Determine the greatest common divisor of polynomials $x^2+1$ and $x^3+1$ in $Bbb Q[X]$.
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Exercise: Determine a gcd of the polynomials $x^2+1$ and $x^3+1$ in $Bbb Q[X]$.. Write the gcd as a combination of the given polynomials.
Is it correct that I keep using long division until the result is $0$, and then the previous result would be a gcd?
x^2+1 / x^3 + 1 x
x^3 + x
________-
1 - x
1 - x / x^2 + 1 -x-1
x^2 - x
________-
x + 1
x - 1
________-
2
2 / 1 - x -1/2x + 1/2
- x
______-
1
1
______-
0
Conclusion: A gcd is $2$?
I'm not sure what's meant with "Write the gcd as a combination of the given polynomials.", so it would be great if someone could point me in the right direction.
linear-algebra polynomials rational-numbers greatest-common-divisor
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up vote
2
down vote
favorite
Exercise: Determine a gcd of the polynomials $x^2+1$ and $x^3+1$ in $Bbb Q[X]$.. Write the gcd as a combination of the given polynomials.
Is it correct that I keep using long division until the result is $0$, and then the previous result would be a gcd?
x^2+1 / x^3 + 1 x
x^3 + x
________-
1 - x
1 - x / x^2 + 1 -x-1
x^2 - x
________-
x + 1
x - 1
________-
2
2 / 1 - x -1/2x + 1/2
- x
______-
1
1
______-
0
Conclusion: A gcd is $2$?
I'm not sure what's meant with "Write the gcd as a combination of the given polynomials.", so it would be great if someone could point me in the right direction.
linear-algebra polynomials rational-numbers greatest-common-divisor
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Exercise: Determine a gcd of the polynomials $x^2+1$ and $x^3+1$ in $Bbb Q[X]$.. Write the gcd as a combination of the given polynomials.
Is it correct that I keep using long division until the result is $0$, and then the previous result would be a gcd?
x^2+1 / x^3 + 1 x
x^3 + x
________-
1 - x
1 - x / x^2 + 1 -x-1
x^2 - x
________-
x + 1
x - 1
________-
2
2 / 1 - x -1/2x + 1/2
- x
______-
1
1
______-
0
Conclusion: A gcd is $2$?
I'm not sure what's meant with "Write the gcd as a combination of the given polynomials.", so it would be great if someone could point me in the right direction.
linear-algebra polynomials rational-numbers greatest-common-divisor
Exercise: Determine a gcd of the polynomials $x^2+1$ and $x^3+1$ in $Bbb Q[X]$.. Write the gcd as a combination of the given polynomials.
Is it correct that I keep using long division until the result is $0$, and then the previous result would be a gcd?
x^2+1 / x^3 + 1 x
x^3 + x
________-
1 - x
1 - x / x^2 + 1 -x-1
x^2 - x
________-
x + 1
x - 1
________-
2
2 / 1 - x -1/2x + 1/2
- x
______-
1
1
______-
0
Conclusion: A gcd is $2$?
I'm not sure what's meant with "Write the gcd as a combination of the given polynomials.", so it would be great if someone could point me in the right direction.
linear-algebra polynomials rational-numbers greatest-common-divisor
linear-algebra polynomials rational-numbers greatest-common-divisor
edited Sep 9 at 11:14
thesmallprint
2,3991617
2,3991617
asked Sep 9 at 10:24
SJ19
204
204
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3 Answers
3
active
oldest
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up vote
1
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accepted
The Euclidean algorithm steps are
beginalign*
x^3+1&=x (x^2+1) + (-x+1)\
x^2+1&=-x (-x+1)+ (x+1)\
-x+1&=-(x+1)+2
endalign*
So the GCD is $1$.
We can go back up
beginalign*
2 &= (-x+1) +(x+1)\
&= (-x+1) + (x^2+1)+x (-x+1)\
&= (-x+1) (x+1)+(x^2+1)\
&= (x^3+1 - x(x^2+1))(x+1)+(x^2+1)\
&= (x+1) (x^3+1) + (-x^2-x+1) (x^2+1)
endalign*
and so $1 = frac12 [(x+1) (x^3+1) + (-x^2-x+1) (x^2+1)]$
Maybe I'm wrong but shouldn't gcd=1?? (It has to be a monic polynomial)
â giannispapav
Sep 9 at 11:16
Since the polynomials are members of $Bbb Q[x]$, wouldn't $$1=frac12(x+1) (x^3+1) + frac12(-x^2-x+1) (x^2+1)$$ work aswell? I fail to see how the gcd is even defined here.
â cansomeonehelpmeout
Sep 9 at 11:19
Thank you very much. Is it correct that 1 and 2 are both possible answers?
â SJ19
Sep 9 at 11:58
@SJ19: As a matter of convention, the greatest common divisor of polynomials is usually taken to mean the monic polynomial with the highest possible degree. So $1$ in this case.
â TonyK
Sep 9 at 12:05
@TonyK Alright. And how are you concluding GCD=1 from "-x+1 = -(x+1)+2"?
â SJ19
Sep 9 at 13:17
 |Â
show 1 more comment
up vote
1
down vote
Notice that $x^2+1$ is irreducible over $mathbbQ$. Let $d = gcd(x^2+1, x^3+1)$. Then $d mid x^2+1$ so $d$ is a nonzero constant, or $d$ is an associate of $x^2+1$.
However, $x^2+1 notmid x^3+1$ so $d$ must be constant. We can choose $d = 1$ since $gcd$ is usually supposed to be monic.
Finally, notice that
$$1=left[frac12(x+1) right](x^3+1) + left[frac12(-x^2-x+1)right] (x^2+1)$$
add a comment |Â
up vote
1
down vote
$$ left( x^3 + 1 right) $$
$$ left( x^2 + 1 right) $$
$$ left( x^3 + 1 right) = left( x^2 + 1 right) cdot colormagenta left( x right) + left( - x + 1 right) $$
$$ left( x^2 + 1 right) = left( - x + 1 right) cdot colormagenta left( - x - 1 right) + left( 2 right) $$
$$ left( - x + 1 right) = left( 2 right) cdot colormagenta left( frac - x + 1 2 right) + left( 0 right) $$
$$ frac 01 $$
$$ frac 10 $$
$$ colormagenta left( x right) Longrightarrow Longrightarrow frac left( x right) left( 1 right) $$
$$ colormagenta left( - x - 1 right) Longrightarrow Longrightarrow frac left( - x^2 - x + 1 right) left( - x - 1 right) $$
$$ colormagenta left( frac - x + 1 2 right) Longrightarrow Longrightarrow frac left( frac x^3 + 1 2 right) left( frac x^2 + 1 2 right) $$
$$ left( x^3 + 1 right) left( frac - x - 1 2 right) - left( x^2 + 1 right) left( frac - x^2 - x + 1 2 right) = left( -1 right) $$
..................................
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The Euclidean algorithm steps are
beginalign*
x^3+1&=x (x^2+1) + (-x+1)\
x^2+1&=-x (-x+1)+ (x+1)\
-x+1&=-(x+1)+2
endalign*
So the GCD is $1$.
We can go back up
beginalign*
2 &= (-x+1) +(x+1)\
&= (-x+1) + (x^2+1)+x (-x+1)\
&= (-x+1) (x+1)+(x^2+1)\
&= (x^3+1 - x(x^2+1))(x+1)+(x^2+1)\
&= (x+1) (x^3+1) + (-x^2-x+1) (x^2+1)
endalign*
and so $1 = frac12 [(x+1) (x^3+1) + (-x^2-x+1) (x^2+1)]$
Maybe I'm wrong but shouldn't gcd=1?? (It has to be a monic polynomial)
â giannispapav
Sep 9 at 11:16
Since the polynomials are members of $Bbb Q[x]$, wouldn't $$1=frac12(x+1) (x^3+1) + frac12(-x^2-x+1) (x^2+1)$$ work aswell? I fail to see how the gcd is even defined here.
â cansomeonehelpmeout
Sep 9 at 11:19
Thank you very much. Is it correct that 1 and 2 are both possible answers?
â SJ19
Sep 9 at 11:58
@SJ19: As a matter of convention, the greatest common divisor of polynomials is usually taken to mean the monic polynomial with the highest possible degree. So $1$ in this case.
â TonyK
Sep 9 at 12:05
@TonyK Alright. And how are you concluding GCD=1 from "-x+1 = -(x+1)+2"?
â SJ19
Sep 9 at 13:17
 |Â
show 1 more comment
up vote
1
down vote
accepted
The Euclidean algorithm steps are
beginalign*
x^3+1&=x (x^2+1) + (-x+1)\
x^2+1&=-x (-x+1)+ (x+1)\
-x+1&=-(x+1)+2
endalign*
So the GCD is $1$.
We can go back up
beginalign*
2 &= (-x+1) +(x+1)\
&= (-x+1) + (x^2+1)+x (-x+1)\
&= (-x+1) (x+1)+(x^2+1)\
&= (x^3+1 - x(x^2+1))(x+1)+(x^2+1)\
&= (x+1) (x^3+1) + (-x^2-x+1) (x^2+1)
endalign*
and so $1 = frac12 [(x+1) (x^3+1) + (-x^2-x+1) (x^2+1)]$
Maybe I'm wrong but shouldn't gcd=1?? (It has to be a monic polynomial)
â giannispapav
Sep 9 at 11:16
Since the polynomials are members of $Bbb Q[x]$, wouldn't $$1=frac12(x+1) (x^3+1) + frac12(-x^2-x+1) (x^2+1)$$ work aswell? I fail to see how the gcd is even defined here.
â cansomeonehelpmeout
Sep 9 at 11:19
Thank you very much. Is it correct that 1 and 2 are both possible answers?
â SJ19
Sep 9 at 11:58
@SJ19: As a matter of convention, the greatest common divisor of polynomials is usually taken to mean the monic polynomial with the highest possible degree. So $1$ in this case.
â TonyK
Sep 9 at 12:05
@TonyK Alright. And how are you concluding GCD=1 from "-x+1 = -(x+1)+2"?
â SJ19
Sep 9 at 13:17
 |Â
show 1 more comment
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The Euclidean algorithm steps are
beginalign*
x^3+1&=x (x^2+1) + (-x+1)\
x^2+1&=-x (-x+1)+ (x+1)\
-x+1&=-(x+1)+2
endalign*
So the GCD is $1$.
We can go back up
beginalign*
2 &= (-x+1) +(x+1)\
&= (-x+1) + (x^2+1)+x (-x+1)\
&= (-x+1) (x+1)+(x^2+1)\
&= (x^3+1 - x(x^2+1))(x+1)+(x^2+1)\
&= (x+1) (x^3+1) + (-x^2-x+1) (x^2+1)
endalign*
and so $1 = frac12 [(x+1) (x^3+1) + (-x^2-x+1) (x^2+1)]$
The Euclidean algorithm steps are
beginalign*
x^3+1&=x (x^2+1) + (-x+1)\
x^2+1&=-x (-x+1)+ (x+1)\
-x+1&=-(x+1)+2
endalign*
So the GCD is $1$.
We can go back up
beginalign*
2 &= (-x+1) +(x+1)\
&= (-x+1) + (x^2+1)+x (-x+1)\
&= (-x+1) (x+1)+(x^2+1)\
&= (x^3+1 - x(x^2+1))(x+1)+(x^2+1)\
&= (x+1) (x^3+1) + (-x^2-x+1) (x^2+1)
endalign*
and so $1 = frac12 [(x+1) (x^3+1) + (-x^2-x+1) (x^2+1)]$
edited Sep 9 at 11:44
answered Sep 9 at 10:47
P. Quinton
848111
848111
Maybe I'm wrong but shouldn't gcd=1?? (It has to be a monic polynomial)
â giannispapav
Sep 9 at 11:16
Since the polynomials are members of $Bbb Q[x]$, wouldn't $$1=frac12(x+1) (x^3+1) + frac12(-x^2-x+1) (x^2+1)$$ work aswell? I fail to see how the gcd is even defined here.
â cansomeonehelpmeout
Sep 9 at 11:19
Thank you very much. Is it correct that 1 and 2 are both possible answers?
â SJ19
Sep 9 at 11:58
@SJ19: As a matter of convention, the greatest common divisor of polynomials is usually taken to mean the monic polynomial with the highest possible degree. So $1$ in this case.
â TonyK
Sep 9 at 12:05
@TonyK Alright. And how are you concluding GCD=1 from "-x+1 = -(x+1)+2"?
â SJ19
Sep 9 at 13:17
 |Â
show 1 more comment
Maybe I'm wrong but shouldn't gcd=1?? (It has to be a monic polynomial)
â giannispapav
Sep 9 at 11:16
Since the polynomials are members of $Bbb Q[x]$, wouldn't $$1=frac12(x+1) (x^3+1) + frac12(-x^2-x+1) (x^2+1)$$ work aswell? I fail to see how the gcd is even defined here.
â cansomeonehelpmeout
Sep 9 at 11:19
Thank you very much. Is it correct that 1 and 2 are both possible answers?
â SJ19
Sep 9 at 11:58
@SJ19: As a matter of convention, the greatest common divisor of polynomials is usually taken to mean the monic polynomial with the highest possible degree. So $1$ in this case.
â TonyK
Sep 9 at 12:05
@TonyK Alright. And how are you concluding GCD=1 from "-x+1 = -(x+1)+2"?
â SJ19
Sep 9 at 13:17
Maybe I'm wrong but shouldn't gcd=1?? (It has to be a monic polynomial)
â giannispapav
Sep 9 at 11:16
Maybe I'm wrong but shouldn't gcd=1?? (It has to be a monic polynomial)
â giannispapav
Sep 9 at 11:16
Since the polynomials are members of $Bbb Q[x]$, wouldn't $$1=frac12(x+1) (x^3+1) + frac12(-x^2-x+1) (x^2+1)$$ work aswell? I fail to see how the gcd is even defined here.
â cansomeonehelpmeout
Sep 9 at 11:19
Since the polynomials are members of $Bbb Q[x]$, wouldn't $$1=frac12(x+1) (x^3+1) + frac12(-x^2-x+1) (x^2+1)$$ work aswell? I fail to see how the gcd is even defined here.
â cansomeonehelpmeout
Sep 9 at 11:19
Thank you very much. Is it correct that 1 and 2 are both possible answers?
â SJ19
Sep 9 at 11:58
Thank you very much. Is it correct that 1 and 2 are both possible answers?
â SJ19
Sep 9 at 11:58
@SJ19: As a matter of convention, the greatest common divisor of polynomials is usually taken to mean the monic polynomial with the highest possible degree. So $1$ in this case.
â TonyK
Sep 9 at 12:05
@SJ19: As a matter of convention, the greatest common divisor of polynomials is usually taken to mean the monic polynomial with the highest possible degree. So $1$ in this case.
â TonyK
Sep 9 at 12:05
@TonyK Alright. And how are you concluding GCD=1 from "-x+1 = -(x+1)+2"?
â SJ19
Sep 9 at 13:17
@TonyK Alright. And how are you concluding GCD=1 from "-x+1 = -(x+1)+2"?
â SJ19
Sep 9 at 13:17
 |Â
show 1 more comment
up vote
1
down vote
Notice that $x^2+1$ is irreducible over $mathbbQ$. Let $d = gcd(x^2+1, x^3+1)$. Then $d mid x^2+1$ so $d$ is a nonzero constant, or $d$ is an associate of $x^2+1$.
However, $x^2+1 notmid x^3+1$ so $d$ must be constant. We can choose $d = 1$ since $gcd$ is usually supposed to be monic.
Finally, notice that
$$1=left[frac12(x+1) right](x^3+1) + left[frac12(-x^2-x+1)right] (x^2+1)$$
add a comment |Â
up vote
1
down vote
Notice that $x^2+1$ is irreducible over $mathbbQ$. Let $d = gcd(x^2+1, x^3+1)$. Then $d mid x^2+1$ so $d$ is a nonzero constant, or $d$ is an associate of $x^2+1$.
However, $x^2+1 notmid x^3+1$ so $d$ must be constant. We can choose $d = 1$ since $gcd$ is usually supposed to be monic.
Finally, notice that
$$1=left[frac12(x+1) right](x^3+1) + left[frac12(-x^2-x+1)right] (x^2+1)$$
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Notice that $x^2+1$ is irreducible over $mathbbQ$. Let $d = gcd(x^2+1, x^3+1)$. Then $d mid x^2+1$ so $d$ is a nonzero constant, or $d$ is an associate of $x^2+1$.
However, $x^2+1 notmid x^3+1$ so $d$ must be constant. We can choose $d = 1$ since $gcd$ is usually supposed to be monic.
Finally, notice that
$$1=left[frac12(x+1) right](x^3+1) + left[frac12(-x^2-x+1)right] (x^2+1)$$
Notice that $x^2+1$ is irreducible over $mathbbQ$. Let $d = gcd(x^2+1, x^3+1)$. Then $d mid x^2+1$ so $d$ is a nonzero constant, or $d$ is an associate of $x^2+1$.
However, $x^2+1 notmid x^3+1$ so $d$ must be constant. We can choose $d = 1$ since $gcd$ is usually supposed to be monic.
Finally, notice that
$$1=left[frac12(x+1) right](x^3+1) + left[frac12(-x^2-x+1)right] (x^2+1)$$
answered Sep 9 at 11:50
mechanodroid
24.7k62245
24.7k62245
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up vote
1
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$$ left( x^3 + 1 right) $$
$$ left( x^2 + 1 right) $$
$$ left( x^3 + 1 right) = left( x^2 + 1 right) cdot colormagenta left( x right) + left( - x + 1 right) $$
$$ left( x^2 + 1 right) = left( - x + 1 right) cdot colormagenta left( - x - 1 right) + left( 2 right) $$
$$ left( - x + 1 right) = left( 2 right) cdot colormagenta left( frac - x + 1 2 right) + left( 0 right) $$
$$ frac 01 $$
$$ frac 10 $$
$$ colormagenta left( x right) Longrightarrow Longrightarrow frac left( x right) left( 1 right) $$
$$ colormagenta left( - x - 1 right) Longrightarrow Longrightarrow frac left( - x^2 - x + 1 right) left( - x - 1 right) $$
$$ colormagenta left( frac - x + 1 2 right) Longrightarrow Longrightarrow frac left( frac x^3 + 1 2 right) left( frac x^2 + 1 2 right) $$
$$ left( x^3 + 1 right) left( frac - x - 1 2 right) - left( x^2 + 1 right) left( frac - x^2 - x + 1 2 right) = left( -1 right) $$
..................................
add a comment |Â
up vote
1
down vote
$$ left( x^3 + 1 right) $$
$$ left( x^2 + 1 right) $$
$$ left( x^3 + 1 right) = left( x^2 + 1 right) cdot colormagenta left( x right) + left( - x + 1 right) $$
$$ left( x^2 + 1 right) = left( - x + 1 right) cdot colormagenta left( - x - 1 right) + left( 2 right) $$
$$ left( - x + 1 right) = left( 2 right) cdot colormagenta left( frac - x + 1 2 right) + left( 0 right) $$
$$ frac 01 $$
$$ frac 10 $$
$$ colormagenta left( x right) Longrightarrow Longrightarrow frac left( x right) left( 1 right) $$
$$ colormagenta left( - x - 1 right) Longrightarrow Longrightarrow frac left( - x^2 - x + 1 right) left( - x - 1 right) $$
$$ colormagenta left( frac - x + 1 2 right) Longrightarrow Longrightarrow frac left( frac x^3 + 1 2 right) left( frac x^2 + 1 2 right) $$
$$ left( x^3 + 1 right) left( frac - x - 1 2 right) - left( x^2 + 1 right) left( frac - x^2 - x + 1 2 right) = left( -1 right) $$
..................................
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$$ left( x^3 + 1 right) $$
$$ left( x^2 + 1 right) $$
$$ left( x^3 + 1 right) = left( x^2 + 1 right) cdot colormagenta left( x right) + left( - x + 1 right) $$
$$ left( x^2 + 1 right) = left( - x + 1 right) cdot colormagenta left( - x - 1 right) + left( 2 right) $$
$$ left( - x + 1 right) = left( 2 right) cdot colormagenta left( frac - x + 1 2 right) + left( 0 right) $$
$$ frac 01 $$
$$ frac 10 $$
$$ colormagenta left( x right) Longrightarrow Longrightarrow frac left( x right) left( 1 right) $$
$$ colormagenta left( - x - 1 right) Longrightarrow Longrightarrow frac left( - x^2 - x + 1 right) left( - x - 1 right) $$
$$ colormagenta left( frac - x + 1 2 right) Longrightarrow Longrightarrow frac left( frac x^3 + 1 2 right) left( frac x^2 + 1 2 right) $$
$$ left( x^3 + 1 right) left( frac - x - 1 2 right) - left( x^2 + 1 right) left( frac - x^2 - x + 1 2 right) = left( -1 right) $$
..................................
$$ left( x^3 + 1 right) $$
$$ left( x^2 + 1 right) $$
$$ left( x^3 + 1 right) = left( x^2 + 1 right) cdot colormagenta left( x right) + left( - x + 1 right) $$
$$ left( x^2 + 1 right) = left( - x + 1 right) cdot colormagenta left( - x - 1 right) + left( 2 right) $$
$$ left( - x + 1 right) = left( 2 right) cdot colormagenta left( frac - x + 1 2 right) + left( 0 right) $$
$$ frac 01 $$
$$ frac 10 $$
$$ colormagenta left( x right) Longrightarrow Longrightarrow frac left( x right) left( 1 right) $$
$$ colormagenta left( - x - 1 right) Longrightarrow Longrightarrow frac left( - x^2 - x + 1 right) left( - x - 1 right) $$
$$ colormagenta left( frac - x + 1 2 right) Longrightarrow Longrightarrow frac left( frac x^3 + 1 2 right) left( frac x^2 + 1 2 right) $$
$$ left( x^3 + 1 right) left( frac - x - 1 2 right) - left( x^2 + 1 right) left( frac - x^2 - x + 1 2 right) = left( -1 right) $$
..................................
answered Sep 9 at 14:33
Will Jagy
98.3k595196
98.3k595196
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