minimal polynomial of $1+alpha^2$ where $alpha$ is a root of $x^3-x-1$
up vote
0
down vote
favorite
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
86.2k5105218
asked Apr 7 '17 at 11:56
Lehostal11
214
add a comment |
up vote
0
down vote
favorite
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
86.2k5105218
asked Apr 7 '17 at 11:56
Lehostal11
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 |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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
86.2k5105218
asked Apr 7 '17 at 11:56
Lehostal11
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
86.2k5105218
asked Apr 7 '17 at 11:56
Lehostal11
214
edited Apr 8 '17 at 7:28
Marc van Leeuwen
86.2k5105218
asked Apr 7 '17 at 11:56
Lehostal11
214
edited Apr 8 '17 at 7:28
Marc van Leeuwen
86.2k5105218
edited Apr 8 '17 at 7:28
Marc van Leeuwen
86.2k5105218
edited Apr 8 '17 at 7:28
Marc van Leeuwen
86.2k5105218
86.2k5105218
asked Apr 7 '17 at 11:56
Lehostal11
214
asked Apr 7 '17 at 11:56
Lehostal11
214
asked Apr 7 '17 at 11:56
Lehostal11
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
up vote
3
down vote
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
lhf
162k9166385
add a comment |
up vote
2
down vote
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 Leeuwen
86.2k5105218
add a comment |
up vote
1
down vote
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
DonAntonio
176k1491224
add a comment |
up vote
1
down vote
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
Bernard
117k637109
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 |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2222267%2fminimal-polynomial-of-1-alpha2-where-alpha-is-a-root-of-x3-x-1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
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
lhf
162k9166385
add a comment |
up vote
3
down vote
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
lhf
162k9166385
add a comment |
up vote
3
down vote
up vote
3
down vote
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
lhf
162k9166385
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
lhf
162k9166385
answered Apr 7 '17 at 13:25
lhf
162k9166385
answered Apr 7 '17 at 13:25
lhf
162k9166385
answered Apr 7 '17 at 13:25
lhf
162k9166385
162k9166385
add a comment |
add a comment |
up vote
2
down vote
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 Leeuwen
86.2k5105218
add a comment |
up vote
2
down vote
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 Leeuwen
86.2k5105218
add a comment |
up vote
2
down vote
up vote
2
down vote
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 Leeuwen
86.2k5105218
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 Leeuwen
86.2k5105218
edited Nov 22 at 10:25
answered Apr 8 '17 at 10:07
Marc van Leeuwen
86.2k5105218
answered Apr 8 '17 at 10:07
Marc van Leeuwen
86.2k5105218
answered Apr 8 '17 at 10:07
Marc van Leeuwen
86.2k5105218
86.2k5105218
add a comment |
add a comment |
up vote
1
down vote
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
DonAntonio
176k1491224
add a comment |
up vote
1
down vote
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
DonAntonio
176k1491224
add a comment |
up vote
1
down vote
up vote
1
down vote
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
DonAntonio
176k1491224
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
DonAntonio
176k1491224
answered Apr 7 '17 at 12:06
DonAntonio
176k1491224
answered Apr 7 '17 at 12:06
DonAntonio
176k1491224
answered Apr 7 '17 at 12:06
DonAntonio
176k1491224
176k1491224
add a comment |
add a comment |
up vote
1
down vote
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
Bernard
117k637109
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 |
up vote
1
down vote
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
Bernard
117k637109
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 |
up vote
1
down vote
up vote
1
down vote
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
Bernard
117k637109
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
Bernard
117k637109
edited Apr 7 '17 at 13:10
answered Apr 7 '17 at 12:06
Bernard
117k637109
answered Apr 7 '17 at 12:06
Bernard
117k637109
answered Apr 7 '17 at 12:06
Bernard
117k637109
117k637109
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2222267%2fminimal-polynomial-of-1-alpha2-where-alpha-is-a-root-of-x3-x-1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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