Showing isomorphism of Quotient Ring to direct product of Complex numbers
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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
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up vote
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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
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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
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
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
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
add a comment |
add a comment |
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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