Mixing
Proving divisibility of a polynomial by a square of a polynomial.
I need to prove that a polynomial $f left( x right) in mathbb{Q} left[ X right]$ is divisible by a square of a polynomial iff $f$ and $f'$ have a greatest common divisor of positive degree.
I have no idea where to start, except that I have to use the properties of Polynomial Rings as Euclidean Domain.
abstract-algebra polynomials ring-theory euclidean-domain
asked Nov 21 '18 at 10:44
Aniruddha Deshmukh
934418
add a comment |
I need to prove that a polynomial $f left( x right) in mathbb{Q} left[ X right]$ is divisible by a square of a polynomial iff $f$ and $f'$ have a greatest common divisor of positive degree.
I have no idea where to start, except that I have to use the properties of Polynomial Rings as Euclidean Domain.
abstract-algebra polynomials ring-theory euclidean-domain
asked Nov 21 '18 at 10:44
Aniruddha Deshmukh
934418
The easy direction: Suppose there is some polynomial $g(x)$, which is not a unit,such that $g(x)^2k_1(x)=f(x)$. Then take derivative on both side of the equation to get $2g(x)g'(x)k_1(x)+g(x)^2k_1'(x)=f'(x)$. Well this says that g(x) is a common factor of $f(x)$ and $f'(x)$.
– mathnoob
Nov 21 '18 at 11:22
add a comment |
I need to prove that a polynomial $f left( x right) in mathbb{Q} left[ X right]$ is divisible by a square of a polynomial iff $f$ and $f'$ have a greatest common divisor of positive degree.
I have no idea where to start, except that I have to use the properties of Polynomial Rings as Euclidean Domain.
abstract-algebra polynomials ring-theory euclidean-domain
asked Nov 21 '18 at 10:44
Aniruddha Deshmukh
934418
I need to prove that a polynomial $f left( x right) in mathbb{Q} left[ X right]$ is divisible by a square of a polynomial iff $f$ and $f'$ have a greatest common divisor of positive degree.
I have no idea where to start, except that I have to use the properties of Polynomial Rings as Euclidean Domain.
abstract-algebra polynomials ring-theory euclidean-domain
abstract-algebra polynomials ring-theory euclidean-domain
asked Nov 21 '18 at 10:44
Aniruddha Deshmukh
934418
asked Nov 21 '18 at 10:44
Aniruddha Deshmukh
934418
asked Nov 21 '18 at 10:44
Aniruddha Deshmukh
934418
asked Nov 21 '18 at 10:44
Aniruddha Deshmukh
934418
asked Nov 21 '18 at 10:44
Aniruddha Deshmukh
934418
934418
The easy direction: Suppose there is some polynomial $g(x)$, which is not a unit,such that $g(x)^2k_1(x)=f(x)$. Then take derivative on both side of the equation to get $2g(x)g'(x)k_1(x)+g(x)^2k_1'(x)=f'(x)$. Well this says that g(x) is a common factor of $f(x)$ and $f'(x)$.
– mathnoob
Nov 21 '18 at 11:22
add a comment |
The easy direction: Suppose there is some polynomial $g(x)$, which is not a unit,such that $g(x)^2k_1(x)=f(x)$. Then take derivative on both side of the equation to get $2g(x)g'(x)k_1(x)+g(x)^2k_1'(x)=f'(x)$. Well this says that g(x) is a common factor of $f(x)$ and $f'(x)$.
– mathnoob
Nov 21 '18 at 11:22
The easy direction: Suppose there is some polynomial $g(x)$, which is not a unit,such that $g(x)^2k_1(x)=f(x)$. Then take derivative on both side of the equation to get $2g(x)g'(x)k_1(x)+g(x)^2k_1'(x)=f'(x)$. Well this says that g(x) is a common factor of $f(x)$ and $f'(x)$.
– mathnoob
Nov 21 '18 at 11:22
The easy direction: Suppose there is some polynomial $g(x)$, which is not a unit,such that $g(x)^2k_1(x)=f(x)$. Then take derivative on both side of the equation to get $2g(x)g'(x)k_1(x)+g(x)^2k_1'(x)=f'(x)$. Well this says that g(x) is a common factor of $f(x)$ and $f'(x)$.
– mathnoob
Nov 21 '18 at 11:22
add a comment |
2 Answers
2
active
oldest
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Let $q(x)$ be an irreducible common divisor of $f$ and $f'$. Division of $f$ by $q^2$ provides
$$
f=kq^2+r, quad text{with},,,deg r<deg q^2
$$
and as $q$ divides $f$, then $q$ divides $r$, and thus
$$
f=kq^2+r_1q, quad text{with},,,deg r_1<deg q.
$$
Differentiating we obtain
$$
f'=k'q^2+(2kq'+r_1')q+r_1q'
$$
But since $q$ divides $f'$, then $q$ divides $r_1q'$, and since $q$ is irreducible, then $q$ divides $q'$or $r_1$. But, $,deg q>deg q'$, and hence $q$ does not divide $q'$, as $q'ne 0$. Hence $q$ divides $r_1$, which implies that $r_1equiv 0$, since $deg r_1<deg q$.
answered Nov 21 '18 at 11:13
Yiorgos S. Smyrlis
62.8k1383163
"Division of $f$ by $q$" should be "Division of $f$ by $q^2$" right? Also, "q divides f', then q' divides $r_1q$' " should be "q divides f' then q divides $r_1q$' "right?
– mathnoob
Nov 21 '18 at 11:25
@mathnoob Yes - I corrected my answer. Thanx!
– Yiorgos S. Smyrlis
Nov 21 '18 at 11:34
add a comment |
Hint: One direction is easy. For the other direction, note that $f$ and $f'$ have a greatest common divisor of positive degree iff $f$ and $f'$ have an irreducible common divisor.
Make sure to pinpoint where working over $mathbb{Q}$ enters in your argument.
answered Nov 21 '18 at 10:53
lhf
163k10167387
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
Let $q(x)$ be an irreducible common divisor of $f$ and $f'$. Division of $f$ by $q^2$ provides
$$
f=kq^2+r, quad text{with},,,deg r<deg q^2
$$
and as $q$ divides $f$, then $q$ divides $r$, and thus
$$
f=kq^2+r_1q, quad text{with},,,deg r_1<deg q.
$$
Differentiating we obtain
$$
f'=k'q^2+(2kq'+r_1')q+r_1q'
$$
But since $q$ divides $f'$, then $q$ divides $r_1q'$, and since $q$ is irreducible, then $q$ divides $q'$or $r_1$. But, $,deg q>deg q'$, and hence $q$ does not divide $q'$, as $q'ne 0$. Hence $q$ divides $r_1$, which implies that $r_1equiv 0$, since $deg r_1<deg q$.
answered Nov 21 '18 at 11:13
Yiorgos S. Smyrlis
62.8k1383163
"Division of $f$ by $q$" should be "Division of $f$ by $q^2$" right? Also, "q divides f', then q' divides $r_1q$' " should be "q divides f' then q divides $r_1q$' "right?
– mathnoob
Nov 21 '18 at 11:25
@mathnoob Yes - I corrected my answer. Thanx!
– Yiorgos S. Smyrlis
Nov 21 '18 at 11:34
add a comment |
Let $q(x)$ be an irreducible common divisor of $f$ and $f'$. Division of $f$ by $q^2$ provides
$$
f=kq^2+r, quad text{with},,,deg r<deg q^2
$$
and as $q$ divides $f$, then $q$ divides $r$, and thus
$$
f=kq^2+r_1q, quad text{with},,,deg r_1<deg q.
$$
Differentiating we obtain
$$
f'=k'q^2+(2kq'+r_1')q+r_1q'
$$
But since $q$ divides $f'$, then $q$ divides $r_1q'$, and since $q$ is irreducible, then $q$ divides $q'$or $r_1$. But, $,deg q>deg q'$, and hence $q$ does not divide $q'$, as $q'ne 0$. Hence $q$ divides $r_1$, which implies that $r_1equiv 0$, since $deg r_1<deg q$.
answered Nov 21 '18 at 11:13
Yiorgos S. Smyrlis
62.8k1383163
"Division of $f$ by $q$" should be "Division of $f$ by $q^2$" right? Also, "q divides f', then q' divides $r_1q$' " should be "q divides f' then q divides $r_1q$' "right?
– mathnoob
Nov 21 '18 at 11:25
@mathnoob Yes - I corrected my answer. Thanx!
– Yiorgos S. Smyrlis
Nov 21 '18 at 11:34
add a comment |
Let $q(x)$ be an irreducible common divisor of $f$ and $f'$. Division of $f$ by $q^2$ provides
$$
f=kq^2+r, quad text{with},,,deg r<deg q^2
$$
and as $q$ divides $f$, then $q$ divides $r$, and thus
$$
f=kq^2+r_1q, quad text{with},,,deg r_1<deg q.
$$
Differentiating we obtain
$$
f'=k'q^2+(2kq'+r_1')q+r_1q'
$$
But since $q$ divides $f'$, then $q$ divides $r_1q'$, and since $q$ is irreducible, then $q$ divides $q'$or $r_1$. But, $,deg q>deg q'$, and hence $q$ does not divide $q'$, as $q'ne 0$. Hence $q$ divides $r_1$, which implies that $r_1equiv 0$, since $deg r_1<deg q$.
answered Nov 21 '18 at 11:13
Yiorgos S. Smyrlis
62.8k1383163
Let $q(x)$ be an irreducible common divisor of $f$ and $f'$. Division of $f$ by $q^2$ provides
$$
f=kq^2+r, quad text{with},,,deg r<deg q^2
$$
and as $q$ divides $f$, then $q$ divides $r$, and thus
$$
f=kq^2+r_1q, quad text{with},,,deg r_1<deg q.
$$
Differentiating we obtain
$$
f'=k'q^2+(2kq'+r_1')q+r_1q'
$$
But since $q$ divides $f'$, then $q$ divides $r_1q'$, and since $q$ is irreducible, then $q$ divides $q'$or $r_1$. But, $,deg q>deg q'$, and hence $q$ does not divide $q'$, as $q'ne 0$. Hence $q$ divides $r_1$, which implies that $r_1equiv 0$, since $deg r_1<deg q$.
answered Nov 21 '18 at 11:13
Yiorgos S. Smyrlis
62.8k1383163
edited Nov 21 '18 at 11:34
answered Nov 21 '18 at 11:13
Yiorgos S. Smyrlis
62.8k1383163
answered Nov 21 '18 at 11:13
Yiorgos S. Smyrlis
62.8k1383163
answered Nov 21 '18 at 11:13
Yiorgos S. Smyrlis
62.8k1383163
62.8k1383163
"Division of $f$ by $q$" should be "Division of $f$ by $q^2$" right? Also, "q divides f', then q' divides $r_1q$' " should be "q divides f' then q divides $r_1q$' "right?
– mathnoob
Nov 21 '18 at 11:25
@mathnoob Yes - I corrected my answer. Thanx!
– Yiorgos S. Smyrlis
Nov 21 '18 at 11:34
add a comment |
"Division of $f$ by $q$" should be "Division of $f$ by $q^2$" right? Also, "q divides f', then q' divides $r_1q$' " should be "q divides f' then q divides $r_1q$' "right?
– mathnoob
Nov 21 '18 at 11:25
@mathnoob Yes - I corrected my answer. Thanx!
– Yiorgos S. Smyrlis
Nov 21 '18 at 11:34
"Division of $f$ by $q$" should be "Division of $f$ by $q^2$" right? Also, "q divides f', then q' divides $r_1q$' " should be "q divides f' then q divides $r_1q$' "right?
– mathnoob
Nov 21 '18 at 11:25
"Division of $f$ by $q$" should be "Division of $f$ by $q^2$" right? Also, "q divides f', then q' divides $r_1q$' " should be "q divides f' then q divides $r_1q$' "right?
– mathnoob
Nov 21 '18 at 11:25
@mathnoob Yes - I corrected my answer. Thanx!
– Yiorgos S. Smyrlis
Nov 21 '18 at 11:34
@mathnoob Yes - I corrected my answer. Thanx!
– Yiorgos S. Smyrlis
Nov 21 '18 at 11:34
add a comment |
Hint: One direction is easy. For the other direction, note that $f$ and $f'$ have a greatest common divisor of positive degree iff $f$ and $f'$ have an irreducible common divisor.
Make sure to pinpoint where working over $mathbb{Q}$ enters in your argument.
answered Nov 21 '18 at 10:53
lhf
163k10167387
add a comment |
Hint: One direction is easy. For the other direction, note that $f$ and $f'$ have a greatest common divisor of positive degree iff $f$ and $f'$ have an irreducible common divisor.
Make sure to pinpoint where working over $mathbb{Q}$ enters in your argument.
answered Nov 21 '18 at 10:53
lhf
163k10167387
add a comment |
Hint: One direction is easy. For the other direction, note that $f$ and $f'$ have a greatest common divisor of positive degree iff $f$ and $f'$ have an irreducible common divisor.
Make sure to pinpoint where working over $mathbb{Q}$ enters in your argument.
answered Nov 21 '18 at 10:53
lhf
163k10167387
Hint: One direction is easy. For the other direction, note that $f$ and $f'$ have a greatest common divisor of positive degree iff $f$ and $f'$ have an irreducible common divisor.
Make sure to pinpoint where working over $mathbb{Q}$ enters in your argument.
answered Nov 21 '18 at 10:53
lhf
163k10167387
edited Nov 21 '18 at 11:00
answered Nov 21 '18 at 10:53
lhf
163k10167387
answered Nov 21 '18 at 10:53
lhf
163k10167387
answered Nov 21 '18 at 10:53
lhf
163k10167387
163k10167387
add a comment |
add a comment |
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The easy direction: Suppose there is some polynomial $g(x)$, which is not a unit,such that $g(x)^2k_1(x)=f(x)$. Then take derivative on both side of the equation to get $2g(x)g'(x)k_1(x)+g(x)^2k_1'(x)=f'(x)$. Well this says that g(x) is a common factor of $f(x)$ and $f'(x)$.
– mathnoob
Nov 21 '18 at 11:22