If $a+b, ab$ algebraic then $a,b$ are algebraic












0














Let $L/K$ be a field extension and $a,bin L $. It can easily be shown that if $a,b$ are algebraic over $K$ then sum and product are, too.



But I read that also the converse is true, say if $a+b$ and $ab$ are algebraic over $K$ then so is $a$ and $b$.



My strategy is to use that, since product and sum are algebraic we have that



$displaystyle [mathbb{Q}(ab,a+b):mathbb{Q}]$ is finite, thus algebraic.



Now by using field operations (denoted $f$, e.g. multiplication, inverse ...) I somehow have to obtain an expression $a=f(a+b,a*b)$ and the same for $b$. Then the conclusion would follow. However, I could not achieve this.






abstract-algebra field-theory







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  • 1




    Note that you don't need $a = f(a+b, ab)$, you only need $g(a) = f(a+b, ab)$ for some non-constant polynomial $g$.
    – Arthur
    Nov 29 '18 at 13:05






  • 1




    If the sum and product are algebraic, then so is $(a+b)^2$, then so is $(a+b)^2-4ab=(a-b)^2$, then so is $a-b$, then so is ${a+bover2}+{a-bover2}$.
    – Ivan Neretin
    Nov 29 '18 at 13:09










  • @Arthur Why would that be? If this is true, I would be done immediately, cause obtaining $a$ as a polynomial is not hard.
    – EpsilonDelta
    Nov 29 '18 at 13:12






  • 1




    Because $f(a+b, ab)$ is algebraic, meaning $g(x) - f(a+b, ab)$ is a polynomial with algebraic coefficients, meaning its roots (one of which is $a$) are algebraic. So technically, you can even weaken it to there being a polynomial $h(x, y, z)$ such that $h(a, a+b, ab) = 0$.
    – Arthur
    Nov 29 '18 at 13:14


















0














Let $L/K$ be a field extension and $a,bin L $. It can easily be shown that if $a,b$ are algebraic over $K$ then sum and product are, too.



But I read that also the converse is true, say if $a+b$ and $ab$ are algebraic over $K$ then so is $a$ and $b$.



My strategy is to use that, since product and sum are algebraic we have that



$displaystyle [mathbb{Q}(ab,a+b):mathbb{Q}]$ is finite, thus algebraic.



Now by using field operations (denoted $f$, e.g. multiplication, inverse ...) I somehow have to obtain an expression $a=f(a+b,a*b)$ and the same for $b$. Then the conclusion would follow. However, I could not achieve this.






abstract-algebra field-theory







share|cite|improve this question


















  • 1




    Note that you don't need $a = f(a+b, ab)$, you only need $g(a) = f(a+b, ab)$ for some non-constant polynomial $g$.
    – Arthur
    Nov 29 '18 at 13:05






  • 1




    If the sum and product are algebraic, then so is $(a+b)^2$, then so is $(a+b)^2-4ab=(a-b)^2$, then so is $a-b$, then so is ${a+bover2}+{a-bover2}$.
    – Ivan Neretin
    Nov 29 '18 at 13:09










  • @Arthur Why would that be? If this is true, I would be done immediately, cause obtaining $a$ as a polynomial is not hard.
    – EpsilonDelta
    Nov 29 '18 at 13:12






  • 1




    Because $f(a+b, ab)$ is algebraic, meaning $g(x) - f(a+b, ab)$ is a polynomial with algebraic coefficients, meaning its roots (one of which is $a$) are algebraic. So technically, you can even weaken it to there being a polynomial $h(x, y, z)$ such that $h(a, a+b, ab) = 0$.
    – Arthur
    Nov 29 '18 at 13:14
















0












0








0


1





Let $L/K$ be a field extension and $a,bin L $. It can easily be shown that if $a,b$ are algebraic over $K$ then sum and product are, too.



But I read that also the converse is true, say if $a+b$ and $ab$ are algebraic over $K$ then so is $a$ and $b$.



My strategy is to use that, since product and sum are algebraic we have that



$displaystyle [mathbb{Q}(ab,a+b):mathbb{Q}]$ is finite, thus algebraic.



Now by using field operations (denoted $f$, e.g. multiplication, inverse ...) I somehow have to obtain an expression $a=f(a+b,a*b)$ and the same for $b$. Then the conclusion would follow. However, I could not achieve this.






abstract-algebra field-theory







share|cite|improve this question













Let $L/K$ be a field extension and $a,bin L $. It can easily be shown that if $a,b$ are algebraic over $K$ then sum and product are, too.



But I read that also the converse is true, say if $a+b$ and $ab$ are algebraic over $K$ then so is $a$ and $b$.



My strategy is to use that, since product and sum are algebraic we have that



$displaystyle [mathbb{Q}(ab,a+b):mathbb{Q}]$ is finite, thus algebraic.



Now by using field operations (denoted $f$, e.g. multiplication, inverse ...) I somehow have to obtain an expression $a=f(a+b,a*b)$ and the same for $b$. Then the conclusion would follow. However, I could not achieve this.






abstract-algebra field-theory




abstract-algebra field-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 29 '18 at 13:03









EpsilonDeltaEpsilonDelta

6281615




6281615








  • 1




    Note that you don't need $a = f(a+b, ab)$, you only need $g(a) = f(a+b, ab)$ for some non-constant polynomial $g$.
    – Arthur
    Nov 29 '18 at 13:05






  • 1




    If the sum and product are algebraic, then so is $(a+b)^2$, then so is $(a+b)^2-4ab=(a-b)^2$, then so is $a-b$, then so is ${a+bover2}+{a-bover2}$.
    – Ivan Neretin
    Nov 29 '18 at 13:09










  • @Arthur Why would that be? If this is true, I would be done immediately, cause obtaining $a$ as a polynomial is not hard.
    – EpsilonDelta
    Nov 29 '18 at 13:12






  • 1




    Because $f(a+b, ab)$ is algebraic, meaning $g(x) - f(a+b, ab)$ is a polynomial with algebraic coefficients, meaning its roots (one of which is $a$) are algebraic. So technically, you can even weaken it to there being a polynomial $h(x, y, z)$ such that $h(a, a+b, ab) = 0$.
    – Arthur
    Nov 29 '18 at 13:14
















  • 1




    Note that you don't need $a = f(a+b, ab)$, you only need $g(a) = f(a+b, ab)$ for some non-constant polynomial $g$.
    – Arthur
    Nov 29 '18 at 13:05






  • 1




    If the sum and product are algebraic, then so is $(a+b)^2$, then so is $(a+b)^2-4ab=(a-b)^2$, then so is $a-b$, then so is ${a+bover2}+{a-bover2}$.
    – Ivan Neretin
    Nov 29 '18 at 13:09










  • @Arthur Why would that be? If this is true, I would be done immediately, cause obtaining $a$ as a polynomial is not hard.
    – EpsilonDelta
    Nov 29 '18 at 13:12






  • 1




    Because $f(a+b, ab)$ is algebraic, meaning $g(x) - f(a+b, ab)$ is a polynomial with algebraic coefficients, meaning its roots (one of which is $a$) are algebraic. So technically, you can even weaken it to there being a polynomial $h(x, y, z)$ such that $h(a, a+b, ab) = 0$.
    – Arthur
    Nov 29 '18 at 13:14










1




1




Note that you don't need $a = f(a+b, ab)$, you only need $g(a) = f(a+b, ab)$ for some non-constant polynomial $g$.
– Arthur
Nov 29 '18 at 13:05




Note that you don't need $a = f(a+b, ab)$, you only need $g(a) = f(a+b, ab)$ for some non-constant polynomial $g$.
– Arthur
Nov 29 '18 at 13:05




1




1




If the sum and product are algebraic, then so is $(a+b)^2$, then so is $(a+b)^2-4ab=(a-b)^2$, then so is $a-b$, then so is ${a+bover2}+{a-bover2}$.
– Ivan Neretin
Nov 29 '18 at 13:09




If the sum and product are algebraic, then so is $(a+b)^2$, then so is $(a+b)^2-4ab=(a-b)^2$, then so is $a-b$, then so is ${a+bover2}+{a-bover2}$.
– Ivan Neretin
Nov 29 '18 at 13:09












@Arthur Why would that be? If this is true, I would be done immediately, cause obtaining $a$ as a polynomial is not hard.
– EpsilonDelta
Nov 29 '18 at 13:12




@Arthur Why would that be? If this is true, I would be done immediately, cause obtaining $a$ as a polynomial is not hard.
– EpsilonDelta
Nov 29 '18 at 13:12




1




1




Because $f(a+b, ab)$ is algebraic, meaning $g(x) - f(a+b, ab)$ is a polynomial with algebraic coefficients, meaning its roots (one of which is $a$) are algebraic. So technically, you can even weaken it to there being a polynomial $h(x, y, z)$ such that $h(a, a+b, ab) = 0$.
– Arthur
Nov 29 '18 at 13:14






Because $f(a+b, ab)$ is algebraic, meaning $g(x) - f(a+b, ab)$ is a polynomial with algebraic coefficients, meaning its roots (one of which is $a$) are algebraic. So technically, you can even weaken it to there being a polynomial $h(x, y, z)$ such that $h(a, a+b, ab) = 0$.
– Arthur
Nov 29 '18 at 13:14












2 Answers
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Hint : $(X-a)(X-b) =X^2 - (a+b)X + ab$






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    0














    If a+b is algebraic,



    $Longrightarrow a = k+l and b = k-l $,for some algebraic number k.



    $Longrightarrow ab=k^2-l^2$, is algebraic (given)



    But, we know that the difference of two algebraic numbers is algebraic,



    $therefore k^2-ab=l^2$ is algebraic.



    $therefore l (= sqrt {l^2})$ is algebraic.



    $therefore a = k+l and b = k-l$ are algebraic numbers.



    Hope it is helpful:)






    share|cite|improve this answer





















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      2 Answers
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      2 Answers
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      1














      Hint : $(X-a)(X-b) =X^2 - (a+b)X + ab$






      share|cite|improve this answer


























        1














        Hint : $(X-a)(X-b) =X^2 - (a+b)X + ab$






        share|cite|improve this answer
























          1












          1








          1






          Hint : $(X-a)(X-b) =X^2 - (a+b)X + ab$






          share|cite|improve this answer












          Hint : $(X-a)(X-b) =X^2 - (a+b)X + ab$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 29 '18 at 13:32









          MaxMax

          12.9k11040




          12.9k11040























              0














              If a+b is algebraic,



              $Longrightarrow a = k+l and b = k-l $,for some algebraic number k.



              $Longrightarrow ab=k^2-l^2$, is algebraic (given)



              But, we know that the difference of two algebraic numbers is algebraic,



              $therefore k^2-ab=l^2$ is algebraic.



              $therefore l (= sqrt {l^2})$ is algebraic.



              $therefore a = k+l and b = k-l$ are algebraic numbers.



              Hope it is helpful:)






              share|cite|improve this answer


























                0














                If a+b is algebraic,



                $Longrightarrow a = k+l and b = k-l $,for some algebraic number k.



                $Longrightarrow ab=k^2-l^2$, is algebraic (given)



                But, we know that the difference of two algebraic numbers is algebraic,



                $therefore k^2-ab=l^2$ is algebraic.



                $therefore l (= sqrt {l^2})$ is algebraic.



                $therefore a = k+l and b = k-l$ are algebraic numbers.



                Hope it is helpful:)






                share|cite|improve this answer
























                  0












                  0








                  0






                  If a+b is algebraic,



                  $Longrightarrow a = k+l and b = k-l $,for some algebraic number k.



                  $Longrightarrow ab=k^2-l^2$, is algebraic (given)



                  But, we know that the difference of two algebraic numbers is algebraic,



                  $therefore k^2-ab=l^2$ is algebraic.



                  $therefore l (= sqrt {l^2})$ is algebraic.



                  $therefore a = k+l and b = k-l$ are algebraic numbers.



                  Hope it is helpful:)






                  share|cite|improve this answer












                  If a+b is algebraic,



                  $Longrightarrow a = k+l and b = k-l $,for some algebraic number k.



                  $Longrightarrow ab=k^2-l^2$, is algebraic (given)



                  But, we know that the difference of two algebraic numbers is algebraic,



                  $therefore k^2-ab=l^2$ is algebraic.



                  $therefore l (= sqrt {l^2})$ is algebraic.



                  $therefore a = k+l and b = k-l$ are algebraic numbers.



                  Hope it is helpful:)







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 29 '18 at 13:42









                  MartundMartund

                  1,407212




                  1,407212






























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