Let $F$ be a field and $f(x) in F[x]$ be a polynomial of degree $> 1$. If $f(a) = 0$ for some $a in F$,...
My Attempt at this is as follows:
Let $f(x) = a_0+a_1x+a_2x^2+...+a_mx^m$ $forall$ $a_i$ $in$ $F[X]$.
Given $f(a)=0$
$Rightarrow$ $$f(x) = (x-a)t(x) label{a}tag{1}$$ such that $deg(t(x))geq1$,
and since $deg(x-alpha)=1$, we can express $f(x)$ as a product of two functions such that their degree is not zero, hence $f(x)$ is reducible.
So for proving the above result is this approach allowed? Or should I use some other way or do I have to prove the equation which I used in $ref{a}$?
abstract-algebra ring-theory
edited Nov 25 '18 at 11:05
the_fox
2,43411431
asked Nov 25 '18 at 10:06
Rohit Bharadwaj
518
add a comment |
My Attempt at this is as follows:
Let $f(x) = a_0+a_1x+a_2x^2+...+a_mx^m$ $forall$ $a_i$ $in$ $F[X]$.
Given $f(a)=0$
$Rightarrow$ $$f(x) = (x-a)t(x) label{a}tag{1}$$ such that $deg(t(x))geq1$,
and since $deg(x-alpha)=1$, we can express $f(x)$ as a product of two functions such that their degree is not zero, hence $f(x)$ is reducible.
So for proving the above result is this approach allowed? Or should I use some other way or do I have to prove the equation which I used in $ref{a}$?
abstract-algebra ring-theory
edited Nov 25 '18 at 11:05
the_fox
2,43411431
asked Nov 25 '18 at 10:06
Rohit Bharadwaj
518
add a comment |
My Attempt at this is as follows:
Let $f(x) = a_0+a_1x+a_2x^2+...+a_mx^m$ $forall$ $a_i$ $in$ $F[X]$.
Given $f(a)=0$
$Rightarrow$ $$f(x) = (x-a)t(x) label{a}tag{1}$$ such that $deg(t(x))geq1$,
and since $deg(x-alpha)=1$, we can express $f(x)$ as a product of two functions such that their degree is not zero, hence $f(x)$ is reducible.
So for proving the above result is this approach allowed? Or should I use some other way or do I have to prove the equation which I used in $ref{a}$?
abstract-algebra ring-theory
edited Nov 25 '18 at 11:05
the_fox
2,43411431
asked Nov 25 '18 at 10:06
Rohit Bharadwaj
518
My Attempt at this is as follows:
Let $f(x) = a_0+a_1x+a_2x^2+...+a_mx^m$ $forall$ $a_i$ $in$ $F[X]$.
Given $f(a)=0$
$Rightarrow$ $$f(x) = (x-a)t(x) label{a}tag{1}$$ such that $deg(t(x))geq1$,
and since $deg(x-alpha)=1$, we can express $f(x)$ as a product of two functions such that their degree is not zero, hence $f(x)$ is reducible.
So for proving the above result is this approach allowed? Or should I use some other way or do I have to prove the equation which I used in $ref{a}$?
abstract-algebra ring-theory
abstract-algebra ring-theory
edited Nov 25 '18 at 11:05
the_fox
2,43411431
asked Nov 25 '18 at 10:06
Rohit Bharadwaj
518
edited Nov 25 '18 at 11:05
the_fox
2,43411431
asked Nov 25 '18 at 10:06
Rohit Bharadwaj
518
edited Nov 25 '18 at 11:05
the_fox
2,43411431
edited Nov 25 '18 at 11:05
the_fox
2,43411431
edited Nov 25 '18 at 11:05
the_fox
2,43411431
2,43411431
asked Nov 25 '18 at 10:06
Rohit Bharadwaj
518
asked Nov 25 '18 at 10:06
Rohit Bharadwaj
518
asked Nov 25 '18 at 10:06
Rohit Bharadwaj
518
518
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
You should use polynomial division, otherwise you're simply stating what you want to prove. The division algorithm states that if $f(x)$ and $g(x)$ are polynomials, with $gne0$, then there exist unique polynomials $t(x)$ and $r(x)$ such that
$f(x)=g(x)t(x)+r(x)$;
$deg r(x)<deg g(x)$ (or $r=0$).
In this case we can take $g(x)=x-alpha$, so
$$
f(x)=(x-alpha)t(x)+r(x)
$$
where $r=0$ or has degree less than $deg(x-alpha)=1$. Therefore $r(x)=c$ is constant.
Evaluate at $alpha$:
$$
0=f(alpha)=(alpha-alpha)t(alpha)+c=c
$$
Hence $f(x)=(x-alpha)t(x)$. Since $deg f(x)>1$, $deg t(x)>0$, so $f$ is reducible.
answered Nov 25 '18 at 10:56
egreg
178k1484201
I think this is a more correct or better proof
– Rohit Bharadwaj
Nov 25 '18 at 11:01
one question, aren't you being specific in this case by taking $f(x)=(x-alpha)t(x)$ I mean you could have took any general polynomial right then why that one..... could we have any more general approach?
– Rohit Bharadwaj
Nov 26 '18 at 4:10
@RohitBharadwaj I stated the general theorem
– egreg
Nov 26 '18 at 8:52
i mean why did you take $(x-alpha)$, in the division algorithm it is $f(x)=g(x)t(x)+r(x)$
– Rohit Bharadwaj
Nov 26 '18 at 17:05
@RohitBharadwaj Where's the problem? You want to show divisibility by $x-alpha$, don't you?
– egreg
Nov 26 '18 at 17:10
|
show 2 more comments
Your proof is correct. I would not write “s.t. $deg t(x)geqslant1$”. It's more specific than that (although what you wrote is correct): $deg t(x)=m-1$. And $f(x)$ doesn't became reducible; it is reducible.
And the equality $(1)$ doesn't have to be proved (as part of this proof), since it is a standard theorem about polynomials.
answered Nov 25 '18 at 10:20
José Carlos Santos
150k22121221
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You should use polynomial division, otherwise you're simply stating what you want to prove. The division algorithm states that if $f(x)$ and $g(x)$ are polynomials, with $gne0$, then there exist unique polynomials $t(x)$ and $r(x)$ such that
$f(x)=g(x)t(x)+r(x)$;
$deg r(x)<deg g(x)$ (or $r=0$).
In this case we can take $g(x)=x-alpha$, so
$$
f(x)=(x-alpha)t(x)+r(x)
$$
where $r=0$ or has degree less than $deg(x-alpha)=1$. Therefore $r(x)=c$ is constant.
Evaluate at $alpha$:
$$
0=f(alpha)=(alpha-alpha)t(alpha)+c=c
$$
Hence $f(x)=(x-alpha)t(x)$. Since $deg f(x)>1$, $deg t(x)>0$, so $f$ is reducible.
answered Nov 25 '18 at 10:56
egreg
178k1484201
I think this is a more correct or better proof
– Rohit Bharadwaj
Nov 25 '18 at 11:01
one question, aren't you being specific in this case by taking $f(x)=(x-alpha)t(x)$ I mean you could have took any general polynomial right then why that one..... could we have any more general approach?
– Rohit Bharadwaj
Nov 26 '18 at 4:10
@RohitBharadwaj I stated the general theorem
– egreg
Nov 26 '18 at 8:52
i mean why did you take $(x-alpha)$, in the division algorithm it is $f(x)=g(x)t(x)+r(x)$
– Rohit Bharadwaj
Nov 26 '18 at 17:05
@RohitBharadwaj Where's the problem? You want to show divisibility by $x-alpha$, don't you?
– egreg
Nov 26 '18 at 17:10
|
show 2 more comments
You should use polynomial division, otherwise you're simply stating what you want to prove. The division algorithm states that if $f(x)$ and $g(x)$ are polynomials, with $gne0$, then there exist unique polynomials $t(x)$ and $r(x)$ such that
$f(x)=g(x)t(x)+r(x)$;
$deg r(x)<deg g(x)$ (or $r=0$).
In this case we can take $g(x)=x-alpha$, so
$$
f(x)=(x-alpha)t(x)+r(x)
$$
where $r=0$ or has degree less than $deg(x-alpha)=1$. Therefore $r(x)=c$ is constant.
Evaluate at $alpha$:
$$
0=f(alpha)=(alpha-alpha)t(alpha)+c=c
$$
Hence $f(x)=(x-alpha)t(x)$. Since $deg f(x)>1$, $deg t(x)>0$, so $f$ is reducible.
answered Nov 25 '18 at 10:56
egreg
178k1484201
I think this is a more correct or better proof
– Rohit Bharadwaj
Nov 25 '18 at 11:01
one question, aren't you being specific in this case by taking $f(x)=(x-alpha)t(x)$ I mean you could have took any general polynomial right then why that one..... could we have any more general approach?
– Rohit Bharadwaj
Nov 26 '18 at 4:10
@RohitBharadwaj I stated the general theorem
– egreg
Nov 26 '18 at 8:52
i mean why did you take $(x-alpha)$, in the division algorithm it is $f(x)=g(x)t(x)+r(x)$
– Rohit Bharadwaj
Nov 26 '18 at 17:05
@RohitBharadwaj Where's the problem? You want to show divisibility by $x-alpha$, don't you?
– egreg
Nov 26 '18 at 17:10
|
show 2 more comments
You should use polynomial division, otherwise you're simply stating what you want to prove. The division algorithm states that if $f(x)$ and $g(x)$ are polynomials, with $gne0$, then there exist unique polynomials $t(x)$ and $r(x)$ such that
$f(x)=g(x)t(x)+r(x)$;
$deg r(x)<deg g(x)$ (or $r=0$).
In this case we can take $g(x)=x-alpha$, so
$$
f(x)=(x-alpha)t(x)+r(x)
$$
where $r=0$ or has degree less than $deg(x-alpha)=1$. Therefore $r(x)=c$ is constant.
Evaluate at $alpha$:
$$
0=f(alpha)=(alpha-alpha)t(alpha)+c=c
$$
Hence $f(x)=(x-alpha)t(x)$. Since $deg f(x)>1$, $deg t(x)>0$, so $f$ is reducible.
answered Nov 25 '18 at 10:56
egreg
178k1484201
You should use polynomial division, otherwise you're simply stating what you want to prove. The division algorithm states that if $f(x)$ and $g(x)$ are polynomials, with $gne0$, then there exist unique polynomials $t(x)$ and $r(x)$ such that
$f(x)=g(x)t(x)+r(x)$;
$deg r(x)<deg g(x)$ (or $r=0$).
In this case we can take $g(x)=x-alpha$, so
$$
f(x)=(x-alpha)t(x)+r(x)
$$
where $r=0$ or has degree less than $deg(x-alpha)=1$. Therefore $r(x)=c$ is constant.
Evaluate at $alpha$:
$$
0=f(alpha)=(alpha-alpha)t(alpha)+c=c
$$
Hence $f(x)=(x-alpha)t(x)$. Since $deg f(x)>1$, $deg t(x)>0$, so $f$ is reducible.
answered Nov 25 '18 at 10:56
egreg
178k1484201
edited Nov 26 '18 at 8:52
answered Nov 25 '18 at 10:56
egreg
178k1484201
answered Nov 25 '18 at 10:56
egreg
178k1484201
answered Nov 25 '18 at 10:56
egreg
178k1484201
178k1484201
I think this is a more correct or better proof
– Rohit Bharadwaj
Nov 25 '18 at 11:01
one question, aren't you being specific in this case by taking $f(x)=(x-alpha)t(x)$ I mean you could have took any general polynomial right then why that one..... could we have any more general approach?
– Rohit Bharadwaj
Nov 26 '18 at 4:10
@RohitBharadwaj I stated the general theorem
– egreg
Nov 26 '18 at 8:52
i mean why did you take $(x-alpha)$, in the division algorithm it is $f(x)=g(x)t(x)+r(x)$
– Rohit Bharadwaj
Nov 26 '18 at 17:05
@RohitBharadwaj Where's the problem? You want to show divisibility by $x-alpha$, don't you?
– egreg
Nov 26 '18 at 17:10
|
show 2 more comments
I think this is a more correct or better proof
– Rohit Bharadwaj
Nov 25 '18 at 11:01
one question, aren't you being specific in this case by taking $f(x)=(x-alpha)t(x)$ I mean you could have took any general polynomial right then why that one..... could we have any more general approach?
– Rohit Bharadwaj
Nov 26 '18 at 4:10
@RohitBharadwaj I stated the general theorem
– egreg
Nov 26 '18 at 8:52
i mean why did you take $(x-alpha)$, in the division algorithm it is $f(x)=g(x)t(x)+r(x)$
– Rohit Bharadwaj
Nov 26 '18 at 17:05
@RohitBharadwaj Where's the problem? You want to show divisibility by $x-alpha$, don't you?
– egreg
Nov 26 '18 at 17:10
I think this is a more correct or better proof
– Rohit Bharadwaj
Nov 25 '18 at 11:01
I think this is a more correct or better proof
– Rohit Bharadwaj
Nov 25 '18 at 11:01
one question, aren't you being specific in this case by taking $f(x)=(x-alpha)t(x)$ I mean you could have took any general polynomial right then why that one..... could we have any more general approach?
– Rohit Bharadwaj
Nov 26 '18 at 4:10
one question, aren't you being specific in this case by taking $f(x)=(x-alpha)t(x)$ I mean you could have took any general polynomial right then why that one..... could we have any more general approach?
– Rohit Bharadwaj
Nov 26 '18 at 4:10
@RohitBharadwaj I stated the general theorem
– egreg
Nov 26 '18 at 8:52
@RohitBharadwaj I stated the general theorem
– egreg
Nov 26 '18 at 8:52
i mean why did you take $(x-alpha)$, in the division algorithm it is $f(x)=g(x)t(x)+r(x)$
– Rohit Bharadwaj
Nov 26 '18 at 17:05
i mean why did you take $(x-alpha)$, in the division algorithm it is $f(x)=g(x)t(x)+r(x)$
– Rohit Bharadwaj
Nov 26 '18 at 17:05
@RohitBharadwaj Where's the problem? You want to show divisibility by $x-alpha$, don't you?
– egreg
Nov 26 '18 at 17:10
@RohitBharadwaj Where's the problem? You want to show divisibility by $x-alpha$, don't you?
– egreg
Nov 26 '18 at 17:10
|
show 2 more comments
Your proof is correct. I would not write “s.t. $deg t(x)geqslant1$”. It's more specific than that (although what you wrote is correct): $deg t(x)=m-1$. And $f(x)$ doesn't became reducible; it is reducible.
And the equality $(1)$ doesn't have to be proved (as part of this proof), since it is a standard theorem about polynomials.
answered Nov 25 '18 at 10:20
José Carlos Santos
150k22121221
add a comment |
Your proof is correct. I would not write “s.t. $deg t(x)geqslant1$”. It's more specific than that (although what you wrote is correct): $deg t(x)=m-1$. And $f(x)$ doesn't became reducible; it is reducible.
And the equality $(1)$ doesn't have to be proved (as part of this proof), since it is a standard theorem about polynomials.
answered Nov 25 '18 at 10:20
José Carlos Santos
150k22121221
add a comment |
Your proof is correct. I would not write “s.t. $deg t(x)geqslant1$”. It's more specific than that (although what you wrote is correct): $deg t(x)=m-1$. And $f(x)$ doesn't became reducible; it is reducible.
And the equality $(1)$ doesn't have to be proved (as part of this proof), since it is a standard theorem about polynomials.
answered Nov 25 '18 at 10:20
José Carlos Santos
150k22121221
Your proof is correct. I would not write “s.t. $deg t(x)geqslant1$”. It's more specific than that (although what you wrote is correct): $deg t(x)=m-1$. And $f(x)$ doesn't became reducible; it is reducible.
And the equality $(1)$ doesn't have to be proved (as part of this proof), since it is a standard theorem about polynomials.
answered Nov 25 '18 at 10:20
José Carlos Santos
150k22121221
answered Nov 25 '18 at 10:20
José Carlos Santos
150k22121221
answered Nov 25 '18 at 10:20
José Carlos Santos
150k22121221
answered Nov 25 '18 at 10:20
José Carlos Santos
150k22121221
150k22121221
add a comment |
add a comment |
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