Polynomial ring isomorphism
$begingroup$
Let $p=x^2+1$ and $I=<p>$.
Proof that
$frac{mathbb{R[x]}}{I} simeq mathbb C$
So, I can't see that the mapping $phi:{mathbb R[x]} rightarrow {mathbb C}$ defined as $(f(x))phi = f(i)$ is surjective.
And can't see that $kerphi = I$ too.
Help me please!
abstract-algebra ring-theory
asked Dec 13 '18 at 2:26
Rodrigo Geaquinto GonçalvesRodrigo Geaquinto Gonçalves
82
$endgroup$
add a comment |
$begingroup$
Let $p=x^2+1$ and $I=<p>$.
Proof that
$frac{mathbb{R[x]}}{I} simeq mathbb C$
So, I can't see that the mapping $phi:{mathbb R[x]} rightarrow {mathbb C}$ defined as $(f(x))phi = f(i)$ is surjective.
And can't see that $kerphi = I$ too.
Help me please!
abstract-algebra ring-theory
asked Dec 13 '18 at 2:26
Rodrigo Geaquinto GonçalvesRodrigo Geaquinto Gonçalves
82
$endgroup$
1
$begingroup$
I recommend plugging $i$ into some polynomials and see what comes out, and when you get zero.
$endgroup$
– Charlie Frohman
Dec 13 '18 at 2:31
$begingroup$
Try to find an $f(x) in mathbb{R}[x]$ such that $phi(f(x)) = a+bi$; if this is difficult, you are overthinking.
$endgroup$
– Morgan Rodgers
Dec 13 '18 at 2:31
add a comment |
$begingroup$
Let $p=x^2+1$ and $I=<p>$.
Proof that
$frac{mathbb{R[x]}}{I} simeq mathbb C$
So, I can't see that the mapping $phi:{mathbb R[x]} rightarrow {mathbb C}$ defined as $(f(x))phi = f(i)$ is surjective.
And can't see that $kerphi = I$ too.
Help me please!
abstract-algebra ring-theory
asked Dec 13 '18 at 2:26
Rodrigo Geaquinto GonçalvesRodrigo Geaquinto Gonçalves
82
$endgroup$
Let $p=x^2+1$ and $I=<p>$.
Proof that
$frac{mathbb{R[x]}}{I} simeq mathbb C$
So, I can't see that the mapping $phi:{mathbb R[x]} rightarrow {mathbb C}$ defined as $(f(x))phi = f(i)$ is surjective.
And can't see that $kerphi = I$ too.
Help me please!
abstract-algebra ring-theory
abstract-algebra ring-theory
asked Dec 13 '18 at 2:26
Rodrigo Geaquinto GonçalvesRodrigo Geaquinto Gonçalves
82
asked Dec 13 '18 at 2:26
Rodrigo Geaquinto GonçalvesRodrigo Geaquinto Gonçalves
82
asked Dec 13 '18 at 2:26
Rodrigo Geaquinto GonçalvesRodrigo Geaquinto Gonçalves
82
asked Dec 13 '18 at 2:26
Rodrigo Geaquinto GonçalvesRodrigo Geaquinto Gonçalves
82
asked Dec 13 '18 at 2:26
Rodrigo Geaquinto GonçalvesRodrigo Geaquinto Gonçalves
82
82
1
$begingroup$
I recommend plugging $i$ into some polynomials and see what comes out, and when you get zero.
$endgroup$
– Charlie Frohman
Dec 13 '18 at 2:31
$begingroup$
Try to find an $f(x) in mathbb{R}[x]$ such that $phi(f(x)) = a+bi$; if this is difficult, you are overthinking.
$endgroup$
– Morgan Rodgers
Dec 13 '18 at 2:31
add a comment |
1
$begingroup$
I recommend plugging $i$ into some polynomials and see what comes out, and when you get zero.
$endgroup$
– Charlie Frohman
Dec 13 '18 at 2:31
$begingroup$
Try to find an $f(x) in mathbb{R}[x]$ such that $phi(f(x)) = a+bi$; if this is difficult, you are overthinking.
$endgroup$
– Morgan Rodgers
Dec 13 '18 at 2:31
1
1
$begingroup$
I recommend plugging $i$ into some polynomials and see what comes out, and when you get zero.
$endgroup$
– Charlie Frohman
Dec 13 '18 at 2:31
$begingroup$
I recommend plugging $i$ into some polynomials and see what comes out, and when you get zero.
$endgroup$
– Charlie Frohman
Dec 13 '18 at 2:31
$begingroup$
Try to find an $f(x) in mathbb{R}[x]$ such that $phi(f(x)) = a+bi$; if this is difficult, you are overthinking.
$endgroup$
– Morgan Rodgers
Dec 13 '18 at 2:31
$begingroup$
Try to find an $f(x) in mathbb{R}[x]$ such that $phi(f(x)) = a+bi$; if this is difficult, you are overthinking.
$endgroup$
– Morgan Rodgers
Dec 13 '18 at 2:31
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We want to prove that $phi$ is surjective. Let $z=a+ib in mathbb{C}$. Now $z=phi(a+bx)$, hence $phi$ is surjective.
Suppose now that $phi (p(x))=p(i)=0$. Since $i$ is a root and our polynomial has real coefficients, we know that also $-i$ is a solution. Therefore we can write $p(x)=(x-i)(x+i)q(x)=(x^2-1)q(x)$ for some $q(x) in mathbb{R}[x]$
But how does look like an element of our ideal?
If $f(x) in I$, then $f(x)=(x^2-1)g(x)$ for some $g(x) in mathbb{R}[x]$. So every polynomial in the kernel is in the ideal and every polynomial in the ideal is in the kernel: $ker(phi)=I$ and by the first isomorphism theorem we get $frac{mathbb{R}[x]}{I}cong mathbb{C}$
answered Dec 14 '18 at 0:44
user289143user289143
1,002313
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
We want to prove that $phi$ is surjective. Let $z=a+ib in mathbb{C}$. Now $z=phi(a+bx)$, hence $phi$ is surjective.
Suppose now that $phi (p(x))=p(i)=0$. Since $i$ is a root and our polynomial has real coefficients, we know that also $-i$ is a solution. Therefore we can write $p(x)=(x-i)(x+i)q(x)=(x^2-1)q(x)$ for some $q(x) in mathbb{R}[x]$
But how does look like an element of our ideal?
If $f(x) in I$, then $f(x)=(x^2-1)g(x)$ for some $g(x) in mathbb{R}[x]$. So every polynomial in the kernel is in the ideal and every polynomial in the ideal is in the kernel: $ker(phi)=I$ and by the first isomorphism theorem we get $frac{mathbb{R}[x]}{I}cong mathbb{C}$
answered Dec 14 '18 at 0:44
user289143user289143
1,002313
$endgroup$
add a comment |
$begingroup$
We want to prove that $phi$ is surjective. Let $z=a+ib in mathbb{C}$. Now $z=phi(a+bx)$, hence $phi$ is surjective.
Suppose now that $phi (p(x))=p(i)=0$. Since $i$ is a root and our polynomial has real coefficients, we know that also $-i$ is a solution. Therefore we can write $p(x)=(x-i)(x+i)q(x)=(x^2-1)q(x)$ for some $q(x) in mathbb{R}[x]$
But how does look like an element of our ideal?
If $f(x) in I$, then $f(x)=(x^2-1)g(x)$ for some $g(x) in mathbb{R}[x]$. So every polynomial in the kernel is in the ideal and every polynomial in the ideal is in the kernel: $ker(phi)=I$ and by the first isomorphism theorem we get $frac{mathbb{R}[x]}{I}cong mathbb{C}$
answered Dec 14 '18 at 0:44
user289143user289143
1,002313
$endgroup$
add a comment |
$begingroup$
We want to prove that $phi$ is surjective. Let $z=a+ib in mathbb{C}$. Now $z=phi(a+bx)$, hence $phi$ is surjective.
Suppose now that $phi (p(x))=p(i)=0$. Since $i$ is a root and our polynomial has real coefficients, we know that also $-i$ is a solution. Therefore we can write $p(x)=(x-i)(x+i)q(x)=(x^2-1)q(x)$ for some $q(x) in mathbb{R}[x]$
But how does look like an element of our ideal?
If $f(x) in I$, then $f(x)=(x^2-1)g(x)$ for some $g(x) in mathbb{R}[x]$. So every polynomial in the kernel is in the ideal and every polynomial in the ideal is in the kernel: $ker(phi)=I$ and by the first isomorphism theorem we get $frac{mathbb{R}[x]}{I}cong mathbb{C}$
answered Dec 14 '18 at 0:44
user289143user289143
1,002313
$endgroup$
We want to prove that $phi$ is surjective. Let $z=a+ib in mathbb{C}$. Now $z=phi(a+bx)$, hence $phi$ is surjective.
Suppose now that $phi (p(x))=p(i)=0$. Since $i$ is a root and our polynomial has real coefficients, we know that also $-i$ is a solution. Therefore we can write $p(x)=(x-i)(x+i)q(x)=(x^2-1)q(x)$ for some $q(x) in mathbb{R}[x]$
But how does look like an element of our ideal?
If $f(x) in I$, then $f(x)=(x^2-1)g(x)$ for some $g(x) in mathbb{R}[x]$. So every polynomial in the kernel is in the ideal and every polynomial in the ideal is in the kernel: $ker(phi)=I$ and by the first isomorphism theorem we get $frac{mathbb{R}[x]}{I}cong mathbb{C}$
answered Dec 14 '18 at 0:44
user289143user289143
1,002313
answered Dec 14 '18 at 0:44
user289143user289143
1,002313
answered Dec 14 '18 at 0:44
user289143user289143
1,002313
answered Dec 14 '18 at 0:44
user289143user289143
1,002313
1,002313
add a comment |
add a comment |
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$begingroup$
I recommend plugging $i$ into some polynomials and see what comes out, and when you get zero.
$endgroup$
– Charlie Frohman
Dec 13 '18 at 2:31
$begingroup$
Try to find an $f(x) in mathbb{R}[x]$ such that $phi(f(x)) = a+bi$; if this is difficult, you are overthinking.
$endgroup$
– Morgan Rodgers
Dec 13 '18 at 2:31