Mixing
Finding the $gcd$ of $a(x)$ and $b(x)$ in field $mathbb{F}$
$begingroup$
I'm trying to find the $gcd$ of $a(x) = x^4 + 2x^3+x^2+4x+2$ and $b(x)=x^2+3x+1$ over $mathbb{F_5}$. I've already tried Euclid's algorithm:
$x^4 + 2x^3+x^2+4x+2 = x^2(x^2+3x+1) - x^3+4x+2$. Now I should express $b(x)$ in terms of the remainder $-x^3+4x+2$, but I'm not sure how to do this since $deg(b(x)) < 3$. Did I do something incorrectly? How do I find the $gcd$?
finite-fields euclidean-algorithm
edited Jan 25 at 17:53
J. W. Tanner
3,4601320
asked Jan 25 at 17:13
ZacharyZachary
1799
$endgroup$
add a comment |
$begingroup$
I'm trying to find the $gcd$ of $a(x) = x^4 + 2x^3+x^2+4x+2$ and $b(x)=x^2+3x+1$ over $mathbb{F_5}$. I've already tried Euclid's algorithm:
$x^4 + 2x^3+x^2+4x+2 = x^2(x^2+3x+1) - x^3+4x+2$. Now I should express $b(x)$ in terms of the remainder $-x^3+4x+2$, but I'm not sure how to do this since $deg(b(x)) < 3$. Did I do something incorrectly? How do I find the $gcd$?
finite-fields euclidean-algorithm
edited Jan 25 at 17:53
J. W. Tanner
3,4601320
asked Jan 25 at 17:13
ZacharyZachary
1799
$endgroup$
$begingroup$
You didn't complete the (Euclidean) division - doing so yields a remainder of smaller degree than the divisor.
$endgroup$
– Bill Dubuque
Jan 25 at 18:00
add a comment |
$begingroup$
I'm trying to find the $gcd$ of $a(x) = x^4 + 2x^3+x^2+4x+2$ and $b(x)=x^2+3x+1$ over $mathbb{F_5}$. I've already tried Euclid's algorithm:
$x^4 + 2x^3+x^2+4x+2 = x^2(x^2+3x+1) - x^3+4x+2$. Now I should express $b(x)$ in terms of the remainder $-x^3+4x+2$, but I'm not sure how to do this since $deg(b(x)) < 3$. Did I do something incorrectly? How do I find the $gcd$?
finite-fields euclidean-algorithm
edited Jan 25 at 17:53
J. W. Tanner
3,4601320
asked Jan 25 at 17:13
ZacharyZachary
1799
$endgroup$
I'm trying to find the $gcd$ of $a(x) = x^4 + 2x^3+x^2+4x+2$ and $b(x)=x^2+3x+1$ over $mathbb{F_5}$. I've already tried Euclid's algorithm:
$x^4 + 2x^3+x^2+4x+2 = x^2(x^2+3x+1) - x^3+4x+2$. Now I should express $b(x)$ in terms of the remainder $-x^3+4x+2$, but I'm not sure how to do this since $deg(b(x)) < 3$. Did I do something incorrectly? How do I find the $gcd$?
finite-fields euclidean-algorithm
finite-fields euclidean-algorithm
edited Jan 25 at 17:53
J. W. Tanner
3,4601320
asked Jan 25 at 17:13
ZacharyZachary
1799
edited Jan 25 at 17:53
J. W. Tanner
3,4601320
asked Jan 25 at 17:13
ZacharyZachary
1799
edited Jan 25 at 17:53
J. W. Tanner
3,4601320
edited Jan 25 at 17:53
J. W. Tanner
3,4601320
edited Jan 25 at 17:53
J. W. Tanner
3,4601320
3,4601320
asked Jan 25 at 17:13
ZacharyZachary
1799
asked Jan 25 at 17:13
ZacharyZachary
1799
asked Jan 25 at 17:13
ZacharyZachary
1799
1799
$begingroup$
You didn't complete the (Euclidean) division - doing so yields a remainder of smaller degree than the divisor.
$endgroup$
– Bill Dubuque
Jan 25 at 18:00
add a comment |
$begingroup$
You didn't complete the (Euclidean) division - doing so yields a remainder of smaller degree than the divisor.
$endgroup$
– Bill Dubuque
Jan 25 at 18:00
$begingroup$
You didn't complete the (Euclidean) division - doing so yields a remainder of smaller degree than the divisor.
$endgroup$
– Bill Dubuque
Jan 25 at 18:00
$begingroup$
You didn't complete the (Euclidean) division - doing so yields a remainder of smaller degree than the divisor.
$endgroup$
– Bill Dubuque
Jan 25 at 18:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It is perhaps easier just to factorize the polynomials over $Bbb F_5$ instead of using the Euclidean algorithm:
$$
x^4+2x^3+x^2+4x+2=(x^3+3x^2+4x+3)(x+4)
$$
$$
x^2+3x+1=(x+4)^2
$$
For this is enough to look for roots. The second polynomial obviously has $x=-4=1$ as a double root. So what about the first polynomial? Clearly it has also a root $1$. What about the cubic polynomial left?
So you see that $gcd(a(x),b(x))=x+4$ over $Bbb F_5$. Note that the gcd is $1$ over $Bbb Z$.
answered Jan 25 at 17:19
Dietrich BurdeDietrich Burde
81.1k648106
$endgroup$
$begingroup$
Thanks! One more question: assume I want to find the coefficients $lambda (x), mu (x)$ such that $gcd(a(x),b(x)) = lambda (x) a(x) + mu (x) b(x)$. Can I find them using your method?
$endgroup$
– Zachary
Jan 25 at 17:55
1
$begingroup$
@Zachary Use the Extended Euclidean algorithm, e.g. see here.
$endgroup$
– Bill Dubuque
Jan 25 at 18:03
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is perhaps easier just to factorize the polynomials over $Bbb F_5$ instead of using the Euclidean algorithm:
$$
x^4+2x^3+x^2+4x+2=(x^3+3x^2+4x+3)(x+4)
$$
$$
x^2+3x+1=(x+4)^2
$$
For this is enough to look for roots. The second polynomial obviously has $x=-4=1$ as a double root. So what about the first polynomial? Clearly it has also a root $1$. What about the cubic polynomial left?
So you see that $gcd(a(x),b(x))=x+4$ over $Bbb F_5$. Note that the gcd is $1$ over $Bbb Z$.
answered Jan 25 at 17:19
Dietrich BurdeDietrich Burde
81.1k648106
$endgroup$
$begingroup$
Thanks! One more question: assume I want to find the coefficients $lambda (x), mu (x)$ such that $gcd(a(x),b(x)) = lambda (x) a(x) + mu (x) b(x)$. Can I find them using your method?
$endgroup$
– Zachary
Jan 25 at 17:55
1
$begingroup$
@Zachary Use the Extended Euclidean algorithm, e.g. see here.
$endgroup$
– Bill Dubuque
Jan 25 at 18:03
add a comment |
$begingroup$
It is perhaps easier just to factorize the polynomials over $Bbb F_5$ instead of using the Euclidean algorithm:
$$
x^4+2x^3+x^2+4x+2=(x^3+3x^2+4x+3)(x+4)
$$
$$
x^2+3x+1=(x+4)^2
$$
For this is enough to look for roots. The second polynomial obviously has $x=-4=1$ as a double root. So what about the first polynomial? Clearly it has also a root $1$. What about the cubic polynomial left?
So you see that $gcd(a(x),b(x))=x+4$ over $Bbb F_5$. Note that the gcd is $1$ over $Bbb Z$.
answered Jan 25 at 17:19
Dietrich BurdeDietrich Burde
81.1k648106
$endgroup$
$begingroup$
Thanks! One more question: assume I want to find the coefficients $lambda (x), mu (x)$ such that $gcd(a(x),b(x)) = lambda (x) a(x) + mu (x) b(x)$. Can I find them using your method?
$endgroup$
– Zachary
Jan 25 at 17:55
1
$begingroup$
@Zachary Use the Extended Euclidean algorithm, e.g. see here.
$endgroup$
– Bill Dubuque
Jan 25 at 18:03
add a comment |
$begingroup$
It is perhaps easier just to factorize the polynomials over $Bbb F_5$ instead of using the Euclidean algorithm:
$$
x^4+2x^3+x^2+4x+2=(x^3+3x^2+4x+3)(x+4)
$$
$$
x^2+3x+1=(x+4)^2
$$
For this is enough to look for roots. The second polynomial obviously has $x=-4=1$ as a double root. So what about the first polynomial? Clearly it has also a root $1$. What about the cubic polynomial left?
So you see that $gcd(a(x),b(x))=x+4$ over $Bbb F_5$. Note that the gcd is $1$ over $Bbb Z$.
answered Jan 25 at 17:19
Dietrich BurdeDietrich Burde
81.1k648106
$endgroup$
It is perhaps easier just to factorize the polynomials over $Bbb F_5$ instead of using the Euclidean algorithm:
$$
x^4+2x^3+x^2+4x+2=(x^3+3x^2+4x+3)(x+4)
$$
$$
x^2+3x+1=(x+4)^2
$$
For this is enough to look for roots. The second polynomial obviously has $x=-4=1$ as a double root. So what about the first polynomial? Clearly it has also a root $1$. What about the cubic polynomial left?
So you see that $gcd(a(x),b(x))=x+4$ over $Bbb F_5$. Note that the gcd is $1$ over $Bbb Z$.
answered Jan 25 at 17:19
Dietrich BurdeDietrich Burde
81.1k648106
answered Jan 25 at 17:19
Dietrich BurdeDietrich Burde
81.1k648106
answered Jan 25 at 17:19
Dietrich BurdeDietrich Burde
81.1k648106
answered Jan 25 at 17:19
Dietrich BurdeDietrich Burde
81.1k648106
81.1k648106
$begingroup$
Thanks! One more question: assume I want to find the coefficients $lambda (x), mu (x)$ such that $gcd(a(x),b(x)) = lambda (x) a(x) + mu (x) b(x)$. Can I find them using your method?
$endgroup$
– Zachary
Jan 25 at 17:55
1
$begingroup$
@Zachary Use the Extended Euclidean algorithm, e.g. see here.
$endgroup$
– Bill Dubuque
Jan 25 at 18:03
add a comment |
$begingroup$
Thanks! One more question: assume I want to find the coefficients $lambda (x), mu (x)$ such that $gcd(a(x),b(x)) = lambda (x) a(x) + mu (x) b(x)$. Can I find them using your method?
$endgroup$
– Zachary
Jan 25 at 17:55
1
$begingroup$
@Zachary Use the Extended Euclidean algorithm, e.g. see here.
$endgroup$
– Bill Dubuque
Jan 25 at 18:03
$begingroup$
Thanks! One more question: assume I want to find the coefficients $lambda (x), mu (x)$ such that $gcd(a(x),b(x)) = lambda (x) a(x) + mu (x) b(x)$. Can I find them using your method?
$endgroup$
– Zachary
Jan 25 at 17:55
$begingroup$
Thanks! One more question: assume I want to find the coefficients $lambda (x), mu (x)$ such that $gcd(a(x),b(x)) = lambda (x) a(x) + mu (x) b(x)$. Can I find them using your method?
$endgroup$
– Zachary
Jan 25 at 17:55
1
1
$begingroup$
@Zachary Use the Extended Euclidean algorithm, e.g. see here.
$endgroup$
– Bill Dubuque
Jan 25 at 18:03
$begingroup$
@Zachary Use the Extended Euclidean algorithm, e.g. see here.
$endgroup$
– Bill Dubuque
Jan 25 at 18:03
add a comment |
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$begingroup$
You didn't complete the (Euclidean) division - doing so yields a remainder of smaller degree than the divisor.
$endgroup$
– Bill Dubuque
Jan 25 at 18:00