Are the Complex Numbers Isomorphic to the Polynomials, mod $x^2+1$?
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My friend told me that the Complex Numbers are Isomorphic to the Polynomials mod $x^2+1$, is this so? And how can this be proved?
polynomials complex-numbers
edited Jul 30 '18 at 15:36
Namaste
1
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add a comment |
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My friend told me that the Complex Numbers are Isomorphic to the Polynomials mod $x^2+1$, is this so? And how can this be proved?
polynomials complex-numbers
edited Jul 30 '18 at 15:36
Namaste
1
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The graph isomorphism tag refers (I think) to a certain problem in Graph Theory.
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– N. S.
Sep 17 '13 at 21:16
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Intuitively: since we are modding out by a quadratic, all our elements are linear, that is $a + bx$. Next, since $x^2 + 1 = 0$ in this ring, $x^2 = -1$, so $x$ behaves like the $i$ from $mathbb{C}$. For a proof, see the answer by @N.S. .
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– Henry Swanson
Sep 17 '13 at 21:22
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@HenrySwanson:that means 3+2i is equivalent to 3+2x
$endgroup$
– P.Styles
Jun 9 '16 at 19:15
add a comment |
$begingroup$
My friend told me that the Complex Numbers are Isomorphic to the Polynomials mod $x^2+1$, is this so? And how can this be proved?
polynomials complex-numbers
edited Jul 30 '18 at 15:36
Namaste
1
$endgroup$
My friend told me that the Complex Numbers are Isomorphic to the Polynomials mod $x^2+1$, is this so? And how can this be proved?
polynomials complex-numbers
polynomials complex-numbers
edited Jul 30 '18 at 15:36
Namaste
1
edited Jul 30 '18 at 15:36
Namaste
1
edited Jul 30 '18 at 15:36
Namaste
1
edited Jul 30 '18 at 15:36
Namaste
1
edited Jul 30 '18 at 15:36
Namaste
1
1
asked Sep 17 '13 at 21:11
user82004
$begingroup$
The graph isomorphism tag refers (I think) to a certain problem in Graph Theory.
$endgroup$
– N. S.
Sep 17 '13 at 21:16
$begingroup$
Intuitively: since we are modding out by a quadratic, all our elements are linear, that is $a + bx$. Next, since $x^2 + 1 = 0$ in this ring, $x^2 = -1$, so $x$ behaves like the $i$ from $mathbb{C}$. For a proof, see the answer by @N.S. .
$endgroup$
– Henry Swanson
Sep 17 '13 at 21:22
$begingroup$
@HenrySwanson:that means 3+2i is equivalent to 3+2x
$endgroup$
– P.Styles
Jun 9 '16 at 19:15
add a comment |
$begingroup$
The graph isomorphism tag refers (I think) to a certain problem in Graph Theory.
$endgroup$
– N. S.
Sep 17 '13 at 21:16
$begingroup$
Intuitively: since we are modding out by a quadratic, all our elements are linear, that is $a + bx$. Next, since $x^2 + 1 = 0$ in this ring, $x^2 = -1$, so $x$ behaves like the $i$ from $mathbb{C}$. For a proof, see the answer by @N.S. .
$endgroup$
– Henry Swanson
Sep 17 '13 at 21:22
$begingroup$
@HenrySwanson:that means 3+2i is equivalent to 3+2x
$endgroup$
– P.Styles
Jun 9 '16 at 19:15
$begingroup$
The graph isomorphism tag refers (I think) to a certain problem in Graph Theory.
$endgroup$
– N. S.
Sep 17 '13 at 21:16
$begingroup$
The graph isomorphism tag refers (I think) to a certain problem in Graph Theory.
$endgroup$
– N. S.
Sep 17 '13 at 21:16
$begingroup$
Intuitively: since we are modding out by a quadratic, all our elements are linear, that is $a + bx$. Next, since $x^2 + 1 = 0$ in this ring, $x^2 = -1$, so $x$ behaves like the $i$ from $mathbb{C}$. For a proof, see the answer by @N.S. .
$endgroup$
– Henry Swanson
Sep 17 '13 at 21:22
$begingroup$
Intuitively: since we are modding out by a quadratic, all our elements are linear, that is $a + bx$. Next, since $x^2 + 1 = 0$ in this ring, $x^2 = -1$, so $x$ behaves like the $i$ from $mathbb{C}$. For a proof, see the answer by @N.S. .
$endgroup$
– Henry Swanson
Sep 17 '13 at 21:22
$begingroup$
@HenrySwanson:that means 3+2i is equivalent to 3+2x
$endgroup$
– P.Styles
Jun 9 '16 at 19:15
$begingroup$
@HenrySwanson:that means 3+2i is equivalent to 3+2x
$endgroup$
– P.Styles
Jun 9 '16 at 19:15
add a comment |
2 Answers
2
active
oldest
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Let $phi : mathbb R[x] to mathbb C$ be defined by $phi(P)=P(i)$.
This can be proven to be ring/group morphism, which is onto and $ker(phi)=langle x^2+1 rangle$. Apply the first isomorphism theorem.
edited Sep 17 '13 at 21:25
mrf
37.6k64786
answered Sep 17 '13 at 21:15
N. S.N. S.
104k7114209
$endgroup$
add a comment |
$begingroup$
$Bbb R[x]$ is a PID so every prime ideal is maximal. $x^2+1$ is irreducible over $Bbb R[x]$, and therefore $(x^2+1)$ is a maximal ideal, so $Bbb R[x]/(x^2+1)$ is a field.
Now, consider the evaluation homomorphism $phi: Bbb R[x]rightarrowBbb C$, such that $phi(f)=f(i)$.
It is clear that $(x^2+1)subseteqker(phi)$, and equality follows since $(x^2+1)$ is maximal. From the 1st isomorphism theorem, we have $$Bbb R[x]/(x^2+1)congBbb C$$
answered Nov 6 '18 at 16:42
cansomeonehelpmeoutcansomeonehelpmeout
7,1273935
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Let $phi : mathbb R[x] to mathbb C$ be defined by $phi(P)=P(i)$.
This can be proven to be ring/group morphism, which is onto and $ker(phi)=langle x^2+1 rangle$. Apply the first isomorphism theorem.
edited Sep 17 '13 at 21:25
mrf
37.6k64786
answered Sep 17 '13 at 21:15
N. S.N. S.
104k7114209
$endgroup$
add a comment |
$begingroup$
Let $phi : mathbb R[x] to mathbb C$ be defined by $phi(P)=P(i)$.
This can be proven to be ring/group morphism, which is onto and $ker(phi)=langle x^2+1 rangle$. Apply the first isomorphism theorem.
edited Sep 17 '13 at 21:25
mrf
37.6k64786
answered Sep 17 '13 at 21:15
N. S.N. S.
104k7114209
$endgroup$
add a comment |
$begingroup$
Let $phi : mathbb R[x] to mathbb C$ be defined by $phi(P)=P(i)$.
This can be proven to be ring/group morphism, which is onto and $ker(phi)=langle x^2+1 rangle$. Apply the first isomorphism theorem.
edited Sep 17 '13 at 21:25
mrf
37.6k64786
answered Sep 17 '13 at 21:15
N. S.N. S.
104k7114209
$endgroup$
Let $phi : mathbb R[x] to mathbb C$ be defined by $phi(P)=P(i)$.
This can be proven to be ring/group morphism, which is onto and $ker(phi)=langle x^2+1 rangle$. Apply the first isomorphism theorem.
edited Sep 17 '13 at 21:25
mrf
37.6k64786
answered Sep 17 '13 at 21:15
N. S.N. S.
104k7114209
edited Sep 17 '13 at 21:25
mrf
37.6k64786
edited Sep 17 '13 at 21:25
mrf
37.6k64786
edited Sep 17 '13 at 21:25
mrf
37.6k64786
37.6k64786
answered Sep 17 '13 at 21:15
N. S.N. S.
104k7114209
answered Sep 17 '13 at 21:15
N. S.N. S.
104k7114209
answered Sep 17 '13 at 21:15
N. S.N. S.
104k7114209
104k7114209
add a comment |
add a comment |
$begingroup$
$Bbb R[x]$ is a PID so every prime ideal is maximal. $x^2+1$ is irreducible over $Bbb R[x]$, and therefore $(x^2+1)$ is a maximal ideal, so $Bbb R[x]/(x^2+1)$ is a field.
Now, consider the evaluation homomorphism $phi: Bbb R[x]rightarrowBbb C$, such that $phi(f)=f(i)$.
It is clear that $(x^2+1)subseteqker(phi)$, and equality follows since $(x^2+1)$ is maximal. From the 1st isomorphism theorem, we have $$Bbb R[x]/(x^2+1)congBbb C$$
answered Nov 6 '18 at 16:42
cansomeonehelpmeoutcansomeonehelpmeout
7,1273935
$endgroup$
add a comment |
$begingroup$
$Bbb R[x]$ is a PID so every prime ideal is maximal. $x^2+1$ is irreducible over $Bbb R[x]$, and therefore $(x^2+1)$ is a maximal ideal, so $Bbb R[x]/(x^2+1)$ is a field.
Now, consider the evaluation homomorphism $phi: Bbb R[x]rightarrowBbb C$, such that $phi(f)=f(i)$.
It is clear that $(x^2+1)subseteqker(phi)$, and equality follows since $(x^2+1)$ is maximal. From the 1st isomorphism theorem, we have $$Bbb R[x]/(x^2+1)congBbb C$$
answered Nov 6 '18 at 16:42
cansomeonehelpmeoutcansomeonehelpmeout
7,1273935
$endgroup$
add a comment |
$begingroup$
$Bbb R[x]$ is a PID so every prime ideal is maximal. $x^2+1$ is irreducible over $Bbb R[x]$, and therefore $(x^2+1)$ is a maximal ideal, so $Bbb R[x]/(x^2+1)$ is a field.
Now, consider the evaluation homomorphism $phi: Bbb R[x]rightarrowBbb C$, such that $phi(f)=f(i)$.
It is clear that $(x^2+1)subseteqker(phi)$, and equality follows since $(x^2+1)$ is maximal. From the 1st isomorphism theorem, we have $$Bbb R[x]/(x^2+1)congBbb C$$
answered Nov 6 '18 at 16:42
cansomeonehelpmeoutcansomeonehelpmeout
7,1273935
$endgroup$
$Bbb R[x]$ is a PID so every prime ideal is maximal. $x^2+1$ is irreducible over $Bbb R[x]$, and therefore $(x^2+1)$ is a maximal ideal, so $Bbb R[x]/(x^2+1)$ is a field.
Now, consider the evaluation homomorphism $phi: Bbb R[x]rightarrowBbb C$, such that $phi(f)=f(i)$.
It is clear that $(x^2+1)subseteqker(phi)$, and equality follows since $(x^2+1)$ is maximal. From the 1st isomorphism theorem, we have $$Bbb R[x]/(x^2+1)congBbb C$$
answered Nov 6 '18 at 16:42
cansomeonehelpmeoutcansomeonehelpmeout
7,1273935
answered Nov 6 '18 at 16:42
cansomeonehelpmeoutcansomeonehelpmeout
7,1273935
answered Nov 6 '18 at 16:42
cansomeonehelpmeoutcansomeonehelpmeout
7,1273935
answered Nov 6 '18 at 16:42
cansomeonehelpmeoutcansomeonehelpmeout
7,1273935
7,1273935
add a comment |
add a comment |
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$begingroup$
The graph isomorphism tag refers (I think) to a certain problem in Graph Theory.
$endgroup$
– N. S.
Sep 17 '13 at 21:16
$begingroup$
Intuitively: since we are modding out by a quadratic, all our elements are linear, that is $a + bx$. Next, since $x^2 + 1 = 0$ in this ring, $x^2 = -1$, so $x$ behaves like the $i$ from $mathbb{C}$. For a proof, see the answer by @N.S. .
$endgroup$
– Henry Swanson
Sep 17 '13 at 21:22
$begingroup$
@HenrySwanson:that means 3+2i is equivalent to 3+2x
$endgroup$
– P.Styles
Jun 9 '16 at 19:15