Polynomial ring isomorphism












-1












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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


















-1












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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
















-1












-1








-1





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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
















  • 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












1 Answer
1






active

oldest

votes


















0












$begingroup$

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 $frac{mathbb{R}[x]}{I}cong mathbb{C}$






share|cite|improve this 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 $frac{mathbb{R}[x]}{I}cong mathbb{C}$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      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 $frac{mathbb{R}[x]}{I}cong mathbb{C}$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        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 $frac{mathbb{R}[x]}{I}cong mathbb{C}$






        share|cite|improve this answer









        $endgroup$



        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 $frac{mathbb{R}[x]}{I}cong mathbb{C}$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 14 '18 at 0:44









        user289143user289143

        1,002313




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