Proof: $ { mathbb{Q}[x] } / (x^2 +3x +1) cong mathbb{Q}(sqrt{5}) $












2














I have a problem with proving this:
$$
mathbb{Q}[x] / (x^2 +3x +1) cong mathbb{Q}(sqrt{5})
$$
My observations:



In $$mathbb{Q}[x] / (x^2 +3x +1)$$ $$x^2 equiv -1 -3x$$ so $$forall f in mathbb{Q}[x]/(x^2 +3x +1) qquad(deg(f) le 1).$$



Also, we can see that
$$x^2 + 3x +1 = (x+frac 32 +sqrt5)(x+frac 32-sqrt5)$$



And then I got stuck.






abstract-algebra ring-theory field-theory ring-isomorphism







share|cite|improve this question





























    2














    I have a problem with proving this:
    $$
    mathbb{Q}[x] / (x^2 +3x +1) cong mathbb{Q}(sqrt{5})
    $$
    My observations:



    In $$mathbb{Q}[x] / (x^2 +3x +1)$$ $$x^2 equiv -1 -3x$$ so $$forall f in mathbb{Q}[x]/(x^2 +3x +1) qquad(deg(f) le 1).$$



    Also, we can see that
    $$x^2 + 3x +1 = (x+frac 32 +sqrt5)(x+frac 32-sqrt5)$$



    And then I got stuck.






    abstract-algebra ring-theory field-theory ring-isomorphism







    share|cite|improve this question



























      2












      2








      2







      I have a problem with proving this:
      $$
      mathbb{Q}[x] / (x^2 +3x +1) cong mathbb{Q}(sqrt{5})
      $$
      My observations:



      In $$mathbb{Q}[x] / (x^2 +3x +1)$$ $$x^2 equiv -1 -3x$$ so $$forall f in mathbb{Q}[x]/(x^2 +3x +1) qquad(deg(f) le 1).$$



      Also, we can see that
      $$x^2 + 3x +1 = (x+frac 32 +sqrt5)(x+frac 32-sqrt5)$$



      And then I got stuck.






      abstract-algebra ring-theory field-theory ring-isomorphism







      share|cite|improve this question















      I have a problem with proving this:
      $$
      mathbb{Q}[x] / (x^2 +3x +1) cong mathbb{Q}(sqrt{5})
      $$
      My observations:



      In $$mathbb{Q}[x] / (x^2 +3x +1)$$ $$x^2 equiv -1 -3x$$ so $$forall f in mathbb{Q}[x]/(x^2 +3x +1) qquad(deg(f) le 1).$$



      Also, we can see that
      $$x^2 + 3x +1 = (x+frac 32 +sqrt5)(x+frac 32-sqrt5)$$



      And then I got stuck.






      abstract-algebra ring-theory field-theory ring-isomorphism




      abstract-algebra ring-theory field-theory ring-isomorphism






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 24 at 23:38









      Clayton

      18.9k33085




      18.9k33085










      asked Nov 24 at 23:12









      Sharlotta Neimor

      846




      846






















          1 Answer
          1






          active

          oldest

          votes


















          2














          A root of $x^2+3x+1$ is $(sqrt{5}-3)/2$, so
          $$
          mathbb{Q}[x]big/(x^2+3x+1)congmathbb{Q}bigl((sqrt{5}-3)/2bigr)
          $$
          Observe that
          $$
          sqrt{5}=3+2frac{sqrt{5}-3}{2}
          $$
          and you're done.




          In general, if $fin F[x]$ is an irreducible polynomial over the field $F$ and $E$ is an extension field of $F$ such that $f(alpha)=0$, for some $alphain K$, then
          $$
          F[x]/(f)cong F(alpha)
          $$
          Consider the homomorphism $varphicolon F[x]to K$ which is the identity on $F$ and sends $x$ to $alpha$. Then the image of $varphi$ is $F[alpha]$, the smallest subring of $K$ containing $F$ and $alpha$.



          Show that $kervarphi=(f)$ and this will give
          $$
          F[x]/(f)cong F[alpha]
          $$
          On the other hand, $(f)$ is a maximal ideal in $F[x]$, so $F[alpha]$ is a field and therefore $F[alpha]=F(alpha)$, the smallest subfield of $K$ containing $F$ and $alpha$.






          share|cite|improve this answer























          • can you explain please, why ${mathbb{Q}[x]/(x^2 + 3x +1) cong mathbb(Q)((sqrt5 -3)/2)$ ?
            – Sharlotta Neimor
            Nov 24 at 23:24








          • 1




            @SharlottaNeimor It's a standard result on field extensions: if $f$ is an irreducible polynomial over a field $F$, then $F[x]/(f)cong F(alpha)$, where $alpha$ is any root of $f$ in some extension field of $F$.
            – egreg
            Nov 24 at 23:28










          • can you recommend me any resources where I can find the proof or may I ask you the theorem title?
            – Sharlotta Neimor
            Nov 24 at 23:30






          • 1




            @SharlottaNeimor I added a sketch of the proof.
            – egreg
            Nov 24 at 23:36










          • thank you so match
            – Sharlotta Neimor
            Nov 24 at 23:59











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          1 Answer
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          active

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          active

          oldest

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          2














          A root of $x^2+3x+1$ is $(sqrt{5}-3)/2$, so
          $$
          mathbb{Q}[x]big/(x^2+3x+1)congmathbb{Q}bigl((sqrt{5}-3)/2bigr)
          $$
          Observe that
          $$
          sqrt{5}=3+2frac{sqrt{5}-3}{2}
          $$
          and you're done.




          In general, if $fin F[x]$ is an irreducible polynomial over the field $F$ and $E$ is an extension field of $F$ such that $f(alpha)=0$, for some $alphain K$, then
          $$
          F[x]/(f)cong F(alpha)
          $$
          Consider the homomorphism $varphicolon F[x]to K$ which is the identity on $F$ and sends $x$ to $alpha$. Then the image of $varphi$ is $F[alpha]$, the smallest subring of $K$ containing $F$ and $alpha$.



          Show that $kervarphi=(f)$ and this will give
          $$
          F[x]/(f)cong F[alpha]
          $$
          On the other hand, $(f)$ is a maximal ideal in $F[x]$, so $F[alpha]$ is a field and therefore $F[alpha]=F(alpha)$, the smallest subfield of $K$ containing $F$ and $alpha$.






          share|cite|improve this answer























          • can you explain please, why ${mathbb{Q}[x]/(x^2 + 3x +1) cong mathbb(Q)((sqrt5 -3)/2)$ ?
            – Sharlotta Neimor
            Nov 24 at 23:24








          • 1




            @SharlottaNeimor It's a standard result on field extensions: if $f$ is an irreducible polynomial over a field $F$, then $F[x]/(f)cong F(alpha)$, where $alpha$ is any root of $f$ in some extension field of $F$.
            – egreg
            Nov 24 at 23:28










          • can you recommend me any resources where I can find the proof or may I ask you the theorem title?
            – Sharlotta Neimor
            Nov 24 at 23:30






          • 1




            @SharlottaNeimor I added a sketch of the proof.
            – egreg
            Nov 24 at 23:36










          • thank you so match
            – Sharlotta Neimor
            Nov 24 at 23:59
















          2














          A root of $x^2+3x+1$ is $(sqrt{5}-3)/2$, so
          $$
          mathbb{Q}[x]big/(x^2+3x+1)congmathbb{Q}bigl((sqrt{5}-3)/2bigr)
          $$
          Observe that
          $$
          sqrt{5}=3+2frac{sqrt{5}-3}{2}
          $$
          and you're done.




          In general, if $fin F[x]$ is an irreducible polynomial over the field $F$ and $E$ is an extension field of $F$ such that $f(alpha)=0$, for some $alphain K$, then
          $$
          F[x]/(f)cong F(alpha)
          $$
          Consider the homomorphism $varphicolon F[x]to K$ which is the identity on $F$ and sends $x$ to $alpha$. Then the image of $varphi$ is $F[alpha]$, the smallest subring of $K$ containing $F$ and $alpha$.



          Show that $kervarphi=(f)$ and this will give
          $$
          F[x]/(f)cong F[alpha]
          $$
          On the other hand, $(f)$ is a maximal ideal in $F[x]$, so $F[alpha]$ is a field and therefore $F[alpha]=F(alpha)$, the smallest subfield of $K$ containing $F$ and $alpha$.






          share|cite|improve this answer























          • can you explain please, why ${mathbb{Q}[x]/(x^2 + 3x +1) cong mathbb(Q)((sqrt5 -3)/2)$ ?
            – Sharlotta Neimor
            Nov 24 at 23:24








          • 1




            @SharlottaNeimor It's a standard result on field extensions: if $f$ is an irreducible polynomial over a field $F$, then $F[x]/(f)cong F(alpha)$, where $alpha$ is any root of $f$ in some extension field of $F$.
            – egreg
            Nov 24 at 23:28










          • can you recommend me any resources where I can find the proof or may I ask you the theorem title?
            – Sharlotta Neimor
            Nov 24 at 23:30






          • 1




            @SharlottaNeimor I added a sketch of the proof.
            – egreg
            Nov 24 at 23:36










          • thank you so match
            – Sharlotta Neimor
            Nov 24 at 23:59














          2












          2








          2






          A root of $x^2+3x+1$ is $(sqrt{5}-3)/2$, so
          $$
          mathbb{Q}[x]big/(x^2+3x+1)congmathbb{Q}bigl((sqrt{5}-3)/2bigr)
          $$
          Observe that
          $$
          sqrt{5}=3+2frac{sqrt{5}-3}{2}
          $$
          and you're done.




          In general, if $fin F[x]$ is an irreducible polynomial over the field $F$ and $E$ is an extension field of $F$ such that $f(alpha)=0$, for some $alphain K$, then
          $$
          F[x]/(f)cong F(alpha)
          $$
          Consider the homomorphism $varphicolon F[x]to K$ which is the identity on $F$ and sends $x$ to $alpha$. Then the image of $varphi$ is $F[alpha]$, the smallest subring of $K$ containing $F$ and $alpha$.



          Show that $kervarphi=(f)$ and this will give
          $$
          F[x]/(f)cong F[alpha]
          $$
          On the other hand, $(f)$ is a maximal ideal in $F[x]$, so $F[alpha]$ is a field and therefore $F[alpha]=F(alpha)$, the smallest subfield of $K$ containing $F$ and $alpha$.






          share|cite|improve this answer














          A root of $x^2+3x+1$ is $(sqrt{5}-3)/2$, so
          $$
          mathbb{Q}[x]big/(x^2+3x+1)congmathbb{Q}bigl((sqrt{5}-3)/2bigr)
          $$
          Observe that
          $$
          sqrt{5}=3+2frac{sqrt{5}-3}{2}
          $$
          and you're done.




          In general, if $fin F[x]$ is an irreducible polynomial over the field $F$ and $E$ is an extension field of $F$ such that $f(alpha)=0$, for some $alphain K$, then
          $$
          F[x]/(f)cong F(alpha)
          $$
          Consider the homomorphism $varphicolon F[x]to K$ which is the identity on $F$ and sends $x$ to $alpha$. Then the image of $varphi$ is $F[alpha]$, the smallest subring of $K$ containing $F$ and $alpha$.



          Show that $kervarphi=(f)$ and this will give
          $$
          F[x]/(f)cong F[alpha]
          $$
          On the other hand, $(f)$ is a maximal ideal in $F[x]$, so $F[alpha]$ is a field and therefore $F[alpha]=F(alpha)$, the smallest subfield of $K$ containing $F$ and $alpha$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 24 at 23:36

























          answered Nov 24 at 23:18









          egreg

          178k1484201




          178k1484201












          • can you explain please, why ${mathbb{Q}[x]/(x^2 + 3x +1) cong mathbb(Q)((sqrt5 -3)/2)$ ?
            – Sharlotta Neimor
            Nov 24 at 23:24








          • 1




            @SharlottaNeimor It's a standard result on field extensions: if $f$ is an irreducible polynomial over a field $F$, then $F[x]/(f)cong F(alpha)$, where $alpha$ is any root of $f$ in some extension field of $F$.
            – egreg
            Nov 24 at 23:28










          • can you recommend me any resources where I can find the proof or may I ask you the theorem title?
            – Sharlotta Neimor
            Nov 24 at 23:30






          • 1




            @SharlottaNeimor I added a sketch of the proof.
            – egreg
            Nov 24 at 23:36










          • thank you so match
            – Sharlotta Neimor
            Nov 24 at 23:59


















          • can you explain please, why ${mathbb{Q}[x]/(x^2 + 3x +1) cong mathbb(Q)((sqrt5 -3)/2)$ ?
            – Sharlotta Neimor
            Nov 24 at 23:24








          • 1




            @SharlottaNeimor It's a standard result on field extensions: if $f$ is an irreducible polynomial over a field $F$, then $F[x]/(f)cong F(alpha)$, where $alpha$ is any root of $f$ in some extension field of $F$.
            – egreg
            Nov 24 at 23:28










          • can you recommend me any resources where I can find the proof or may I ask you the theorem title?
            – Sharlotta Neimor
            Nov 24 at 23:30






          • 1




            @SharlottaNeimor I added a sketch of the proof.
            – egreg
            Nov 24 at 23:36










          • thank you so match
            – Sharlotta Neimor
            Nov 24 at 23:59
















          can you explain please, why ${mathbb{Q}[x]/(x^2 + 3x +1) cong mathbb(Q)((sqrt5 -3)/2)$ ?
          – Sharlotta Neimor
          Nov 24 at 23:24






          can you explain please, why ${mathbb{Q}[x]/(x^2 + 3x +1) cong mathbb(Q)((sqrt5 -3)/2)$ ?
          – Sharlotta Neimor
          Nov 24 at 23:24






          1




          1




          @SharlottaNeimor It's a standard result on field extensions: if $f$ is an irreducible polynomial over a field $F$, then $F[x]/(f)cong F(alpha)$, where $alpha$ is any root of $f$ in some extension field of $F$.
          – egreg
          Nov 24 at 23:28




          @SharlottaNeimor It's a standard result on field extensions: if $f$ is an irreducible polynomial over a field $F$, then $F[x]/(f)cong F(alpha)$, where $alpha$ is any root of $f$ in some extension field of $F$.
          – egreg
          Nov 24 at 23:28












          can you recommend me any resources where I can find the proof or may I ask you the theorem title?
          – Sharlotta Neimor
          Nov 24 at 23:30




          can you recommend me any resources where I can find the proof or may I ask you the theorem title?
          – Sharlotta Neimor
          Nov 24 at 23:30




          1




          1




          @SharlottaNeimor I added a sketch of the proof.
          – egreg
          Nov 24 at 23:36




          @SharlottaNeimor I added a sketch of the proof.
          – egreg
          Nov 24 at 23:36












          thank you so match
          – Sharlotta Neimor
          Nov 24 at 23:59




          thank you so match
          – Sharlotta Neimor
          Nov 24 at 23:59


















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