found solutions where it shows the converse. I am not really sure where to start... do I prove the contrapositive and assume that $alpha$ is not algebraic? I haven't learned much about the degrees of extensions yet.






abstract-algebra extension-field







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  • 1




    Write down the definition of what it means for $alpha^2$ to be algebraic.
    – Eric Wofsey
    Nov 26 at 0:51






  • 1




    So you have $pleft(alpha^2right) = 0$ for some monic polynomial $p$, and you want to prove that $qleft(alpharight) = 0$ for some monic polynomial $q$. What would you take for $q$?
    – darij grinberg
    Nov 26 at 0:51


















1















If $alpha ^{2}$ is algebraic over $F$, then show that $alpha$ is algebraic over $F$.




I have only found solutions where it shows the converse. I am not really sure where to start... do I prove the contrapositive and assume that $alpha$ is not algebraic? I haven't learned much about the degrees of extensions yet.






abstract-algebra extension-field







share|cite|improve this question


















  • 1




    Write down the definition of what it means for $alpha^2$ to be algebraic.
    – Eric Wofsey
    Nov 26 at 0:51






  • 1




    So you have $pleft(alpha^2right) = 0$ for some monic polynomial $p$, and you want to prove that $qleft(alpharight) = 0$ for some monic polynomial $q$. What would you take for $q$?
    – darij grinberg
    Nov 26 at 0:51
















1












1








1








If $alpha ^{2}$ is algebraic over $F$, then show that $alpha$ is algebraic over $F$.




I have only found solutions where it shows the converse. I am not really sure where to start... do I prove the contrapositive and assume that $alpha$ is not algebraic? I haven't learned much about the degrees of extensions yet.






abstract-algebra extension-field







share|cite|improve this question














If $alpha ^{2}$ is algebraic over $F$, then show that $alpha$ is algebraic over $F$.




I have only found solutions where it shows the converse. I am not really sure where to start... do I prove the contrapositive and assume that $alpha$ is not algebraic? I haven't learned much about the degrees of extensions yet.






abstract-algebra extension-field




abstract-algebra extension-field






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 26 at 0:48









numericalorange

1,719311




1,719311








  • 1




    Write down the definition of what it means for $alpha^2$ to be algebraic.
    – Eric Wofsey
    Nov 26 at 0:51






  • 1




    So you have $pleft(alpha^2right) = 0$ for some monic polynomial $p$, and you want to prove that $qleft(alpharight) = 0$ for some monic polynomial $q$. What would you take for $q$?
    – darij grinberg
    Nov 26 at 0:51
















  • 1




    Write down the definition of what it means for $alpha^2$ to be algebraic.
    – Eric Wofsey
    Nov 26 at 0:51






  • 1




    So you have $pleft(alpha^2right) = 0$ for some monic polynomial $p$, and you want to prove that $qleft(alpharight) = 0$ for some monic polynomial $q$. What would you take for $q$?
    – darij grinberg
    Nov 26 at 0:51










1




1




Write down the definition of what it means for $alpha^2$ to be algebraic.
– Eric Wofsey
Nov 26 at 0:51




Write down the definition of what it means for $alpha^2$ to be algebraic.
– Eric Wofsey
Nov 26 at 0:51




1




1




So you have $pleft(alpha^2right) = 0$ for some monic polynomial $p$, and you want to prove that $qleft(alpharight) = 0$ for some monic polynomial $q$. What would you take for $q$?
– darij grinberg
Nov 26 at 0:51






So you have $pleft(alpha^2right) = 0$ for some monic polynomial $p$, and you want to prove that $qleft(alpharight) = 0$ for some monic polynomial $q$. What would you take for $q$?
– darij grinberg
Nov 26 at 0:51












1 Answer
1






active

oldest

votes


















4














If $alpha^2$ is a root of $p(t)in F[t]$, $alpha$ is a root of $p(t^2)$, which has degree $2deg p$.






share|cite|improve this answer





















  • Thank you for taking the time to reply and help me. In general, if $p(alpha)$ is algebraic, then should I let $g(x)=(p(x)-p(alpha))h(x)$ to show that $alpha$ is algebraic?
    – numericalorange
    Nov 26 at 1:55






  • 1




    I don't know what you denote $g(x)$. What is quite easy to see is that if $p(alpha)$ is a root of $g(x)$, by definition $gbigl(p(alpha)bigr)=0$, so that $alpha$ is a root of the composition $(gcirc p)(x)$.
    – Bernard
    Nov 26 at 9:36











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









4














If $alpha^2$ is a root of $p(t)in F[t]$, $alpha$ is a root of $p(t^2)$, which has degree $2deg p$.






share|cite|improve this answer





















  • Thank you for taking the time to reply and help me. In general, if $p(alpha)$ is algebraic, then should I let $g(x)=(p(x)-p(alpha))h(x)$ to show that $alpha$ is algebraic?
    – numericalorange
    Nov 26 at 1:55






  • 1




    I don't know what you denote $g(x)$. What is quite easy to see is that if $p(alpha)$ is a root of $g(x)$, by definition $gbigl(p(alpha)bigr)=0$, so that $alpha$ is a root of the composition $(gcirc p)(x)$.
    – Bernard
    Nov 26 at 9:36
















4














If $alpha^2$ is a root of $p(t)in F[t]$, $alpha$ is a root of $p(t^2)$, which has degree $2deg p$.






share|cite|improve this answer





















  • Thank you for taking the time to reply and help me. In general, if $p(alpha)$ is algebraic, then should I let $g(x)=(p(x)-p(alpha))h(x)$ to show that $alpha$ is algebraic?
    – numericalorange
    Nov 26 at 1:55






  • 1




    I don't know what you denote $g(x)$. What is quite easy to see is that if $p(alpha)$ is a root of $g(x)$, by definition $gbigl(p(alpha)bigr)=0$, so that $alpha$ is a root of the composition $(gcirc p)(x)$.
    – Bernard
    Nov 26 at 9:36














4












4








4






If $alpha^2$ is a root of $p(t)in F[t]$, $alpha$ is a root of $p(t^2)$, which has degree $2deg p$.






share|cite|improve this answer












If $alpha^2$ is a root of $p(t)in F[t]$, $alpha$ is a root of $p(t^2)$, which has degree $2deg p$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 26 at 0:55









Bernard

118k638112




118k638112












  • Thank you for taking the time to reply and help me. In general, if $p(alpha)$ is algebraic, then should I let $g(x)=(p(x)-p(alpha))h(x)$ to show that $alpha$ is algebraic?
    – numericalorange
    Nov 26 at 1:55






  • 1




    I don't know what you denote $g(x)$. What is quite easy to see is that if $p(alpha)$ is a root of $g(x)$, by definition $gbigl(p(alpha)bigr)=0$, so that $alpha$ is a root of the composition $(gcirc p)(x)$.
    – Bernard
    Nov 26 at 9:36


















  • Thank you for taking the time to reply and help me. In general, if $p(alpha)$ is algebraic, then should I let $g(x)=(p(x)-p(alpha))h(x)$ to show that $alpha$ is algebraic?
    – numericalorange
    Nov 26 at 1:55






  • 1




    I don't know what you denote $g(x)$. What is quite easy to see is that if $p(alpha)$ is a root of $g(x)$, by definition $gbigl(p(alpha)bigr)=0$, so that $alpha$ is a root of the composition $(gcirc p)(x)$.
    – Bernard
    Nov 26 at 9:36
















Thank you for taking the time to reply and help me. In general, if $p(alpha)$ is algebraic, then should I let $g(x)=(p(x)-p(alpha))h(x)$ to show that $alpha$ is algebraic?
– numericalorange
Nov 26 at 1:55




Thank you for taking the time to reply and help me. In general, if $p(alpha)$ is algebraic, then should I let $g(x)=(p(x)-p(alpha))h(x)$ to show that $alpha$ is algebraic?
– numericalorange
Nov 26 at 1:55




1




1




I don't know what you denote $g(x)$. What is quite easy to see is that if $p(alpha)$ is a root of $g(x)$, by definition $gbigl(p(alpha)bigr)=0$, so that $alpha$ is a root of the composition $(gcirc p)(x)$.
– Bernard
Nov 26 at 9:36




I don't know what you denote $g(x)$. What is quite easy to see is that if $p(alpha)$ is a root of $g(x)$, by definition $gbigl(p(alpha)bigr)=0$, so that $alpha$ is a root of the composition $(gcirc p)(x)$.
– Bernard
Nov 26 at 9:36


















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