What is $mathbb{R}[x]$ quotiented by a polynomial $f(x) in mathbb{R}[x]$ isomorphic to?












1












$begingroup$


By the CRT, we have that: $$mathbb{R}[x]/(x^2-2) simeq mathbb{R}[x]/(x-sqrt{2}) times mathbb{R}[x]/(x+ sqrt{2}) simeq mathbb{R} times mathbb{R}.$$



By considering the evaluation map $mathbb{R}[x] to mathbb{C}$ at $2i$, it is also clear that:
$$mathbb{R}[x]/(x^2+2) simeq mathbb{C}.$$



Now, I am wondering if it is true that
$mathbb{R}[x]/(x-2)^2 simeq mathbb{R}[x]/(x)^2,$ and if so how I would prove it. My idea was simply considering the quotient explicitly as a set of the form ${a_0 + a_1x + f(x)(x^2)}$ and identifying it with ${a_0 + a_1x + g(x)(x^2 - 4x + 4)}$. But if that works, what prevents me from using this same method to show that this quotient is isomorphic to the quotients above — which is obviously false?






abstract-algebra ring-theory







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$endgroup$












  • $begingroup$
    $Bbb R[x] = Bbb R[x - 2].$ From another point of view, you can Taylor expand polynomials about $x = 2$ and write them $p(x) = sum a_i (x - 2)^i.$
    $endgroup$
    – Stahl
    Dec 12 '18 at 0:28


















1












$begingroup$


By the CRT, we have that: $$mathbb{R}[x]/(x^2-2) simeq mathbb{R}[x]/(x-sqrt{2}) times mathbb{R}[x]/(x+ sqrt{2}) simeq mathbb{R} times mathbb{R}.$$



By considering the evaluation map $mathbb{R}[x] to mathbb{C}$ at $2i$, it is also clear that:
$$mathbb{R}[x]/(x^2+2) simeq mathbb{C}.$$



Now, I am wondering if it is true that
$mathbb{R}[x]/(x-2)^2 simeq mathbb{R}[x]/(x)^2,$ and if so how I would prove it. My idea was simply considering the quotient explicitly as a set of the form ${a_0 + a_1x + f(x)(x^2)}$ and identifying it with ${a_0 + a_1x + g(x)(x^2 - 4x + 4)}$. But if that works, what prevents me from using this same method to show that this quotient is isomorphic to the quotients above — which is obviously false?






abstract-algebra ring-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    $Bbb R[x] = Bbb R[x - 2].$ From another point of view, you can Taylor expand polynomials about $x = 2$ and write them $p(x) = sum a_i (x - 2)^i.$
    $endgroup$
    – Stahl
    Dec 12 '18 at 0:28
















1












1








1





$begingroup$


By the CRT, we have that: $$mathbb{R}[x]/(x^2-2) simeq mathbb{R}[x]/(x-sqrt{2}) times mathbb{R}[x]/(x+ sqrt{2}) simeq mathbb{R} times mathbb{R}.$$



By considering the evaluation map $mathbb{R}[x] to mathbb{C}$ at $2i$, it is also clear that:
$$mathbb{R}[x]/(x^2+2) simeq mathbb{C}.$$



Now, I am wondering if it is true that
$mathbb{R}[x]/(x-2)^2 simeq mathbb{R}[x]/(x)^2,$ and if so how I would prove it. My idea was simply considering the quotient explicitly as a set of the form ${a_0 + a_1x + f(x)(x^2)}$ and identifying it with ${a_0 + a_1x + g(x)(x^2 - 4x + 4)}$. But if that works, what prevents me from using this same method to show that this quotient is isomorphic to the quotients above — which is obviously false?






abstract-algebra ring-theory







share|cite|improve this question









$endgroup$




By the CRT, we have that: $$mathbb{R}[x]/(x^2-2) simeq mathbb{R}[x]/(x-sqrt{2}) times mathbb{R}[x]/(x+ sqrt{2}) simeq mathbb{R} times mathbb{R}.$$



By considering the evaluation map $mathbb{R}[x] to mathbb{C}$ at $2i$, it is also clear that:
$$mathbb{R}[x]/(x^2+2) simeq mathbb{C}.$$



Now, I am wondering if it is true that
$mathbb{R}[x]/(x-2)^2 simeq mathbb{R}[x]/(x)^2,$ and if so how I would prove it. My idea was simply considering the quotient explicitly as a set of the form ${a_0 + a_1x + f(x)(x^2)}$ and identifying it with ${a_0 + a_1x + g(x)(x^2 - 4x + 4)}$. But if that works, what prevents me from using this same method to show that this quotient is isomorphic to the quotients above — which is obviously false?






abstract-algebra ring-theory




abstract-algebra ring-theory






share|cite|improve this question













share|cite|improve this question











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asked Dec 12 '18 at 0:18









H. LöwH. Löw

1312




1312












  • $begingroup$
    $Bbb R[x] = Bbb R[x - 2].$ From another point of view, you can Taylor expand polynomials about $x = 2$ and write them $p(x) = sum a_i (x - 2)^i.$
    $endgroup$
    – Stahl
    Dec 12 '18 at 0:28




















  • $begingroup$
    $Bbb R[x] = Bbb R[x - 2].$ From another point of view, you can Taylor expand polynomials about $x = 2$ and write them $p(x) = sum a_i (x - 2)^i.$
    $endgroup$
    – Stahl
    Dec 12 '18 at 0:28


















$begingroup$
$Bbb R[x] = Bbb R[x - 2].$ From another point of view, you can Taylor expand polynomials about $x = 2$ and write them $p(x) = sum a_i (x - 2)^i.$
$endgroup$
– Stahl
Dec 12 '18 at 0:28






$begingroup$
$Bbb R[x] = Bbb R[x - 2].$ From another point of view, you can Taylor expand polynomials about $x = 2$ and write them $p(x) = sum a_i (x - 2)^i.$
$endgroup$
– Stahl
Dec 12 '18 at 0:28












2 Answers
2






active

oldest

votes


















2












$begingroup$

I suggest familiarizing yourself with two universal principles. Look them up elsewhere, I’ll just mention an example:



(1) The universal property of polynomial rings. Homomorphisms $mathbb R[x] to mathbb R[u]/(u^2)$ are determined by where you send $x$, and any choice is ok.



(2) The universal property of quotients. A homomorphism $Bbb R[x] to Bbb R[u]/(u^2)$ factors through $Bbb R[x]/(f(x))$ iff $f(x) mapsto 0$. In the ring $Bbb R[u]/(u^2)$ being 0 is the same as being divisible by $u^2$ as a polynomial.



Thus to check if $varphi: xmapsto x-2$ is a homomorphism $Bbb R[x]/(x^2)to Bbb R[x]/(x-2)^2$ , you just need to make sure $varphi(x^2) = 0$, which it is since $(x-2)^2 = 0$.



In the case of $Bbb R[x]/(x^2-2)$ you would send $xmapsto p(x)$ and you would need to check that $p(x)^2$ is 0, in other words it is divisible by $x^2-2$. This implies that $(x^2-2)|p(x)$ hence the homomorphism sends $xmapsto 0$ and is not an isomorphism.



Similar logic works for $x^2+2$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    When you are identifying ${a_0 + a_1x + f(x)(x^2)}$ with ${a_0 + a_1x + g(x)(x^2 - 4x + 4)}$, you are mapping $x^2$ to $x^2 - 4x + 4$, which is a ring homomorphism only because you're implicitly mapping $x$ to $x-2$, which is the isomorphism that shows $mathbb{R}[x]/(x)^2 simeq mathbb{R}[x]/(x-2)^2 $. There is no corresponding ring homomorphism that maps $x^2$ to $x^2+2$ or $x^2$ to $x^2-2$ (think about where $x$ would have to go).






    share|cite|improve this answer









    $endgroup$













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      2 Answers
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      2 Answers
      2






      active

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      active

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      2












      $begingroup$

      I suggest familiarizing yourself with two universal principles. Look them up elsewhere, I’ll just mention an example:



      (1) The universal property of polynomial rings. Homomorphisms $mathbb R[x] to mathbb R[u]/(u^2)$ are determined by where you send $x$, and any choice is ok.



      (2) The universal property of quotients. A homomorphism $Bbb R[x] to Bbb R[u]/(u^2)$ factors through $Bbb R[x]/(f(x))$ iff $f(x) mapsto 0$. In the ring $Bbb R[u]/(u^2)$ being 0 is the same as being divisible by $u^2$ as a polynomial.



      Thus to check if $varphi: xmapsto x-2$ is a homomorphism $Bbb R[x]/(x^2)to Bbb R[x]/(x-2)^2$ , you just need to make sure $varphi(x^2) = 0$, which it is since $(x-2)^2 = 0$.



      In the case of $Bbb R[x]/(x^2-2)$ you would send $xmapsto p(x)$ and you would need to check that $p(x)^2$ is 0, in other words it is divisible by $x^2-2$. This implies that $(x^2-2)|p(x)$ hence the homomorphism sends $xmapsto 0$ and is not an isomorphism.



      Similar logic works for $x^2+2$.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        I suggest familiarizing yourself with two universal principles. Look them up elsewhere, I’ll just mention an example:



        (1) The universal property of polynomial rings. Homomorphisms $mathbb R[x] to mathbb R[u]/(u^2)$ are determined by where you send $x$, and any choice is ok.



        (2) The universal property of quotients. A homomorphism $Bbb R[x] to Bbb R[u]/(u^2)$ factors through $Bbb R[x]/(f(x))$ iff $f(x) mapsto 0$. In the ring $Bbb R[u]/(u^2)$ being 0 is the same as being divisible by $u^2$ as a polynomial.



        Thus to check if $varphi: xmapsto x-2$ is a homomorphism $Bbb R[x]/(x^2)to Bbb R[x]/(x-2)^2$ , you just need to make sure $varphi(x^2) = 0$, which it is since $(x-2)^2 = 0$.



        In the case of $Bbb R[x]/(x^2-2)$ you would send $xmapsto p(x)$ and you would need to check that $p(x)^2$ is 0, in other words it is divisible by $x^2-2$. This implies that $(x^2-2)|p(x)$ hence the homomorphism sends $xmapsto 0$ and is not an isomorphism.



        Similar logic works for $x^2+2$.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          I suggest familiarizing yourself with two universal principles. Look them up elsewhere, I’ll just mention an example:



          (1) The universal property of polynomial rings. Homomorphisms $mathbb R[x] to mathbb R[u]/(u^2)$ are determined by where you send $x$, and any choice is ok.



          (2) The universal property of quotients. A homomorphism $Bbb R[x] to Bbb R[u]/(u^2)$ factors through $Bbb R[x]/(f(x))$ iff $f(x) mapsto 0$. In the ring $Bbb R[u]/(u^2)$ being 0 is the same as being divisible by $u^2$ as a polynomial.



          Thus to check if $varphi: xmapsto x-2$ is a homomorphism $Bbb R[x]/(x^2)to Bbb R[x]/(x-2)^2$ , you just need to make sure $varphi(x^2) = 0$, which it is since $(x-2)^2 = 0$.



          In the case of $Bbb R[x]/(x^2-2)$ you would send $xmapsto p(x)$ and you would need to check that $p(x)^2$ is 0, in other words it is divisible by $x^2-2$. This implies that $(x^2-2)|p(x)$ hence the homomorphism sends $xmapsto 0$ and is not an isomorphism.



          Similar logic works for $x^2+2$.






          share|cite|improve this answer









          $endgroup$



          I suggest familiarizing yourself with two universal principles. Look them up elsewhere, I’ll just mention an example:



          (1) The universal property of polynomial rings. Homomorphisms $mathbb R[x] to mathbb R[u]/(u^2)$ are determined by where you send $x$, and any choice is ok.



          (2) The universal property of quotients. A homomorphism $Bbb R[x] to Bbb R[u]/(u^2)$ factors through $Bbb R[x]/(f(x))$ iff $f(x) mapsto 0$. In the ring $Bbb R[u]/(u^2)$ being 0 is the same as being divisible by $u^2$ as a polynomial.



          Thus to check if $varphi: xmapsto x-2$ is a homomorphism $Bbb R[x]/(x^2)to Bbb R[x]/(x-2)^2$ , you just need to make sure $varphi(x^2) = 0$, which it is since $(x-2)^2 = 0$.



          In the case of $Bbb R[x]/(x^2-2)$ you would send $xmapsto p(x)$ and you would need to check that $p(x)^2$ is 0, in other words it is divisible by $x^2-2$. This implies that $(x^2-2)|p(x)$ hence the homomorphism sends $xmapsto 0$ and is not an isomorphism.



          Similar logic works for $x^2+2$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 12 '18 at 1:02









          BenBen

          3,716616




          3,716616























              0












              $begingroup$

              When you are identifying ${a_0 + a_1x + f(x)(x^2)}$ with ${a_0 + a_1x + g(x)(x^2 - 4x + 4)}$, you are mapping $x^2$ to $x^2 - 4x + 4$, which is a ring homomorphism only because you're implicitly mapping $x$ to $x-2$, which is the isomorphism that shows $mathbb{R}[x]/(x)^2 simeq mathbb{R}[x]/(x-2)^2 $. There is no corresponding ring homomorphism that maps $x^2$ to $x^2+2$ or $x^2$ to $x^2-2$ (think about where $x$ would have to go).






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                When you are identifying ${a_0 + a_1x + f(x)(x^2)}$ with ${a_0 + a_1x + g(x)(x^2 - 4x + 4)}$, you are mapping $x^2$ to $x^2 - 4x + 4$, which is a ring homomorphism only because you're implicitly mapping $x$ to $x-2$, which is the isomorphism that shows $mathbb{R}[x]/(x)^2 simeq mathbb{R}[x]/(x-2)^2 $. There is no corresponding ring homomorphism that maps $x^2$ to $x^2+2$ or $x^2$ to $x^2-2$ (think about where $x$ would have to go).






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  When you are identifying ${a_0 + a_1x + f(x)(x^2)}$ with ${a_0 + a_1x + g(x)(x^2 - 4x + 4)}$, you are mapping $x^2$ to $x^2 - 4x + 4$, which is a ring homomorphism only because you're implicitly mapping $x$ to $x-2$, which is the isomorphism that shows $mathbb{R}[x]/(x)^2 simeq mathbb{R}[x]/(x-2)^2 $. There is no corresponding ring homomorphism that maps $x^2$ to $x^2+2$ or $x^2$ to $x^2-2$ (think about where $x$ would have to go).






                  share|cite|improve this answer









                  $endgroup$



                  When you are identifying ${a_0 + a_1x + f(x)(x^2)}$ with ${a_0 + a_1x + g(x)(x^2 - 4x + 4)}$, you are mapping $x^2$ to $x^2 - 4x + 4$, which is a ring homomorphism only because you're implicitly mapping $x$ to $x-2$, which is the isomorphism that shows $mathbb{R}[x]/(x)^2 simeq mathbb{R}[x]/(x-2)^2 $. There is no corresponding ring homomorphism that maps $x^2$ to $x^2+2$ or $x^2$ to $x^2-2$ (think about where $x$ would have to go).







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 12 '18 at 0:34









                  zoidbergzoidberg

                  1,065113




                  1,065113






























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