I want to know how to prove ideal
I'm just learning about ideal and struggling with these problem.
So I want to know how to prove these.
About ℚ[x] ⊂ C[X], a ∈ C,
$I_a = {f(X) ∈ ℚ[x] | f(a) = 0}$
(1) Show $I_a$ is an ideal of $ℚ[x]$
(2) show $I_{2^{1/2}} = left{(X^2-2)g(X) | g(X) ∈ ℚ[x]right}$
(you can use $2^{1/2}$ is irrational number without condition)
To solve this problem,
can I use under these conditions?
(a) I ≠ ø
(b) f(X), g(X) ∈ I, f(X)+ g(X) ∈ I
(c) f(X) ∈ I and a(X)∈ K[X], f(X)a(X) ∈ I
In addition, to prove problem(1) using (a), is it correct to use constant(the element of ℚ)is included in ℚ[x]?
+) To solve problem(2),
I tried to use GCD(f(x), (x^2-2)) = GCD((x^2-2), r(x))
so, I have to consider when r(x)=0 and deg(r(x)) < 2
when r(x)= 0,
GCD(f(x), (x^2-2)) = GCD((x^2-2), 0)
GCD(f(x), (x^2-2)) = (x^2-2)
however, I got stuck after this.
abstract-algebra polynomials ring-theory ideals
edited Dec 13 '18 at 14:37
Davide Giraudo
125k16150261
asked Dec 13 '18 at 9:02
J.CJ.C
113
add a comment |
I'm just learning about ideal and struggling with these problem.
So I want to know how to prove these.
About ℚ[x] ⊂ C[X], a ∈ C,
$I_a = {f(X) ∈ ℚ[x] | f(a) = 0}$
(1) Show $I_a$ is an ideal of $ℚ[x]$
(2) show $I_{2^{1/2}} = left{(X^2-2)g(X) | g(X) ∈ ℚ[x]right}$
(you can use $2^{1/2}$ is irrational number without condition)
To solve this problem,
can I use under these conditions?
(a) I ≠ ø
(b) f(X), g(X) ∈ I, f(X)+ g(X) ∈ I
(c) f(X) ∈ I and a(X)∈ K[X], f(X)a(X) ∈ I
In addition, to prove problem(1) using (a), is it correct to use constant(the element of ℚ)is included in ℚ[x]?
+) To solve problem(2),
I tried to use GCD(f(x), (x^2-2)) = GCD((x^2-2), r(x))
so, I have to consider when r(x)=0 and deg(r(x)) < 2
when r(x)= 0,
GCD(f(x), (x^2-2)) = GCD((x^2-2), 0)
GCD(f(x), (x^2-2)) = (x^2-2)
however, I got stuck after this.
abstract-algebra polynomials ring-theory ideals
edited Dec 13 '18 at 14:37
Davide Giraudo
125k16150261
asked Dec 13 '18 at 9:02
J.CJ.C
113
1
Heya, it would be better if you included what you've attempted, what definitions you know and have to use and last but not least, format the question properly!
– asdf
Dec 13 '18 at 9:05
I got it! thank you:)
– J.C
Dec 13 '18 at 10:37
add a comment |
I'm just learning about ideal and struggling with these problem.
So I want to know how to prove these.
About ℚ[x] ⊂ C[X], a ∈ C,
$I_a = {f(X) ∈ ℚ[x] | f(a) = 0}$
(1) Show $I_a$ is an ideal of $ℚ[x]$
(2) show $I_{2^{1/2}} = left{(X^2-2)g(X) | g(X) ∈ ℚ[x]right}$
(you can use $2^{1/2}$ is irrational number without condition)
To solve this problem,
can I use under these conditions?
(a) I ≠ ø
(b) f(X), g(X) ∈ I, f(X)+ g(X) ∈ I
(c) f(X) ∈ I and a(X)∈ K[X], f(X)a(X) ∈ I
In addition, to prove problem(1) using (a), is it correct to use constant(the element of ℚ)is included in ℚ[x]?
+) To solve problem(2),
I tried to use GCD(f(x), (x^2-2)) = GCD((x^2-2), r(x))
so, I have to consider when r(x)=0 and deg(r(x)) < 2
when r(x)= 0,
GCD(f(x), (x^2-2)) = GCD((x^2-2), 0)
GCD(f(x), (x^2-2)) = (x^2-2)
however, I got stuck after this.
abstract-algebra polynomials ring-theory ideals
edited Dec 13 '18 at 14:37
Davide Giraudo
125k16150261
asked Dec 13 '18 at 9:02
J.CJ.C
113
I'm just learning about ideal and struggling with these problem.
So I want to know how to prove these.
About ℚ[x] ⊂ C[X], a ∈ C,
$I_a = {f(X) ∈ ℚ[x] | f(a) = 0}$
(1) Show $I_a$ is an ideal of $ℚ[x]$
(2) show $I_{2^{1/2}} = left{(X^2-2)g(X) | g(X) ∈ ℚ[x]right}$
(you can use $2^{1/2}$ is irrational number without condition)
To solve this problem,
can I use under these conditions?
(a) I ≠ ø
(b) f(X), g(X) ∈ I, f(X)+ g(X) ∈ I
(c) f(X) ∈ I and a(X)∈ K[X], f(X)a(X) ∈ I
In addition, to prove problem(1) using (a), is it correct to use constant(the element of ℚ)is included in ℚ[x]?
+) To solve problem(2),
I tried to use GCD(f(x), (x^2-2)) = GCD((x^2-2), r(x))
so, I have to consider when r(x)=0 and deg(r(x)) < 2
when r(x)= 0,
GCD(f(x), (x^2-2)) = GCD((x^2-2), 0)
GCD(f(x), (x^2-2)) = (x^2-2)
however, I got stuck after this.
abstract-algebra polynomials ring-theory ideals
abstract-algebra polynomials ring-theory ideals
edited Dec 13 '18 at 14:37
Davide Giraudo
125k16150261
asked Dec 13 '18 at 9:02
J.CJ.C
113
edited Dec 13 '18 at 14:37
Davide Giraudo
125k16150261
asked Dec 13 '18 at 9:02
J.CJ.C
113
edited Dec 13 '18 at 14:37
Davide Giraudo
125k16150261
edited Dec 13 '18 at 14:37
Davide Giraudo
125k16150261
edited Dec 13 '18 at 14:37
Davide Giraudo
125k16150261
125k16150261
asked Dec 13 '18 at 9:02
J.CJ.C
113
asked Dec 13 '18 at 9:02
J.CJ.C
113
asked Dec 13 '18 at 9:02
J.CJ.C
113
113
1
Heya, it would be better if you included what you've attempted, what definitions you know and have to use and last but not least, format the question properly!
– asdf
Dec 13 '18 at 9:05
I got it! thank you:)
– J.C
Dec 13 '18 at 10:37
add a comment |
1
Heya, it would be better if you included what you've attempted, what definitions you know and have to use and last but not least, format the question properly!
– asdf
Dec 13 '18 at 9:05
I got it! thank you:)
– J.C
Dec 13 '18 at 10:37
1
1
Heya, it would be better if you included what you've attempted, what definitions you know and have to use and last but not least, format the question properly!
– asdf
Dec 13 '18 at 9:05
Heya, it would be better if you included what you've attempted, what definitions you know and have to use and last but not least, format the question properly!
– asdf
Dec 13 '18 at 9:05
I got it! thank you:)
– J.C
Dec 13 '18 at 10:37
I got it! thank you:)
– J.C
Dec 13 '18 at 10:37
add a comment |
1 Answer
1
active
oldest
votes
1) To prove that $I_a$ is an ideal you just need to prove that is closed under addition and it absorbs the multiplication with element of the ring.
Thus, if $f(x), g(x) in I_a$ then $(f+g)(a)=f(a)+g(a)=0$, so $(f+g)(x) in I_a$
Let $f(x) in I_a$ and $p(x) in mathbb{Q}[x]$. Then, $(fp)(a)=f(a)p(a)=0$, i.e. $f(x)p(x) in I_a$. We proved that $I_a$ is an ideal.
2) Now let $f(x), g(x) in I_{2^{1/2}}$. We can write $f(x)$ and $g(x)$ as $f(x)=(x^2-2)f'(x)$ and $g(x)=(x^2-2)g'(x)$. Now $f(x)+g(x)=(x^2-2)f'(x)+(x^2-2)g'(x)=(x^2-2)(f'(x)+g'(x))$ and since $f'(x)+g'(x) in mathbb{Q}[x]$ we have $f(x)+g(x) in I_{2^{1/2}}$.
Now let $f(x)=(x^2-2)f'(x) in I_{2^{1/2}}$ and $p(x) in mathbb{Q}[x]$. We compute $f(x)p(x)=(x^2-2)(f'(x)p(x))$ and since $f'(x)p(x) in mathbb{Q}[x]$ we have $f(x)p(x) in I_{2^{1/2}}$. We proved that also $ I_{2^{1/2}}$ is an ideal.
answered Dec 14 '18 at 0:15
user289143user289143
37018
This proof of (2) starts out by begging the question (and the hardest part of the question): why is it that if $f(2^{1/2}) = 0$ then necessarily you can write $f(x) = (x^2-2) f'(x)$?
– Daniel Schepler
Dec 14 '18 at 0:21
Because $x^2-2$ is the minimal polynomial for $2^{1/2}$ over $mathbb{Q}$, i.e. the smallest monic polynomial such that ${2^{1/2}}$ is a root. It cannot have degree 1 since ${2^{1/2}} notin mathbb{Q}$
– user289143
Dec 14 '18 at 0:24
Thank you for your explanation:)
– J.C
Dec 14 '18 at 3:20
add a comment |
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1) To prove that $I_a$ is an ideal you just need to prove that is closed under addition and it absorbs the multiplication with element of the ring.
Thus, if $f(x), g(x) in I_a$ then $(f+g)(a)=f(a)+g(a)=0$, so $(f+g)(x) in I_a$
Let $f(x) in I_a$ and $p(x) in mathbb{Q}[x]$. Then, $(fp)(a)=f(a)p(a)=0$, i.e. $f(x)p(x) in I_a$. We proved that $I_a$ is an ideal.
2) Now let $f(x), g(x) in I_{2^{1/2}}$. We can write $f(x)$ and $g(x)$ as $f(x)=(x^2-2)f'(x)$ and $g(x)=(x^2-2)g'(x)$. Now $f(x)+g(x)=(x^2-2)f'(x)+(x^2-2)g'(x)=(x^2-2)(f'(x)+g'(x))$ and since $f'(x)+g'(x) in mathbb{Q}[x]$ we have $f(x)+g(x) in I_{2^{1/2}}$.
Now let $f(x)=(x^2-2)f'(x) in I_{2^{1/2}}$ and $p(x) in mathbb{Q}[x]$. We compute $f(x)p(x)=(x^2-2)(f'(x)p(x))$ and since $f'(x)p(x) in mathbb{Q}[x]$ we have $f(x)p(x) in I_{2^{1/2}}$. We proved that also $ I_{2^{1/2}}$ is an ideal.
answered Dec 14 '18 at 0:15
user289143user289143
37018
This proof of (2) starts out by begging the question (and the hardest part of the question): why is it that if $f(2^{1/2}) = 0$ then necessarily you can write $f(x) = (x^2-2) f'(x)$?
– Daniel Schepler
Dec 14 '18 at 0:21
Because $x^2-2$ is the minimal polynomial for $2^{1/2}$ over $mathbb{Q}$, i.e. the smallest monic polynomial such that ${2^{1/2}}$ is a root. It cannot have degree 1 since ${2^{1/2}} notin mathbb{Q}$
– user289143
Dec 14 '18 at 0:24
Thank you for your explanation:)
– J.C
Dec 14 '18 at 3:20
add a comment |
1) To prove that $I_a$ is an ideal you just need to prove that is closed under addition and it absorbs the multiplication with element of the ring.
Thus, if $f(x), g(x) in I_a$ then $(f+g)(a)=f(a)+g(a)=0$, so $(f+g)(x) in I_a$
Let $f(x) in I_a$ and $p(x) in mathbb{Q}[x]$. Then, $(fp)(a)=f(a)p(a)=0$, i.e. $f(x)p(x) in I_a$. We proved that $I_a$ is an ideal.
2) Now let $f(x), g(x) in I_{2^{1/2}}$. We can write $f(x)$ and $g(x)$ as $f(x)=(x^2-2)f'(x)$ and $g(x)=(x^2-2)g'(x)$. Now $f(x)+g(x)=(x^2-2)f'(x)+(x^2-2)g'(x)=(x^2-2)(f'(x)+g'(x))$ and since $f'(x)+g'(x) in mathbb{Q}[x]$ we have $f(x)+g(x) in I_{2^{1/2}}$.
Now let $f(x)=(x^2-2)f'(x) in I_{2^{1/2}}$ and $p(x) in mathbb{Q}[x]$. We compute $f(x)p(x)=(x^2-2)(f'(x)p(x))$ and since $f'(x)p(x) in mathbb{Q}[x]$ we have $f(x)p(x) in I_{2^{1/2}}$. We proved that also $ I_{2^{1/2}}$ is an ideal.
answered Dec 14 '18 at 0:15
user289143user289143
37018
This proof of (2) starts out by begging the question (and the hardest part of the question): why is it that if $f(2^{1/2}) = 0$ then necessarily you can write $f(x) = (x^2-2) f'(x)$?
– Daniel Schepler
Dec 14 '18 at 0:21
Because $x^2-2$ is the minimal polynomial for $2^{1/2}$ over $mathbb{Q}$, i.e. the smallest monic polynomial such that ${2^{1/2}}$ is a root. It cannot have degree 1 since ${2^{1/2}} notin mathbb{Q}$
– user289143
Dec 14 '18 at 0:24
Thank you for your explanation:)
– J.C
Dec 14 '18 at 3:20
add a comment |
1) To prove that $I_a$ is an ideal you just need to prove that is closed under addition and it absorbs the multiplication with element of the ring.
Thus, if $f(x), g(x) in I_a$ then $(f+g)(a)=f(a)+g(a)=0$, so $(f+g)(x) in I_a$
Let $f(x) in I_a$ and $p(x) in mathbb{Q}[x]$. Then, $(fp)(a)=f(a)p(a)=0$, i.e. $f(x)p(x) in I_a$. We proved that $I_a$ is an ideal.
2) Now let $f(x), g(x) in I_{2^{1/2}}$. We can write $f(x)$ and $g(x)$ as $f(x)=(x^2-2)f'(x)$ and $g(x)=(x^2-2)g'(x)$. Now $f(x)+g(x)=(x^2-2)f'(x)+(x^2-2)g'(x)=(x^2-2)(f'(x)+g'(x))$ and since $f'(x)+g'(x) in mathbb{Q}[x]$ we have $f(x)+g(x) in I_{2^{1/2}}$.
Now let $f(x)=(x^2-2)f'(x) in I_{2^{1/2}}$ and $p(x) in mathbb{Q}[x]$. We compute $f(x)p(x)=(x^2-2)(f'(x)p(x))$ and since $f'(x)p(x) in mathbb{Q}[x]$ we have $f(x)p(x) in I_{2^{1/2}}$. We proved that also $ I_{2^{1/2}}$ is an ideal.
answered Dec 14 '18 at 0:15
user289143user289143
37018
1) To prove that $I_a$ is an ideal you just need to prove that is closed under addition and it absorbs the multiplication with element of the ring.
Thus, if $f(x), g(x) in I_a$ then $(f+g)(a)=f(a)+g(a)=0$, so $(f+g)(x) in I_a$
Let $f(x) in I_a$ and $p(x) in mathbb{Q}[x]$. Then, $(fp)(a)=f(a)p(a)=0$, i.e. $f(x)p(x) in I_a$. We proved that $I_a$ is an ideal.
2) Now let $f(x), g(x) in I_{2^{1/2}}$. We can write $f(x)$ and $g(x)$ as $f(x)=(x^2-2)f'(x)$ and $g(x)=(x^2-2)g'(x)$. Now $f(x)+g(x)=(x^2-2)f'(x)+(x^2-2)g'(x)=(x^2-2)(f'(x)+g'(x))$ and since $f'(x)+g'(x) in mathbb{Q}[x]$ we have $f(x)+g(x) in I_{2^{1/2}}$.
Now let $f(x)=(x^2-2)f'(x) in I_{2^{1/2}}$ and $p(x) in mathbb{Q}[x]$. We compute $f(x)p(x)=(x^2-2)(f'(x)p(x))$ and since $f'(x)p(x) in mathbb{Q}[x]$ we have $f(x)p(x) in I_{2^{1/2}}$. We proved that also $ I_{2^{1/2}}$ is an ideal.
answered Dec 14 '18 at 0:15
user289143user289143
37018
answered Dec 14 '18 at 0:15
user289143user289143
37018
answered Dec 14 '18 at 0:15
user289143user289143
37018
answered Dec 14 '18 at 0:15
user289143user289143
37018
37018
This proof of (2) starts out by begging the question (and the hardest part of the question): why is it that if $f(2^{1/2}) = 0$ then necessarily you can write $f(x) = (x^2-2) f'(x)$?
– Daniel Schepler
Dec 14 '18 at 0:21
Because $x^2-2$ is the minimal polynomial for $2^{1/2}$ over $mathbb{Q}$, i.e. the smallest monic polynomial such that ${2^{1/2}}$ is a root. It cannot have degree 1 since ${2^{1/2}} notin mathbb{Q}$
– user289143
Dec 14 '18 at 0:24
Thank you for your explanation:)
– J.C
Dec 14 '18 at 3:20
add a comment |
This proof of (2) starts out by begging the question (and the hardest part of the question): why is it that if $f(2^{1/2}) = 0$ then necessarily you can write $f(x) = (x^2-2) f'(x)$?
– Daniel Schepler
Dec 14 '18 at 0:21
Because $x^2-2$ is the minimal polynomial for $2^{1/2}$ over $mathbb{Q}$, i.e. the smallest monic polynomial such that ${2^{1/2}}$ is a root. It cannot have degree 1 since ${2^{1/2}} notin mathbb{Q}$
– user289143
Dec 14 '18 at 0:24
Thank you for your explanation:)
– J.C
Dec 14 '18 at 3:20
This proof of (2) starts out by begging the question (and the hardest part of the question): why is it that if $f(2^{1/2}) = 0$ then necessarily you can write $f(x) = (x^2-2) f'(x)$?
– Daniel Schepler
Dec 14 '18 at 0:21
This proof of (2) starts out by begging the question (and the hardest part of the question): why is it that if $f(2^{1/2}) = 0$ then necessarily you can write $f(x) = (x^2-2) f'(x)$?
– Daniel Schepler
Dec 14 '18 at 0:21
Because $x^2-2$ is the minimal polynomial for $2^{1/2}$ over $mathbb{Q}$, i.e. the smallest monic polynomial such that ${2^{1/2}}$ is a root. It cannot have degree 1 since ${2^{1/2}} notin mathbb{Q}$
– user289143
Dec 14 '18 at 0:24
Because $x^2-2$ is the minimal polynomial for $2^{1/2}$ over $mathbb{Q}$, i.e. the smallest monic polynomial such that ${2^{1/2}}$ is a root. It cannot have degree 1 since ${2^{1/2}} notin mathbb{Q}$
– user289143
Dec 14 '18 at 0:24
Thank you for your explanation:)
– J.C
Dec 14 '18 at 3:20
Thank you for your explanation:)
– J.C
Dec 14 '18 at 3:20
add a comment |
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1
Heya, it would be better if you included what you've attempted, what definitions you know and have to use and last but not least, format the question properly!
– asdf
Dec 13 '18 at 9:05
I got it! thank you:)
– J.C
Dec 13 '18 at 10:37