Showing isomorphism of Quotient Ring to direct product of Complex numbers
up vote
0
down vote
favorite
So I need to prove $$mathbb{C}[x]/(x^3+1)$$ is isomorphic to $mathbb{C} times mathbb{C} times mathbb{C}$, where $mathbb{C}$ is the field of complex numbers. Based on an example in the book I feel like the answer might have to do with representing the elements of the field using the division algorithm of $mathbb{C}$, but I'm honestly not sure and don't really now how to start here. Quotient groups tend to be kind of unintuitive for me.
abstract-algebra ring-theory ring-isomorphism
edited Nov 18 at 0:34
Bias of Priene
286112
asked Nov 17 at 22:53
Mason
12
add a comment |
up vote
0
down vote
favorite
So I need to prove $$mathbb{C}[x]/(x^3+1)$$ is isomorphic to $mathbb{C} times mathbb{C} times mathbb{C}$, where $mathbb{C}$ is the field of complex numbers. Based on an example in the book I feel like the answer might have to do with representing the elements of the field using the division algorithm of $mathbb{C}$, but I'm honestly not sure and don't really now how to start here. Quotient groups tend to be kind of unintuitive for me.
abstract-algebra ring-theory ring-isomorphism
edited Nov 18 at 0:34
Bias of Priene
286112
asked Nov 17 at 22:53
Mason
12
The quotient of the polynomial ring $C[x]$ by a polynomial of degree $n$ is always an $n$-dimensional vector space over $C$ (with $1, x, x^2, ldots, x^{n-1}$ as a basis). (I assume that by "C X C X C" you mean $C times C times C$, i.e., $C^3$.)
– Rob Arthan
Nov 17 at 23:19
@Rob: the isomorphism here is presumably one of rings, given the tags. Mason, are you familiar with the Chinese remainder theorem?
– Qiaochu Yuan
Nov 18 at 0:43
We've worked with it a bit in class, but I honestly don't understand the ring version very well.
– Mason
Nov 18 at 0:48
1
Here's one way to move toward the answer: what are all the homomorphisms $mathbb{C}[x]/(x^3 + 1) to mathbb{C}$ of $mathbb{C}$-algebras? You should find that there are exactly three; use all of them.
– Qiaochu Yuan
Nov 18 at 1:44
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
So I need to prove $$mathbb{C}[x]/(x^3+1)$$ is isomorphic to $mathbb{C} times mathbb{C} times mathbb{C}$, where $mathbb{C}$ is the field of complex numbers. Based on an example in the book I feel like the answer might have to do with representing the elements of the field using the division algorithm of $mathbb{C}$, but I'm honestly not sure and don't really now how to start here. Quotient groups tend to be kind of unintuitive for me.
abstract-algebra ring-theory ring-isomorphism
edited Nov 18 at 0:34
Bias of Priene
286112
asked Nov 17 at 22:53
Mason
12
So I need to prove $$mathbb{C}[x]/(x^3+1)$$ is isomorphic to $mathbb{C} times mathbb{C} times mathbb{C}$, where $mathbb{C}$ is the field of complex numbers. Based on an example in the book I feel like the answer might have to do with representing the elements of the field using the division algorithm of $mathbb{C}$, but I'm honestly not sure and don't really now how to start here. Quotient groups tend to be kind of unintuitive for me.
abstract-algebra ring-theory ring-isomorphism
abstract-algebra ring-theory ring-isomorphism
edited Nov 18 at 0:34
Bias of Priene
286112
asked Nov 17 at 22:53
Mason
12
edited Nov 18 at 0:34
Bias of Priene
286112
asked Nov 17 at 22:53
Mason
12
edited Nov 18 at 0:34
Bias of Priene
286112
edited Nov 18 at 0:34
Bias of Priene
286112
edited Nov 18 at 0:34
Bias of Priene
286112
286112
asked Nov 17 at 22:53
Mason
12
asked Nov 17 at 22:53
Mason
12
asked Nov 17 at 22:53
Mason
12
12
The quotient of the polynomial ring $C[x]$ by a polynomial of degree $n$ is always an $n$-dimensional vector space over $C$ (with $1, x, x^2, ldots, x^{n-1}$ as a basis). (I assume that by "C X C X C" you mean $C times C times C$, i.e., $C^3$.)
– Rob Arthan
Nov 17 at 23:19
@Rob: the isomorphism here is presumably one of rings, given the tags. Mason, are you familiar with the Chinese remainder theorem?
– Qiaochu Yuan
Nov 18 at 0:43
We've worked with it a bit in class, but I honestly don't understand the ring version very well.
– Mason
Nov 18 at 0:48
1
Here's one way to move toward the answer: what are all the homomorphisms $mathbb{C}[x]/(x^3 + 1) to mathbb{C}$ of $mathbb{C}$-algebras? You should find that there are exactly three; use all of them.
– Qiaochu Yuan
Nov 18 at 1:44
add a comment |
The quotient of the polynomial ring $C[x]$ by a polynomial of degree $n$ is always an $n$-dimensional vector space over $C$ (with $1, x, x^2, ldots, x^{n-1}$ as a basis). (I assume that by "C X C X C" you mean $C times C times C$, i.e., $C^3$.)
– Rob Arthan
Nov 17 at 23:19
@Rob: the isomorphism here is presumably one of rings, given the tags. Mason, are you familiar with the Chinese remainder theorem?
– Qiaochu Yuan
Nov 18 at 0:43
We've worked with it a bit in class, but I honestly don't understand the ring version very well.
– Mason
Nov 18 at 0:48
1
Here's one way to move toward the answer: what are all the homomorphisms $mathbb{C}[x]/(x^3 + 1) to mathbb{C}$ of $mathbb{C}$-algebras? You should find that there are exactly three; use all of them.
– Qiaochu Yuan
Nov 18 at 1:44
The quotient of the polynomial ring $C[x]$ by a polynomial of degree $n$ is always an $n$-dimensional vector space over $C$ (with $1, x, x^2, ldots, x^{n-1}$ as a basis). (I assume that by "C X C X C" you mean $C times C times C$, i.e., $C^3$.)
– Rob Arthan
Nov 17 at 23:19
The quotient of the polynomial ring $C[x]$ by a polynomial of degree $n$ is always an $n$-dimensional vector space over $C$ (with $1, x, x^2, ldots, x^{n-1}$ as a basis). (I assume that by "C X C X C" you mean $C times C times C$, i.e., $C^3$.)
– Rob Arthan
Nov 17 at 23:19
@Rob: the isomorphism here is presumably one of rings, given the tags. Mason, are you familiar with the Chinese remainder theorem?
– Qiaochu Yuan
Nov 18 at 0:43
@Rob: the isomorphism here is presumably one of rings, given the tags. Mason, are you familiar with the Chinese remainder theorem?
– Qiaochu Yuan
Nov 18 at 0:43
We've worked with it a bit in class, but I honestly don't understand the ring version very well.
– Mason
Nov 18 at 0:48
We've worked with it a bit in class, but I honestly don't understand the ring version very well.
– Mason
Nov 18 at 0:48
1
1
Here's one way to move toward the answer: what are all the homomorphisms $mathbb{C}[x]/(x^3 + 1) to mathbb{C}$ of $mathbb{C}$-algebras? You should find that there are exactly three; use all of them.
– Qiaochu Yuan
Nov 18 at 1:44
Here's one way to move toward the answer: what are all the homomorphisms $mathbb{C}[x]/(x^3 + 1) to mathbb{C}$ of $mathbb{C}$-algebras? You should find that there are exactly three; use all of them.
– Qiaochu Yuan
Nov 18 at 1:44
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
I don't know if you still need the answer, but here it goes:
First notice that $x = -1$ is a root of $x^3 + 1$ and then you can divide $x^3 + 1$ by $x+1$ using your favorite method to obtain:
$$x^3 + 1 = (x+1)left(x-frac{1+isqrt{3}}{2}right)left(x - frac{1-isqrt{3}}{2}right)$$
Now notice that you wrote your polynomial as product of irreducible/prime elements, which means that the ideals $langle x+1 rangle$, $leftlangle x-frac{1+isqrt{3}}{2} rightrangle$ and $leftlangle x - frac{1-isqrt{3}}{2} rightrangle$ are coprimes and you can use the Chinese remainder theorem to obtain:
begin{align}
frac{mathbb{C}[x]}{langle x^3+1 rangle} & = frac{mathbb{C}[x]}{leftlangle (x+1)left(x-frac{1+isqrt{3}}{2}right)left(x - frac{1-isqrt{3}}{2}right)rightrangle} \
& = frac{mathbb{C}[x]}{leftlangle x+1rightrangle leftlangle x-frac{1+isqrt{3}}{2} rightrangle leftlangle x - frac{1-isqrt{3}}{2} rightrangle} \
& overset{CRT}{cong} frac{mathbb{C}[x]}{langle x+1 rangle} times frac{mathbb{C}[x]}{leftlangle x-frac{1+isqrt{3}}{2} rightrangle} times frac{mathbb{C}[x]}{leftlangle x-frac{1-isqrt{3}}{2} rightrangle} \
& overset{ev}{cong} mathbb{C} times mathbb{C} times mathbb{C}
end{align}
Here $ev_{alpha} : mathbb{C}[x] to mathbb{C}$ is the evaluation map, ie, $ev_{alpha}(p(x)) = p(alpha)$. Can you see that $ev_{alpha}$ induces an isomorphism between $mathbb{C}[x]/langle x-alpha rangle$ and $mathbb{C}$?
answered Nov 21 at 2:08
Bias of Priene
286112
I did eventually get it, I think my main problem was I just didn't remember the CRT at all. Thank you for the help thou, this will probably also help anybody else with a similar problem.
– Mason
Dec 4 at 16:41
It's just practice. I also didn't remember until Qiaochu commented, but I did the general case for any polynomial once. Could you accept the answer?
– Bias of Priene
Dec 4 at 23:48
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
I don't know if you still need the answer, but here it goes:
First notice that $x = -1$ is a root of $x^3 + 1$ and then you can divide $x^3 + 1$ by $x+1$ using your favorite method to obtain:
$$x^3 + 1 = (x+1)left(x-frac{1+isqrt{3}}{2}right)left(x - frac{1-isqrt{3}}{2}right)$$
Now notice that you wrote your polynomial as product of irreducible/prime elements, which means that the ideals $langle x+1 rangle$, $leftlangle x-frac{1+isqrt{3}}{2} rightrangle$ and $leftlangle x - frac{1-isqrt{3}}{2} rightrangle$ are coprimes and you can use the Chinese remainder theorem to obtain:
begin{align}
frac{mathbb{C}[x]}{langle x^3+1 rangle} & = frac{mathbb{C}[x]}{leftlangle (x+1)left(x-frac{1+isqrt{3}}{2}right)left(x - frac{1-isqrt{3}}{2}right)rightrangle} \
& = frac{mathbb{C}[x]}{leftlangle x+1rightrangle leftlangle x-frac{1+isqrt{3}}{2} rightrangle leftlangle x - frac{1-isqrt{3}}{2} rightrangle} \
& overset{CRT}{cong} frac{mathbb{C}[x]}{langle x+1 rangle} times frac{mathbb{C}[x]}{leftlangle x-frac{1+isqrt{3}}{2} rightrangle} times frac{mathbb{C}[x]}{leftlangle x-frac{1-isqrt{3}}{2} rightrangle} \
& overset{ev}{cong} mathbb{C} times mathbb{C} times mathbb{C}
end{align}
Here $ev_{alpha} : mathbb{C}[x] to mathbb{C}$ is the evaluation map, ie, $ev_{alpha}(p(x)) = p(alpha)$. Can you see that $ev_{alpha}$ induces an isomorphism between $mathbb{C}[x]/langle x-alpha rangle$ and $mathbb{C}$?
answered Nov 21 at 2:08
Bias of Priene
286112
I did eventually get it, I think my main problem was I just didn't remember the CRT at all. Thank you for the help thou, this will probably also help anybody else with a similar problem.
– Mason
Dec 4 at 16:41
It's just practice. I also didn't remember until Qiaochu commented, but I did the general case for any polynomial once. Could you accept the answer?
– Bias of Priene
Dec 4 at 23:48
add a comment |
up vote
1
down vote
I don't know if you still need the answer, but here it goes:
First notice that $x = -1$ is a root of $x^3 + 1$ and then you can divide $x^3 + 1$ by $x+1$ using your favorite method to obtain:
$$x^3 + 1 = (x+1)left(x-frac{1+isqrt{3}}{2}right)left(x - frac{1-isqrt{3}}{2}right)$$
Now notice that you wrote your polynomial as product of irreducible/prime elements, which means that the ideals $langle x+1 rangle$, $leftlangle x-frac{1+isqrt{3}}{2} rightrangle$ and $leftlangle x - frac{1-isqrt{3}}{2} rightrangle$ are coprimes and you can use the Chinese remainder theorem to obtain:
begin{align}
frac{mathbb{C}[x]}{langle x^3+1 rangle} & = frac{mathbb{C}[x]}{leftlangle (x+1)left(x-frac{1+isqrt{3}}{2}right)left(x - frac{1-isqrt{3}}{2}right)rightrangle} \
& = frac{mathbb{C}[x]}{leftlangle x+1rightrangle leftlangle x-frac{1+isqrt{3}}{2} rightrangle leftlangle x - frac{1-isqrt{3}}{2} rightrangle} \
& overset{CRT}{cong} frac{mathbb{C}[x]}{langle x+1 rangle} times frac{mathbb{C}[x]}{leftlangle x-frac{1+isqrt{3}}{2} rightrangle} times frac{mathbb{C}[x]}{leftlangle x-frac{1-isqrt{3}}{2} rightrangle} \
& overset{ev}{cong} mathbb{C} times mathbb{C} times mathbb{C}
end{align}
Here $ev_{alpha} : mathbb{C}[x] to mathbb{C}$ is the evaluation map, ie, $ev_{alpha}(p(x)) = p(alpha)$. Can you see that $ev_{alpha}$ induces an isomorphism between $mathbb{C}[x]/langle x-alpha rangle$ and $mathbb{C}$?
answered Nov 21 at 2:08
Bias of Priene
286112
I did eventually get it, I think my main problem was I just didn't remember the CRT at all. Thank you for the help thou, this will probably also help anybody else with a similar problem.
– Mason
Dec 4 at 16:41
It's just practice. I also didn't remember until Qiaochu commented, but I did the general case for any polynomial once. Could you accept the answer?
– Bias of Priene
Dec 4 at 23:48
add a comment |
up vote
1
down vote
up vote
1
down vote
I don't know if you still need the answer, but here it goes:
First notice that $x = -1$ is a root of $x^3 + 1$ and then you can divide $x^3 + 1$ by $x+1$ using your favorite method to obtain:
$$x^3 + 1 = (x+1)left(x-frac{1+isqrt{3}}{2}right)left(x - frac{1-isqrt{3}}{2}right)$$
Now notice that you wrote your polynomial as product of irreducible/prime elements, which means that the ideals $langle x+1 rangle$, $leftlangle x-frac{1+isqrt{3}}{2} rightrangle$ and $leftlangle x - frac{1-isqrt{3}}{2} rightrangle$ are coprimes and you can use the Chinese remainder theorem to obtain:
begin{align}
frac{mathbb{C}[x]}{langle x^3+1 rangle} & = frac{mathbb{C}[x]}{leftlangle (x+1)left(x-frac{1+isqrt{3}}{2}right)left(x - frac{1-isqrt{3}}{2}right)rightrangle} \
& = frac{mathbb{C}[x]}{leftlangle x+1rightrangle leftlangle x-frac{1+isqrt{3}}{2} rightrangle leftlangle x - frac{1-isqrt{3}}{2} rightrangle} \
& overset{CRT}{cong} frac{mathbb{C}[x]}{langle x+1 rangle} times frac{mathbb{C}[x]}{leftlangle x-frac{1+isqrt{3}}{2} rightrangle} times frac{mathbb{C}[x]}{leftlangle x-frac{1-isqrt{3}}{2} rightrangle} \
& overset{ev}{cong} mathbb{C} times mathbb{C} times mathbb{C}
end{align}
Here $ev_{alpha} : mathbb{C}[x] to mathbb{C}$ is the evaluation map, ie, $ev_{alpha}(p(x)) = p(alpha)$. Can you see that $ev_{alpha}$ induces an isomorphism between $mathbb{C}[x]/langle x-alpha rangle$ and $mathbb{C}$?
answered Nov 21 at 2:08
Bias of Priene
286112
I don't know if you still need the answer, but here it goes:
First notice that $x = -1$ is a root of $x^3 + 1$ and then you can divide $x^3 + 1$ by $x+1$ using your favorite method to obtain:
$$x^3 + 1 = (x+1)left(x-frac{1+isqrt{3}}{2}right)left(x - frac{1-isqrt{3}}{2}right)$$
Now notice that you wrote your polynomial as product of irreducible/prime elements, which means that the ideals $langle x+1 rangle$, $leftlangle x-frac{1+isqrt{3}}{2} rightrangle$ and $leftlangle x - frac{1-isqrt{3}}{2} rightrangle$ are coprimes and you can use the Chinese remainder theorem to obtain:
begin{align}
frac{mathbb{C}[x]}{langle x^3+1 rangle} & = frac{mathbb{C}[x]}{leftlangle (x+1)left(x-frac{1+isqrt{3}}{2}right)left(x - frac{1-isqrt{3}}{2}right)rightrangle} \
& = frac{mathbb{C}[x]}{leftlangle x+1rightrangle leftlangle x-frac{1+isqrt{3}}{2} rightrangle leftlangle x - frac{1-isqrt{3}}{2} rightrangle} \
& overset{CRT}{cong} frac{mathbb{C}[x]}{langle x+1 rangle} times frac{mathbb{C}[x]}{leftlangle x-frac{1+isqrt{3}}{2} rightrangle} times frac{mathbb{C}[x]}{leftlangle x-frac{1-isqrt{3}}{2} rightrangle} \
& overset{ev}{cong} mathbb{C} times mathbb{C} times mathbb{C}
end{align}
Here $ev_{alpha} : mathbb{C}[x] to mathbb{C}$ is the evaluation map, ie, $ev_{alpha}(p(x)) = p(alpha)$. Can you see that $ev_{alpha}$ induces an isomorphism between $mathbb{C}[x]/langle x-alpha rangle$ and $mathbb{C}$?
answered Nov 21 at 2:08
Bias of Priene
286112
answered Nov 21 at 2:08
Bias of Priene
286112
answered Nov 21 at 2:08
Bias of Priene
286112
answered Nov 21 at 2:08
Bias of Priene
286112
286112
I did eventually get it, I think my main problem was I just didn't remember the CRT at all. Thank you for the help thou, this will probably also help anybody else with a similar problem.
– Mason
Dec 4 at 16:41
It's just practice. I also didn't remember until Qiaochu commented, but I did the general case for any polynomial once. Could you accept the answer?
– Bias of Priene
Dec 4 at 23:48
add a comment |
I did eventually get it, I think my main problem was I just didn't remember the CRT at all. Thank you for the help thou, this will probably also help anybody else with a similar problem.
– Mason
Dec 4 at 16:41
It's just practice. I also didn't remember until Qiaochu commented, but I did the general case for any polynomial once. Could you accept the answer?
– Bias of Priene
Dec 4 at 23:48
I did eventually get it, I think my main problem was I just didn't remember the CRT at all. Thank you for the help thou, this will probably also help anybody else with a similar problem.
– Mason
Dec 4 at 16:41
I did eventually get it, I think my main problem was I just didn't remember the CRT at all. Thank you for the help thou, this will probably also help anybody else with a similar problem.
– Mason
Dec 4 at 16:41
It's just practice. I also didn't remember until Qiaochu commented, but I did the general case for any polynomial once. Could you accept the answer?
– Bias of Priene
Dec 4 at 23:48
It's just practice. I also didn't remember until Qiaochu commented, but I did the general case for any polynomial once. Could you accept the answer?
– Bias of Priene
Dec 4 at 23:48
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%2f3002916%2fshowing-isomorphism-of-quotient-ring-to-direct-product-of-complex-numbers%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
The quotient of the polynomial ring $C[x]$ by a polynomial of degree $n$ is always an $n$-dimensional vector space over $C$ (with $1, x, x^2, ldots, x^{n-1}$ as a basis). (I assume that by "C X C X C" you mean $C times C times C$, i.e., $C^3$.)
– Rob Arthan
Nov 17 at 23:19
@Rob: the isomorphism here is presumably one of rings, given the tags. Mason, are you familiar with the Chinese remainder theorem?
– Qiaochu Yuan
Nov 18 at 0:43
We've worked with it a bit in class, but I honestly don't understand the ring version very well.
– Mason
Nov 18 at 0:48
1
Here's one way to move toward the answer: what are all the homomorphisms $mathbb{C}[x]/(x^3 + 1) to mathbb{C}$ of $mathbb{C}$-algebras? You should find that there are exactly three; use all of them.
– Qiaochu Yuan
Nov 18 at 1:44