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Showing the minimal polynomial for an element in an extension field is the same as the minimal polynomial of a linear transformation.
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This is problem 31b) from Dummit and Foote chapter 14.2. I am looking for a hint on how to attack the problem, since I have been thinking about it for a couple of hours but I don't even know where to start. The problem says:
Let $K$ be a finite extension of $F$ of degree $n$. Let $alpha$ be an element of $K$. Prove that the minimal polynomial for $alpha$ over $F$ is the same as the minimal polynomial for the linear transformation $T_alpha$. In this problem, $T_alpha$ is an $F$-linear transformation of $K$ that arises from $alpha$ acting by left multiplication on $K$.
I appreciate any helpful suggestions on how to start attacking the problem or any possible hint. Thanks!
field-theory galois-theory minimal-polynomials
asked Jul 28 at 16:30
user110320
41027
add a comment |Â
up vote
2
down vote
favorite
This is problem 31b) from Dummit and Foote chapter 14.2. I am looking for a hint on how to attack the problem, since I have been thinking about it for a couple of hours but I don't even know where to start. The problem says:
Let $K$ be a finite extension of $F$ of degree $n$. Let $alpha$ be an element of $K$. Prove that the minimal polynomial for $alpha$ over $F$ is the same as the minimal polynomial for the linear transformation $T_alpha$. In this problem, $T_alpha$ is an $F$-linear transformation of $K$ that arises from $alpha$ acting by left multiplication on $K$.
I appreciate any helpful suggestions on how to start attacking the problem or any possible hint. Thanks!
field-theory galois-theory minimal-polynomials
asked Jul 28 at 16:30
user110320
41027
1
what exactly is your definition for $T_alpha$?
– Pink Panther
Jul 28 at 16:36
Sorry for the lack of details and thanks for the comment! I added that info to the problem.
– user110320
Jul 28 at 16:40
so first you can take two minimal polynomials $m_alpha$ and $m_T_alpha$ and then you see that $m_T_alpha(alpha)=m_T_alpha(1cdotalpha)=0$, since we assume that $m_T_alpha(T_alpha(x))=m_T_alpha(alphacdot x)=0$ for all $xin F$ (if i recall correctly, please tell me if i'm wrong). Then $deg(m_T_alpha)geqdeg(m_alpha)$, since $m_T_alpha$ is already a candidate for $m_alpha$. Now you can assume that the polynomials are not equal. If that is the case, then $m_alpha(x)-m_T_alpha(x)neq 0$ for some $xin F$. I'll let you try from here. Hope this helps!
– Pink Panther
Jul 28 at 17:15
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
This is problem 31b) from Dummit and Foote chapter 14.2. I am looking for a hint on how to attack the problem, since I have been thinking about it for a couple of hours but I don't even know where to start. The problem says:
Let $K$ be a finite extension of $F$ of degree $n$. Let $alpha$ be an element of $K$. Prove that the minimal polynomial for $alpha$ over $F$ is the same as the minimal polynomial for the linear transformation $T_alpha$. In this problem, $T_alpha$ is an $F$-linear transformation of $K$ that arises from $alpha$ acting by left multiplication on $K$.
I appreciate any helpful suggestions on how to start attacking the problem or any possible hint. Thanks!
field-theory galois-theory minimal-polynomials
asked Jul 28 at 16:30
user110320
41027
This is problem 31b) from Dummit and Foote chapter 14.2. I am looking for a hint on how to attack the problem, since I have been thinking about it for a couple of hours but I don't even know where to start. The problem says:
Let $K$ be a finite extension of $F$ of degree $n$. Let $alpha$ be an element of $K$. Prove that the minimal polynomial for $alpha$ over $F$ is the same as the minimal polynomial for the linear transformation $T_alpha$. In this problem, $T_alpha$ is an $F$-linear transformation of $K$ that arises from $alpha$ acting by left multiplication on $K$.
I appreciate any helpful suggestions on how to start attacking the problem or any possible hint. Thanks!
field-theory galois-theory minimal-polynomials
asked Jul 28 at 16:30
user110320
41027
edited Jul 28 at 16:39
asked Jul 28 at 16:30
user110320
41027
asked Jul 28 at 16:30
user110320
41027
asked Jul 28 at 16:30
user110320
41027
41027
1
what exactly is your definition for $T_alpha$?
– Pink Panther
Jul 28 at 16:36
Sorry for the lack of details and thanks for the comment! I added that info to the problem.
– user110320
Jul 28 at 16:40
so first you can take two minimal polynomials $m_alpha$ and $m_T_alpha$ and then you see that $m_T_alpha(alpha)=m_T_alpha(1cdotalpha)=0$, since we assume that $m_T_alpha(T_alpha(x))=m_T_alpha(alphacdot x)=0$ for all $xin F$ (if i recall correctly, please tell me if i'm wrong). Then $deg(m_T_alpha)geqdeg(m_alpha)$, since $m_T_alpha$ is already a candidate for $m_alpha$. Now you can assume that the polynomials are not equal. If that is the case, then $m_alpha(x)-m_T_alpha(x)neq 0$ for some $xin F$. I'll let you try from here. Hope this helps!
– Pink Panther
Jul 28 at 17:15
add a comment |Â
1
what exactly is your definition for $T_alpha$?
– Pink Panther
Jul 28 at 16:36
Sorry for the lack of details and thanks for the comment! I added that info to the problem.
– user110320
Jul 28 at 16:40
so first you can take two minimal polynomials $m_alpha$ and $m_T_alpha$ and then you see that $m_T_alpha(alpha)=m_T_alpha(1cdotalpha)=0$, since we assume that $m_T_alpha(T_alpha(x))=m_T_alpha(alphacdot x)=0$ for all $xin F$ (if i recall correctly, please tell me if i'm wrong). Then $deg(m_T_alpha)geqdeg(m_alpha)$, since $m_T_alpha$ is already a candidate for $m_alpha$. Now you can assume that the polynomials are not equal. If that is the case, then $m_alpha(x)-m_T_alpha(x)neq 0$ for some $xin F$. I'll let you try from here. Hope this helps!
– Pink Panther
Jul 28 at 17:15
1
1
what exactly is your definition for $T_alpha$?
– Pink Panther
Jul 28 at 16:36
what exactly is your definition for $T_alpha$?
– Pink Panther
Jul 28 at 16:36
Sorry for the lack of details and thanks for the comment! I added that info to the problem.
– user110320
Jul 28 at 16:40
Sorry for the lack of details and thanks for the comment! I added that info to the problem.
– user110320
Jul 28 at 16:40
so first you can take two minimal polynomials $m_alpha$ and $m_T_alpha$ and then you see that $m_T_alpha(alpha)=m_T_alpha(1cdotalpha)=0$, since we assume that $m_T_alpha(T_alpha(x))=m_T_alpha(alphacdot x)=0$ for all $xin F$ (if i recall correctly, please tell me if i'm wrong). Then $deg(m_T_alpha)geqdeg(m_alpha)$, since $m_T_alpha$ is already a candidate for $m_alpha$. Now you can assume that the polynomials are not equal. If that is the case, then $m_alpha(x)-m_T_alpha(x)neq 0$ for some $xin F$. I'll let you try from here. Hope this helps!
– Pink Panther
Jul 28 at 17:15
so first you can take two minimal polynomials $m_alpha$ and $m_T_alpha$ and then you see that $m_T_alpha(alpha)=m_T_alpha(1cdotalpha)=0$, since we assume that $m_T_alpha(T_alpha(x))=m_T_alpha(alphacdot x)=0$ for all $xin F$ (if i recall correctly, please tell me if i'm wrong). Then $deg(m_T_alpha)geqdeg(m_alpha)$, since $m_T_alpha$ is already a candidate for $m_alpha$. Now you can assume that the polynomials are not equal. If that is the case, then $m_alpha(x)-m_T_alpha(x)neq 0$ for some $xin F$. I'll let you try from here. Hope this helps!
– Pink Panther
Jul 28 at 17:15
add a comment |Â
1 Answer
1
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oldest
votes
up vote
1
down vote
Hints:
- Polynomials $f$ and $g$ are equal if both have leading coefficient $1$ and they divide each other.
- We have $f(a)=0$ iff the minimal polynomial of $a$ (let it be field element or linear transformation) divides $f$.
Hence we have to prove $f(alpha)=0 iff f(T_alpha) =0$ for any polynomial $fin F[x]$.
For this, observe $f(T_alpha)=T_f(alpha)$.
answered Jul 28 at 17:17
Berci
56.4k23570
Your question has been already asked and answered, see e.g. math.stackexchange.com/a/2710883/300700
– nguyen quang do
Jul 30 at 12:30
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Hints:
- Polynomials $f$ and $g$ are equal if both have leading coefficient $1$ and they divide each other.
- We have $f(a)=0$ iff the minimal polynomial of $a$ (let it be field element or linear transformation) divides $f$.
Hence we have to prove $f(alpha)=0 iff f(T_alpha) =0$ for any polynomial $fin F[x]$.
For this, observe $f(T_alpha)=T_f(alpha)$.
answered Jul 28 at 17:17
Berci
56.4k23570
Your question has been already asked and answered, see e.g. math.stackexchange.com/a/2710883/300700
– nguyen quang do
Jul 30 at 12:30
add a comment |Â
up vote
1
down vote
Hints:
- Polynomials $f$ and $g$ are equal if both have leading coefficient $1$ and they divide each other.
- We have $f(a)=0$ iff the minimal polynomial of $a$ (let it be field element or linear transformation) divides $f$.
Hence we have to prove $f(alpha)=0 iff f(T_alpha) =0$ for any polynomial $fin F[x]$.
For this, observe $f(T_alpha)=T_f(alpha)$.
answered Jul 28 at 17:17
Berci
56.4k23570
Your question has been already asked and answered, see e.g. math.stackexchange.com/a/2710883/300700
– nguyen quang do
Jul 30 at 12:30
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hints:
- Polynomials $f$ and $g$ are equal if both have leading coefficient $1$ and they divide each other.
- We have $f(a)=0$ iff the minimal polynomial of $a$ (let it be field element or linear transformation) divides $f$.
Hence we have to prove $f(alpha)=0 iff f(T_alpha) =0$ for any polynomial $fin F[x]$.
For this, observe $f(T_alpha)=T_f(alpha)$.
answered Jul 28 at 17:17
Berci
56.4k23570
Hints:
- Polynomials $f$ and $g$ are equal if both have leading coefficient $1$ and they divide each other.
- We have $f(a)=0$ iff the minimal polynomial of $a$ (let it be field element or linear transformation) divides $f$.
Hence we have to prove $f(alpha)=0 iff f(T_alpha) =0$ for any polynomial $fin F[x]$.
For this, observe $f(T_alpha)=T_f(alpha)$.
answered Jul 28 at 17:17
Berci
56.4k23570
answered Jul 28 at 17:17
Berci
56.4k23570
answered Jul 28 at 17:17
Berci
56.4k23570
answered Jul 28 at 17:17
Berci
56.4k23570
56.4k23570
Your question has been already asked and answered, see e.g. math.stackexchange.com/a/2710883/300700
– nguyen quang do
Jul 30 at 12:30
add a comment |Â
Your question has been already asked and answered, see e.g. math.stackexchange.com/a/2710883/300700
– nguyen quang do
Jul 30 at 12:30
Your question has been already asked and answered, see e.g. math.stackexchange.com/a/2710883/300700
– nguyen quang do
Jul 30 at 12:30
Your question has been already asked and answered, see e.g. math.stackexchange.com/a/2710883/300700
– nguyen quang do
Jul 30 at 12:30
add a comment |Â
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1
what exactly is your definition for $T_alpha$?
– Pink Panther
Jul 28 at 16:36
Sorry for the lack of details and thanks for the comment! I added that info to the problem.
– user110320
Jul 28 at 16:40
so first you can take two minimal polynomials $m_alpha$ and $m_T_alpha$ and then you see that $m_T_alpha(alpha)=m_T_alpha(1cdotalpha)=0$, since we assume that $m_T_alpha(T_alpha(x))=m_T_alpha(alphacdot x)=0$ for all $xin F$ (if i recall correctly, please tell me if i'm wrong). Then $deg(m_T_alpha)geqdeg(m_alpha)$, since $m_T_alpha$ is already a candidate for $m_alpha$. Now you can assume that the polynomials are not equal. If that is the case, then $m_alpha(x)-m_T_alpha(x)neq 0$ for some $xin F$. I'll let you try from here. Hope this helps!
– Pink Panther
Jul 28 at 17:15