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minimal polynomial of $1+alpha^2$ where $alpha$ is a root of $x^3-x-1$
Suppose $alpha$ is a root of $x^3-x-1$, and $gamma=1+alpha^2$.
What is the minimal polynomial of $gamma$ over $mathbb Q$?
abstract-algebra roots minimal-polynomials
edited Apr 8 '17 at 7:28
Marc van Leeuwen
86.5k5106220
asked Apr 7 '17 at 11:56
Lehostal11Lehostal11
214
add a comment |
Suppose $alpha$ is a root of $x^3-x-1$, and $gamma=1+alpha^2$.
What is the minimal polynomial of $gamma$ over $mathbb Q$?
abstract-algebra roots minimal-polynomials
edited Apr 8 '17 at 7:28
Marc van Leeuwen
86.5k5106220
asked Apr 7 '17 at 11:56
Lehostal11Lehostal11
214
1
What have you done so far? If you show your working then people are more willing and able to help on the parts you are stuck on
– lioness99a
Apr 7 '17 at 12:00
Hint: first find the minimum polynomial of $alpha^2$ by writing $alpha^3 - alpha = 1$ and squaring both sides.
– Magdiragdag
Apr 8 '17 at 7:34
add a comment |
Suppose $alpha$ is a root of $x^3-x-1$, and $gamma=1+alpha^2$.
What is the minimal polynomial of $gamma$ over $mathbb Q$?
abstract-algebra roots minimal-polynomials
edited Apr 8 '17 at 7:28
Marc van Leeuwen
86.5k5106220
asked Apr 7 '17 at 11:56
Lehostal11Lehostal11
214
Suppose $alpha$ is a root of $x^3-x-1$, and $gamma=1+alpha^2$.
What is the minimal polynomial of $gamma$ over $mathbb Q$?
abstract-algebra roots minimal-polynomials
abstract-algebra roots minimal-polynomials
edited Apr 8 '17 at 7:28
Marc van Leeuwen
86.5k5106220
asked Apr 7 '17 at 11:56
Lehostal11Lehostal11
214
edited Apr 8 '17 at 7:28
Marc van Leeuwen
86.5k5106220
asked Apr 7 '17 at 11:56
Lehostal11Lehostal11
214
edited Apr 8 '17 at 7:28
Marc van Leeuwen
86.5k5106220
edited Apr 8 '17 at 7:28
Marc van Leeuwen
86.5k5106220
edited Apr 8 '17 at 7:28
Marc van Leeuwen
86.5k5106220
86.5k5106220
asked Apr 7 '17 at 11:56
Lehostal11Lehostal11
214
asked Apr 7 '17 at 11:56
Lehostal11Lehostal11
214
asked Apr 7 '17 at 11:56
Lehostal11Lehostal11
214
214
1
What have you done so far? If you show your working then people are more willing and able to help on the parts you are stuck on
– lioness99a
Apr 7 '17 at 12:00
Hint: first find the minimum polynomial of $alpha^2$ by writing $alpha^3 - alpha = 1$ and squaring both sides.
– Magdiragdag
Apr 8 '17 at 7:34
add a comment |
1
What have you done so far? If you show your working then people are more willing and able to help on the parts you are stuck on
– lioness99a
Apr 7 '17 at 12:00
Hint: first find the minimum polynomial of $alpha^2$ by writing $alpha^3 - alpha = 1$ and squaring both sides.
– Magdiragdag
Apr 8 '17 at 7:34
1
1
What have you done so far? If you show your working then people are more willing and able to help on the parts you are stuck on
– lioness99a
Apr 7 '17 at 12:00
What have you done so far? If you show your working then people are more willing and able to help on the parts you are stuck on
– lioness99a
Apr 7 '17 at 12:00
Hint: first find the minimum polynomial of $alpha^2$ by writing $alpha^3 - alpha = 1$ and squaring both sides.
– Magdiragdag
Apr 8 '17 at 7:34
Hint: first find the minimum polynomial of $alpha^2$ by writing $alpha^3 - alpha = 1$ and squaring both sides.
– Magdiragdag
Apr 8 '17 at 7:34
add a comment |
4 Answers
4
active
oldest
votes
Write the matrix of the map $x mapsto gamma x$ in the basis $1,alpha,alpha^2$.
The minimal polynomial of this matrix is the minimal polynomial of $gamma$.
answered Apr 7 '17 at 13:25
lhflhf
163k10168388
add a comment |
A bit of linear algebra makes this very straightforward. To expand on the answer by lhf, the minimal polynomial over$defQ{Bbb Q}~Q$ of an element $beta$ is also the minimal polynomial of the $Q$-linear map of mulitpliciation by$~beta$. The (sub)field $Q[alpha]$ is isomorphic to $Q[X]/(X^3-X-1)$, and in that $Q$-vector space the matrix, on the basis $[1,alpha,alpha^2]$ of images of $1,X,X^2$ is the companion matrix of $X^3-X-1$, which is
$$ M=pmatrix{0&0&1\1&0&1\0&1&0}.
$$
The minimal polynomial of $gamma=alpha^2+1$ is of degree$~3$ (the only other possibility of an element of $Q[alpha]$ is degree$~1$, but that clearly is not the case here since $X^2+1$ is not divisible by $X^3-X-1$), so it will also be the characteristic polynomial of multiplication by$~gamma$, which has matrix $$ M^2+I=pmatrix{1&1&0\0&2&1\1&0&2}.$$
The characteristic polynomial of that matrix is easily computed to be $X^3-5X^2+8X-5$.
answered Apr 8 '17 at 10:07
Marc van LeeuwenMarc van Leeuwen
86.5k5106220
add a comment |
Hints:
Observe that $;alpha^3=alpha+1;$ , and from here
$$gamma^2=1+2alpha^2+alpha^2+alpha=gamma+3(gamma-1);ldots$$
answered Apr 7 '17 at 12:06
DonAntonioDonAntonio
177k1492225
add a comment |
Hint:
Find the minimal polynomial $p(x)$ of $alpha^2$, using Vieta's relations between the roots of $x^3-x-1$.
The minimal polynomial of $1+alpha^2$ will be $q(x)=p(x-1)$.
Some details:
Explicitely, denotin $beta$ and $gamma$ the other roots of $x^3-x-1$, you have to calculate $$S=alpha^2+beta^2+gamma^2,quad S_2=alpha^2beta^2+beta^2gamma^2+gamma^2alpha^2,quad P=alpha^2beta^2gamma^2,$$ knowing
$$s=alpha+beta+gamma,quad s_2=alphabeta+betagamma+gammaalpha,quad p=alphabetagamma.$$
This is possible by Newton's theorem, since $S, S_2$ and $P$ are symmetric functions in $alpha,beta,gamma$.
Example:
$$S=(alpha+beta+gamma)^2-2(alphabeta+betagamma+gammaalpha)=s^2-2s_2$$
answered Apr 7 '17 at 12:06
BernardBernard
118k639112
I try it this way. But it is still hard. Can you give me more hints?
– Lehostal11
Apr 7 '17 at 12:33
@Lehostal: I've added some details.
– Bernard
Apr 7 '17 at 13:10
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
Write the matrix of the map $x mapsto gamma x$ in the basis $1,alpha,alpha^2$.
The minimal polynomial of this matrix is the minimal polynomial of $gamma$.
answered Apr 7 '17 at 13:25
lhflhf
163k10168388
add a comment |
Write the matrix of the map $x mapsto gamma x$ in the basis $1,alpha,alpha^2$.
The minimal polynomial of this matrix is the minimal polynomial of $gamma$.
answered Apr 7 '17 at 13:25
lhflhf
163k10168388
add a comment |
Write the matrix of the map $x mapsto gamma x$ in the basis $1,alpha,alpha^2$.
The minimal polynomial of this matrix is the minimal polynomial of $gamma$.
answered Apr 7 '17 at 13:25
lhflhf
163k10168388
Write the matrix of the map $x mapsto gamma x$ in the basis $1,alpha,alpha^2$.
The minimal polynomial of this matrix is the minimal polynomial of $gamma$.
answered Apr 7 '17 at 13:25
lhflhf
163k10168388
answered Apr 7 '17 at 13:25
lhflhf
163k10168388
answered Apr 7 '17 at 13:25
lhflhf
163k10168388
answered Apr 7 '17 at 13:25
lhflhf
163k10168388
163k10168388
add a comment |
add a comment |
A bit of linear algebra makes this very straightforward. To expand on the answer by lhf, the minimal polynomial over$defQ{Bbb Q}~Q$ of an element $beta$ is also the minimal polynomial of the $Q$-linear map of mulitpliciation by$~beta$. The (sub)field $Q[alpha]$ is isomorphic to $Q[X]/(X^3-X-1)$, and in that $Q$-vector space the matrix, on the basis $[1,alpha,alpha^2]$ of images of $1,X,X^2$ is the companion matrix of $X^3-X-1$, which is
$$ M=pmatrix{0&0&1\1&0&1\0&1&0}.
$$
The minimal polynomial of $gamma=alpha^2+1$ is of degree$~3$ (the only other possibility of an element of $Q[alpha]$ is degree$~1$, but that clearly is not the case here since $X^2+1$ is not divisible by $X^3-X-1$), so it will also be the characteristic polynomial of multiplication by$~gamma$, which has matrix $$ M^2+I=pmatrix{1&1&0\0&2&1\1&0&2}.$$
The characteristic polynomial of that matrix is easily computed to be $X^3-5X^2+8X-5$.
answered Apr 8 '17 at 10:07
Marc van LeeuwenMarc van Leeuwen
86.5k5106220
add a comment |
A bit of linear algebra makes this very straightforward. To expand on the answer by lhf, the minimal polynomial over$defQ{Bbb Q}~Q$ of an element $beta$ is also the minimal polynomial of the $Q$-linear map of mulitpliciation by$~beta$. The (sub)field $Q[alpha]$ is isomorphic to $Q[X]/(X^3-X-1)$, and in that $Q$-vector space the matrix, on the basis $[1,alpha,alpha^2]$ of images of $1,X,X^2$ is the companion matrix of $X^3-X-1$, which is
$$ M=pmatrix{0&0&1\1&0&1\0&1&0}.
$$
The minimal polynomial of $gamma=alpha^2+1$ is of degree$~3$ (the only other possibility of an element of $Q[alpha]$ is degree$~1$, but that clearly is not the case here since $X^2+1$ is not divisible by $X^3-X-1$), so it will also be the characteristic polynomial of multiplication by$~gamma$, which has matrix $$ M^2+I=pmatrix{1&1&0\0&2&1\1&0&2}.$$
The characteristic polynomial of that matrix is easily computed to be $X^3-5X^2+8X-5$.
answered Apr 8 '17 at 10:07
Marc van LeeuwenMarc van Leeuwen
86.5k5106220
add a comment |
A bit of linear algebra makes this very straightforward. To expand on the answer by lhf, the minimal polynomial over$defQ{Bbb Q}~Q$ of an element $beta$ is also the minimal polynomial of the $Q$-linear map of mulitpliciation by$~beta$. The (sub)field $Q[alpha]$ is isomorphic to $Q[X]/(X^3-X-1)$, and in that $Q$-vector space the matrix, on the basis $[1,alpha,alpha^2]$ of images of $1,X,X^2$ is the companion matrix of $X^3-X-1$, which is
$$ M=pmatrix{0&0&1\1&0&1\0&1&0}.
$$
The minimal polynomial of $gamma=alpha^2+1$ is of degree$~3$ (the only other possibility of an element of $Q[alpha]$ is degree$~1$, but that clearly is not the case here since $X^2+1$ is not divisible by $X^3-X-1$), so it will also be the characteristic polynomial of multiplication by$~gamma$, which has matrix $$ M^2+I=pmatrix{1&1&0\0&2&1\1&0&2}.$$
The characteristic polynomial of that matrix is easily computed to be $X^3-5X^2+8X-5$.
answered Apr 8 '17 at 10:07
Marc van LeeuwenMarc van Leeuwen
86.5k5106220
A bit of linear algebra makes this very straightforward. To expand on the answer by lhf, the minimal polynomial over$defQ{Bbb Q}~Q$ of an element $beta$ is also the minimal polynomial of the $Q$-linear map of mulitpliciation by$~beta$. The (sub)field $Q[alpha]$ is isomorphic to $Q[X]/(X^3-X-1)$, and in that $Q$-vector space the matrix, on the basis $[1,alpha,alpha^2]$ of images of $1,X,X^2$ is the companion matrix of $X^3-X-1$, which is
$$ M=pmatrix{0&0&1\1&0&1\0&1&0}.
$$
The minimal polynomial of $gamma=alpha^2+1$ is of degree$~3$ (the only other possibility of an element of $Q[alpha]$ is degree$~1$, but that clearly is not the case here since $X^2+1$ is not divisible by $X^3-X-1$), so it will also be the characteristic polynomial of multiplication by$~gamma$, which has matrix $$ M^2+I=pmatrix{1&1&0\0&2&1\1&0&2}.$$
The characteristic polynomial of that matrix is easily computed to be $X^3-5X^2+8X-5$.
answered Apr 8 '17 at 10:07
Marc van LeeuwenMarc van Leeuwen
86.5k5106220
edited Nov 22 '18 at 10:25
answered Apr 8 '17 at 10:07
Marc van LeeuwenMarc van Leeuwen
86.5k5106220
answered Apr 8 '17 at 10:07
Marc van LeeuwenMarc van Leeuwen
86.5k5106220
answered Apr 8 '17 at 10:07
Marc van LeeuwenMarc van Leeuwen
86.5k5106220
86.5k5106220
add a comment |
add a comment |
Hints:
Observe that $;alpha^3=alpha+1;$ , and from here
$$gamma^2=1+2alpha^2+alpha^2+alpha=gamma+3(gamma-1);ldots$$
answered Apr 7 '17 at 12:06
DonAntonioDonAntonio
177k1492225
add a comment |
Hints:
Observe that $;alpha^3=alpha+1;$ , and from here
$$gamma^2=1+2alpha^2+alpha^2+alpha=gamma+3(gamma-1);ldots$$
answered Apr 7 '17 at 12:06
DonAntonioDonAntonio
177k1492225
add a comment |
Hints:
Observe that $;alpha^3=alpha+1;$ , and from here
$$gamma^2=1+2alpha^2+alpha^2+alpha=gamma+3(gamma-1);ldots$$
answered Apr 7 '17 at 12:06
DonAntonioDonAntonio
177k1492225
Hints:
Observe that $;alpha^3=alpha+1;$ , and from here
$$gamma^2=1+2alpha^2+alpha^2+alpha=gamma+3(gamma-1);ldots$$
answered Apr 7 '17 at 12:06
DonAntonioDonAntonio
177k1492225
answered Apr 7 '17 at 12:06
DonAntonioDonAntonio
177k1492225
answered Apr 7 '17 at 12:06
DonAntonioDonAntonio
177k1492225
answered Apr 7 '17 at 12:06
DonAntonioDonAntonio
177k1492225
177k1492225
add a comment |
add a comment |
Hint:
Find the minimal polynomial $p(x)$ of $alpha^2$, using Vieta's relations between the roots of $x^3-x-1$.
The minimal polynomial of $1+alpha^2$ will be $q(x)=p(x-1)$.
Some details:
Explicitely, denotin $beta$ and $gamma$ the other roots of $x^3-x-1$, you have to calculate $$S=alpha^2+beta^2+gamma^2,quad S_2=alpha^2beta^2+beta^2gamma^2+gamma^2alpha^2,quad P=alpha^2beta^2gamma^2,$$ knowing
$$s=alpha+beta+gamma,quad s_2=alphabeta+betagamma+gammaalpha,quad p=alphabetagamma.$$
This is possible by Newton's theorem, since $S, S_2$ and $P$ are symmetric functions in $alpha,beta,gamma$.
Example:
$$S=(alpha+beta+gamma)^2-2(alphabeta+betagamma+gammaalpha)=s^2-2s_2$$
answered Apr 7 '17 at 12:06
BernardBernard
118k639112
I try it this way. But it is still hard. Can you give me more hints?
– Lehostal11
Apr 7 '17 at 12:33
@Lehostal: I've added some details.
– Bernard
Apr 7 '17 at 13:10
add a comment |
Hint:
Find the minimal polynomial $p(x)$ of $alpha^2$, using Vieta's relations between the roots of $x^3-x-1$.
The minimal polynomial of $1+alpha^2$ will be $q(x)=p(x-1)$.
Some details:
Explicitely, denotin $beta$ and $gamma$ the other roots of $x^3-x-1$, you have to calculate $$S=alpha^2+beta^2+gamma^2,quad S_2=alpha^2beta^2+beta^2gamma^2+gamma^2alpha^2,quad P=alpha^2beta^2gamma^2,$$ knowing
$$s=alpha+beta+gamma,quad s_2=alphabeta+betagamma+gammaalpha,quad p=alphabetagamma.$$
This is possible by Newton's theorem, since $S, S_2$ and $P$ are symmetric functions in $alpha,beta,gamma$.
Example:
$$S=(alpha+beta+gamma)^2-2(alphabeta+betagamma+gammaalpha)=s^2-2s_2$$
answered Apr 7 '17 at 12:06
BernardBernard
118k639112
I try it this way. But it is still hard. Can you give me more hints?
– Lehostal11
Apr 7 '17 at 12:33
@Lehostal: I've added some details.
– Bernard
Apr 7 '17 at 13:10
add a comment |
Hint:
Find the minimal polynomial $p(x)$ of $alpha^2$, using Vieta's relations between the roots of $x^3-x-1$.
The minimal polynomial of $1+alpha^2$ will be $q(x)=p(x-1)$.
Some details:
Explicitely, denotin $beta$ and $gamma$ the other roots of $x^3-x-1$, you have to calculate $$S=alpha^2+beta^2+gamma^2,quad S_2=alpha^2beta^2+beta^2gamma^2+gamma^2alpha^2,quad P=alpha^2beta^2gamma^2,$$ knowing
$$s=alpha+beta+gamma,quad s_2=alphabeta+betagamma+gammaalpha,quad p=alphabetagamma.$$
This is possible by Newton's theorem, since $S, S_2$ and $P$ are symmetric functions in $alpha,beta,gamma$.
Example:
$$S=(alpha+beta+gamma)^2-2(alphabeta+betagamma+gammaalpha)=s^2-2s_2$$
answered Apr 7 '17 at 12:06
BernardBernard
118k639112
Hint:
Find the minimal polynomial $p(x)$ of $alpha^2$, using Vieta's relations between the roots of $x^3-x-1$.
The minimal polynomial of $1+alpha^2$ will be $q(x)=p(x-1)$.
Some details:
Explicitely, denotin $beta$ and $gamma$ the other roots of $x^3-x-1$, you have to calculate $$S=alpha^2+beta^2+gamma^2,quad S_2=alpha^2beta^2+beta^2gamma^2+gamma^2alpha^2,quad P=alpha^2beta^2gamma^2,$$ knowing
$$s=alpha+beta+gamma,quad s_2=alphabeta+betagamma+gammaalpha,quad p=alphabetagamma.$$
This is possible by Newton's theorem, since $S, S_2$ and $P$ are symmetric functions in $alpha,beta,gamma$.
Example:
$$S=(alpha+beta+gamma)^2-2(alphabeta+betagamma+gammaalpha)=s^2-2s_2$$
answered Apr 7 '17 at 12:06
BernardBernard
118k639112
edited Apr 7 '17 at 13:10
answered Apr 7 '17 at 12:06
BernardBernard
118k639112
answered Apr 7 '17 at 12:06
BernardBernard
118k639112
answered Apr 7 '17 at 12:06
BernardBernard
118k639112
118k639112
I try it this way. But it is still hard. Can you give me more hints?
– Lehostal11
Apr 7 '17 at 12:33
@Lehostal: I've added some details.
– Bernard
Apr 7 '17 at 13:10
add a comment |
I try it this way. But it is still hard. Can you give me more hints?
– Lehostal11
Apr 7 '17 at 12:33
@Lehostal: I've added some details.
– Bernard
Apr 7 '17 at 13:10
I try it this way. But it is still hard. Can you give me more hints?
– Lehostal11
Apr 7 '17 at 12:33
I try it this way. But it is still hard. Can you give me more hints?
– Lehostal11
Apr 7 '17 at 12:33
@Lehostal: I've added some details.
– Bernard
Apr 7 '17 at 13:10
@Lehostal: I've added some details.
– Bernard
Apr 7 '17 at 13:10
add a comment |
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1
What have you done so far? If you show your working then people are more willing and able to help on the parts you are stuck on
– lioness99a
Apr 7 '17 at 12:00
Hint: first find the minimum polynomial of $alpha^2$ by writing $alpha^3 - alpha = 1$ and squaring both sides.
– Magdiragdag
Apr 8 '17 at 7:34