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 $degleft(f_K(alpha)^betaright)| degleft(f_K^betaright)$?
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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 $degleft(f_K(alpha)^betaright)| degleft(f_K^betaright)$?
$begingroup$
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 $p(x)=alpha(x)f(x)+beta(x)g(x)$?
field-theory irreducible-polynomials
edited 2 days ago
Sil
5,16421643
asked Feb 17 at 21:18
MT mathMT math
11
$endgroup$
add a comment |
$begingroup$
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 $p(x)=alpha(x)f(x)+beta(x)g(x)$?
field-theory irreducible-polynomials
edited 2 days ago
Sil
5,16421643
asked Feb 17 at 21:18
MT mathMT math
11
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $p(x)=alpha(x)f(x)+beta(x)g(x)$?
field-theory irreducible-polynomials
edited 2 days ago
Sil
5,16421643
asked Feb 17 at 21:18
MT mathMT math
11
$endgroup$
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
field-theory irreducible-polynomials
edited 2 days ago
Sil
5,16421643
asked Feb 17 at 21:18
MT mathMT math
11
edited 2 days ago
Sil
5,16421643
asked Feb 17 at 21:18
MT mathMT math
11
edited 2 days ago
Sil
5,16421643
edited 2 days ago
Sil
5,16421643
edited 2 days ago
Sil
5,16421643
5,16421643
asked Feb 17 at 21:18
MT mathMT math
11
asked Feb 17 at 21:18
MT mathMT math
11
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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).
answered Feb 17 at 21:29
Robert IsraelRobert Israel
327k23216470
$endgroup$
add a comment |
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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).
answered Feb 17 at 21:29
Robert IsraelRobert Israel
327k23216470
$endgroup$
add a comment |
$begingroup$
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).
answered Feb 17 at 21:29
Robert IsraelRobert Israel
327k23216470
$endgroup$
add a comment |
$begingroup$
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).
answered Feb 17 at 21:29
Robert IsraelRobert Israel
327k23216470
$endgroup$
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).
answered Feb 17 at 21:29
Robert IsraelRobert Israel
327k23216470
answered Feb 17 at 21:29
Robert IsraelRobert Israel
327k23216470
answered Feb 17 at 21:29
Robert IsraelRobert Israel
327k23216470
answered Feb 17 at 21:29
Robert IsraelRobert Israel
327k23216470
327k23216470
add a comment |
add a comment |
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6,G0mn,hI,dBrwR0,BHtrEmxl9y7H3CQ3KTQiJgloq4IKpT,BNQewDhKDFtGNndQhMTOyUznjBilIg6iH6z8 5FO w1HsRRxKnhOG
$begingroup$
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