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Prove that $p(x) = alpha (x)f(x) + beta (x)g(x)$ for some irreducible $f(x)$, $g(x)$.


Determine the irreducible polynomial for $alpha=sqrt3+sqrt5$ over $mathbbQ(sqrt10)$Prove that $k(alpha+beta)=k(alpha,beta)$Prove that $Bbb Z_2(alpha)=Bbb Z_2(beta)$ .$[K(beta):K] degf_i = [K(alpha): K] degg_i$Show that $F(alpha)=F(beta)$ if $alpha$ and $beta$ are roots of the same irreducible polynomialWhen does $[Bbb Q(alpha,beta):Bbb Q]=[Bbb Q(alpha):Bbb Q][Bbb Q(beta):Bbb Q]$Show that $F [alpha,beta]$ is a field if $alpha$ and $beta$ are algebraic over $F$.Isomorphism between two extensions $Bbb F_2(alpha)$ and $Bbb F_2(beta)$A field with an irreducible, separable polynomial with roots $alpha$ and $alpha + 1$ must have positive characteristic.If $f_K(alpha)^betabig| f_K^beta$, then $degleft(f_K(alpha)^betaright)| degleft(f_K^betaright)$?













0












$begingroup$


Let $f(x)$ and $g(x)$ be irreducible polynomials of different degrees in field $F[x]$. Prove that for all $p(x) in F[x]$ that there exists some $alpha(x), beta(x)in F[x]$ such that $p(x) = alpha (x)f(x) + beta (x)g(x).$



I think I have a legitimate proof for this, however I would like to check my thinking.



First, $f(x)$ and $g(x)$ are irreducible and have different degrees, then they are the polynomial equivalent to relatively prime numbers and therefore have a GCD of 1. I did use my definition about irreducible polynomials from the text, but just want to soldify thinking with something I'm already comfortable with.



Second, since the above GCD = 1, then we can say that $1 = a(x)f(x)+b(x)g(x)$.



Can I then multiply both sides of the equation by $p(x)$ to result in $p(x) = p(x)a(x)f(x) + p(x)b(x)g(x)$?



Does that then allow me to say that as long as we define $alpha (x) = p(x)a(x)$ and $beta(x) = p(x)g(x)$ and we know that $F$ is a field so is multiplicatively closed, then $alpha(x), beta(x) in F$, that $p(x)=alpha(x)f(x)+beta(x)g(x)$?










share|cite|improve this question











$endgroup$











  • $begingroup$
    I think you mean for every polynomial $p(x)$ there exist $alpha(x)$ and $beta(x)$ such that...
    $endgroup$
    – Robert Israel
    Feb 17 at 21:23










  • $begingroup$
    The quantifiers in your statement aren't what you intend. What you mean to say, I believe, is that, given $p(x)in F[x]$ there exist $alpha(x),beta (x)in F[x]$ which satisfy the equation you write. That statement is what you (correctly) argue, starting from the result that you can solve the problem for $p(x)=1$.
    $endgroup$
    – lulu
    Feb 17 at 21:24










  • $begingroup$
    I corrected the original problem. Sorry, in an effort to shorten the statement I changed the meaning a little.
    $endgroup$
    – MT math
    Feb 17 at 21:36















0












$begingroup$


Let $f(x)$ and $g(x)$ be irreducible polynomials of different degrees in field $F[x]$. Prove that for all $p(x) in F[x]$ that there exists some $alpha(x), beta(x)in F[x]$ such that $p(x) = alpha (x)f(x) + beta (x)g(x).$



I think I have a legitimate proof for this, however I would like to check my thinking.



First, $f(x)$ and $g(x)$ are irreducible and have different degrees, then they are the polynomial equivalent to relatively prime numbers and therefore have a GCD of 1. I did use my definition about irreducible polynomials from the text, but just want to soldify thinking with something I'm already comfortable with.



Second, since the above GCD = 1, then we can say that $1 = a(x)f(x)+b(x)g(x)$.



Can I then multiply both sides of the equation by $p(x)$ to result in $p(x) = p(x)a(x)f(x) + p(x)b(x)g(x)$?



Does that then allow me to say that as long as we define $alpha (x) = p(x)a(x)$ and $beta(x) = p(x)g(x)$ and we know that $F$ is a field so is multiplicatively closed, then $alpha(x), beta(x) in F$, that $p(x)=alpha(x)f(x)+beta(x)g(x)$?










share|cite|improve this question











$endgroup$











  • $begingroup$
    I think you mean for every polynomial $p(x)$ there exist $alpha(x)$ and $beta(x)$ such that...
    $endgroup$
    – Robert Israel
    Feb 17 at 21:23










  • $begingroup$
    The quantifiers in your statement aren't what you intend. What you mean to say, I believe, is that, given $p(x)in F[x]$ there exist $alpha(x),beta (x)in F[x]$ which satisfy the equation you write. That statement is what you (correctly) argue, starting from the result that you can solve the problem for $p(x)=1$.
    $endgroup$
    – lulu
    Feb 17 at 21:24










  • $begingroup$
    I corrected the original problem. Sorry, in an effort to shorten the statement I changed the meaning a little.
    $endgroup$
    – MT math
    Feb 17 at 21:36













0












0








0





$begingroup$


Let $f(x)$ and $g(x)$ be irreducible polynomials of different degrees in field $F[x]$. Prove that for all $p(x) in F[x]$ that there exists some $alpha(x), beta(x)in F[x]$ such that $p(x) = alpha (x)f(x) + beta (x)g(x).$



I think I have a legitimate proof for this, however I would like to check my thinking.



First, $f(x)$ and $g(x)$ are irreducible and have different degrees, then they are the polynomial equivalent to relatively prime numbers and therefore have a GCD of 1. I did use my definition about irreducible polynomials from the text, but just want to soldify thinking with something I'm already comfortable with.



Second, since the above GCD = 1, then we can say that $1 = a(x)f(x)+b(x)g(x)$.



Can I then multiply both sides of the equation by $p(x)$ to result in $p(x) = p(x)a(x)f(x) + p(x)b(x)g(x)$?



Does that then allow me to say that as long as we define $alpha (x) = p(x)a(x)$ and $beta(x) = p(x)g(x)$ and we know that $F$ is a field so is multiplicatively closed, then $alpha(x), beta(x) in F$, that $p(x)=alpha(x)f(x)+beta(x)g(x)$?










share|cite|improve this question











$endgroup$




Let $f(x)$ and $g(x)$ be irreducible polynomials of different degrees in field $F[x]$. Prove that for all $p(x) in F[x]$ that there exists some $alpha(x), beta(x)in F[x]$ such that $p(x) = alpha (x)f(x) + beta (x)g(x).$



I think I have a legitimate proof for this, however I would like to check my thinking.



First, $f(x)$ and $g(x)$ are irreducible and have different degrees, then they are the polynomial equivalent to relatively prime numbers and therefore have a GCD of 1. I did use my definition about irreducible polynomials from the text, but just want to soldify thinking with something I'm already comfortable with.



Second, since the above GCD = 1, then we can say that $1 = a(x)f(x)+b(x)g(x)$.



Can I then multiply both sides of the equation by $p(x)$ to result in $p(x) = p(x)a(x)f(x) + p(x)b(x)g(x)$?



Does that then allow me to say that as long as we define $alpha (x) = p(x)a(x)$ and $beta(x) = p(x)g(x)$ and we know that $F$ is a field so is multiplicatively closed, then $alpha(x), beta(x) in F$, that $p(x)=alpha(x)f(x)+beta(x)g(x)$?







field-theory irreducible-polynomials






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago









Sil

5,16421643




5,16421643










asked Feb 17 at 21:18









MT mathMT math

11




11











  • $begingroup$
    I think you mean for every polynomial $p(x)$ there exist $alpha(x)$ and $beta(x)$ such that...
    $endgroup$
    – Robert Israel
    Feb 17 at 21:23










  • $begingroup$
    The quantifiers in your statement aren't what you intend. What you mean to say, I believe, is that, given $p(x)in F[x]$ there exist $alpha(x),beta (x)in F[x]$ which satisfy the equation you write. That statement is what you (correctly) argue, starting from the result that you can solve the problem for $p(x)=1$.
    $endgroup$
    – lulu
    Feb 17 at 21:24










  • $begingroup$
    I corrected the original problem. Sorry, in an effort to shorten the statement I changed the meaning a little.
    $endgroup$
    – MT math
    Feb 17 at 21:36
















  • $begingroup$
    I think you mean for every polynomial $p(x)$ there exist $alpha(x)$ and $beta(x)$ such that...
    $endgroup$
    – Robert Israel
    Feb 17 at 21:23










  • $begingroup$
    The quantifiers in your statement aren't what you intend. What you mean to say, I believe, is that, given $p(x)in F[x]$ there exist $alpha(x),beta (x)in F[x]$ which satisfy the equation you write. That statement is what you (correctly) argue, starting from the result that you can solve the problem for $p(x)=1$.
    $endgroup$
    – lulu
    Feb 17 at 21:24










  • $begingroup$
    I corrected the original problem. Sorry, in an effort to shorten the statement I changed the meaning a little.
    $endgroup$
    – MT math
    Feb 17 at 21:36















$begingroup$
I think you mean for every polynomial $p(x)$ there exist $alpha(x)$ and $beta(x)$ such that...
$endgroup$
– Robert Israel
Feb 17 at 21:23




$begingroup$
I think you mean for every polynomial $p(x)$ there exist $alpha(x)$ and $beta(x)$ such that...
$endgroup$
– Robert Israel
Feb 17 at 21:23












$begingroup$
The quantifiers in your statement aren't what you intend. What you mean to say, I believe, is that, given $p(x)in F[x]$ there exist $alpha(x),beta (x)in F[x]$ which satisfy the equation you write. That statement is what you (correctly) argue, starting from the result that you can solve the problem for $p(x)=1$.
$endgroup$
– lulu
Feb 17 at 21:24




$begingroup$
The quantifiers in your statement aren't what you intend. What you mean to say, I believe, is that, given $p(x)in F[x]$ there exist $alpha(x),beta (x)in F[x]$ which satisfy the equation you write. That statement is what you (correctly) argue, starting from the result that you can solve the problem for $p(x)=1$.
$endgroup$
– lulu
Feb 17 at 21:24












$begingroup$
I corrected the original problem. Sorry, in an effort to shorten the statement I changed the meaning a little.
$endgroup$
– MT math
Feb 17 at 21:36




$begingroup$
I corrected the original problem. Sorry, in an effort to shorten the statement I changed the meaning a little.
$endgroup$
– MT math
Feb 17 at 21:36










1 Answer
1






active

oldest

votes


















1












$begingroup$

Yes, if $p(x)$ and $q(x)$ are irreducible over the field $F$ and neither is a scalar multiple of the other their gcd is $1$, and then since $F[x]$ is a principal ideal domain you have Bézout's identity. In fact you just need $p(x)$ to be irreducible and not to divide $q(x)$ (or the same with $p$ and $q$ interchanged).






share|cite|improve this answer









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    $begingroup$

    Yes, if $p(x)$ and $q(x)$ are irreducible over the field $F$ and neither is a scalar multiple of the other their gcd is $1$, and then since $F[x]$ is a principal ideal domain you have Bézout's identity. In fact you just need $p(x)$ to be irreducible and not to divide $q(x)$ (or the same with $p$ and $q$ interchanged).






    share|cite|improve this answer









    $endgroup$

















      1












      $begingroup$

      Yes, if $p(x)$ and $q(x)$ are irreducible over the field $F$ and neither is a scalar multiple of the other their gcd is $1$, and then since $F[x]$ is a principal ideal domain you have Bézout's identity. In fact you just need $p(x)$ to be irreducible and not to divide $q(x)$ (or the same with $p$ and $q$ interchanged).






      share|cite|improve this answer









      $endgroup$















        1












        1








        1





        $begingroup$

        Yes, if $p(x)$ and $q(x)$ are irreducible over the field $F$ and neither is a scalar multiple of the other their gcd is $1$, and then since $F[x]$ is a principal ideal domain you have Bézout's identity. In fact you just need $p(x)$ to be irreducible and not to divide $q(x)$ (or the same with $p$ and $q$ interchanged).






        share|cite|improve this answer









        $endgroup$



        Yes, if $p(x)$ and $q(x)$ are irreducible over the field $F$ and neither is a scalar multiple of the other their gcd is $1$, and then since $F[x]$ is a principal ideal domain you have Bézout's identity. In fact you just need $p(x)$ to be irreducible and not to divide $q(x)$ (or the same with $p$ and $q$ interchanged).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Feb 17 at 21:29









        Robert IsraelRobert Israel

        327k23216470




        327k23216470



























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