Determine the splitting field for a polynomial












2















Let $P(X)=X^4+1 in mathbb{Q[X]}$. Find the splitting field for $P$
over $mathbb{C}$ and determine the degree of it over $mathbb{C}$.




My attempt: Roots of $P$ are $alpha_1 = sqrt{i},alpha_2=-sqrt{i},alpha_3=i^{3/2},alpha_4=-i^{3/2}$



Now the splitting field is $mathbb{Q}(sqrt{i},i)$. Since, $i$ has minimal polynomial of degree 2, $sqrt{i}$ has minimal polynomial of degree 4, thus $[mathbb{Q}(sqrt{i},i):mathbb{Q}]=[mathbb{Q}(sqrt{i}):mathbb{Q}]=4$



Is there a more elegant argument? Can the roots of $P$ be expressed in a better form (analogue to roots of unity for $X^n-1$)?






abstract-algebra extension-field







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  • 3




    The roots of $P$ are roots of unity. They are the zeroes of $X^8-1$ which aren't roots of $X^2+1$, $X+1$ or $X-1$. In other words, they are the primitive 8th roots of unity.
    – Arthur
    Dec 6 at 11:02








  • 2




    You may want to use that $$x^4+1=(x^2+sqrt2,x+1)(x^2-sqrt2,x+1)$$
    – DonAntonio
    Dec 6 at 11:04






  • 1




    What is $sqrt i$? The number $i$ has two square roots.
    – José Carlos Santos
    Dec 6 at 11:11










  • Hint to connect above two comments: calculate $(1+i)^2$ and find (both values of) $sqrt i$ explicitly.
    – Berci
    Dec 6 at 11:23










  • Over $Bbb C$??
    – ajotatxe
    Dec 6 at 11:33
















2















Let $P(X)=X^4+1 in mathbb{Q[X]}$. Find the splitting field for $P$
over $mathbb{C}$ and determine the degree of it over $mathbb{C}$.




My attempt: Roots of $P$ are $alpha_1 = sqrt{i},alpha_2=-sqrt{i},alpha_3=i^{3/2},alpha_4=-i^{3/2}$



Now the splitting field is $mathbb{Q}(sqrt{i},i)$. Since, $i$ has minimal polynomial of degree 2, $sqrt{i}$ has minimal polynomial of degree 4, thus $[mathbb{Q}(sqrt{i},i):mathbb{Q}]=[mathbb{Q}(sqrt{i}):mathbb{Q}]=4$



Is there a more elegant argument? Can the roots of $P$ be expressed in a better form (analogue to roots of unity for $X^n-1$)?






abstract-algebra extension-field







share|cite|improve this question




















  • 3




    The roots of $P$ are roots of unity. They are the zeroes of $X^8-1$ which aren't roots of $X^2+1$, $X+1$ or $X-1$. In other words, they are the primitive 8th roots of unity.
    – Arthur
    Dec 6 at 11:02








  • 2




    You may want to use that $$x^4+1=(x^2+sqrt2,x+1)(x^2-sqrt2,x+1)$$
    – DonAntonio
    Dec 6 at 11:04






  • 1




    What is $sqrt i$? The number $i$ has two square roots.
    – José Carlos Santos
    Dec 6 at 11:11










  • Hint to connect above two comments: calculate $(1+i)^2$ and find (both values of) $sqrt i$ explicitly.
    – Berci
    Dec 6 at 11:23










  • Over $Bbb C$??
    – ajotatxe
    Dec 6 at 11:33














2












2








2








Let $P(X)=X^4+1 in mathbb{Q[X]}$. Find the splitting field for $P$
over $mathbb{C}$ and determine the degree of it over $mathbb{C}$.




My attempt: Roots of $P$ are $alpha_1 = sqrt{i},alpha_2=-sqrt{i},alpha_3=i^{3/2},alpha_4=-i^{3/2}$



Now the splitting field is $mathbb{Q}(sqrt{i},i)$. Since, $i$ has minimal polynomial of degree 2, $sqrt{i}$ has minimal polynomial of degree 4, thus $[mathbb{Q}(sqrt{i},i):mathbb{Q}]=[mathbb{Q}(sqrt{i}):mathbb{Q}]=4$



Is there a more elegant argument? Can the roots of $P$ be expressed in a better form (analogue to roots of unity for $X^n-1$)?






abstract-algebra extension-field







share|cite|improve this question
















Let $P(X)=X^4+1 in mathbb{Q[X]}$. Find the splitting field for $P$
over $mathbb{C}$ and determine the degree of it over $mathbb{C}$.




My attempt: Roots of $P$ are $alpha_1 = sqrt{i},alpha_2=-sqrt{i},alpha_3=i^{3/2},alpha_4=-i^{3/2}$



Now the splitting field is $mathbb{Q}(sqrt{i},i)$. Since, $i$ has minimal polynomial of degree 2, $sqrt{i}$ has minimal polynomial of degree 4, thus $[mathbb{Q}(sqrt{i},i):mathbb{Q}]=[mathbb{Q}(sqrt{i}):mathbb{Q}]=4$



Is there a more elegant argument? Can the roots of $P$ be expressed in a better form (analogue to roots of unity for $X^n-1$)?






abstract-algebra extension-field




abstract-algebra extension-field






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 at 11:27









ajotatxe

53.1k23890




53.1k23890










asked Dec 6 at 10:59









EpsilonDelta

6211615




6211615








  • 3




    The roots of $P$ are roots of unity. They are the zeroes of $X^8-1$ which aren't roots of $X^2+1$, $X+1$ or $X-1$. In other words, they are the primitive 8th roots of unity.
    – Arthur
    Dec 6 at 11:02








  • 2




    You may want to use that $$x^4+1=(x^2+sqrt2,x+1)(x^2-sqrt2,x+1)$$
    – DonAntonio
    Dec 6 at 11:04






  • 1




    What is $sqrt i$? The number $i$ has two square roots.
    – José Carlos Santos
    Dec 6 at 11:11










  • Hint to connect above two comments: calculate $(1+i)^2$ and find (both values of) $sqrt i$ explicitly.
    – Berci
    Dec 6 at 11:23










  • Over $Bbb C$??
    – ajotatxe
    Dec 6 at 11:33














  • 3




    The roots of $P$ are roots of unity. They are the zeroes of $X^8-1$ which aren't roots of $X^2+1$, $X+1$ or $X-1$. In other words, they are the primitive 8th roots of unity.
    – Arthur
    Dec 6 at 11:02








  • 2




    You may want to use that $$x^4+1=(x^2+sqrt2,x+1)(x^2-sqrt2,x+1)$$
    – DonAntonio
    Dec 6 at 11:04






  • 1




    What is $sqrt i$? The number $i$ has two square roots.
    – José Carlos Santos
    Dec 6 at 11:11










  • Hint to connect above two comments: calculate $(1+i)^2$ and find (both values of) $sqrt i$ explicitly.
    – Berci
    Dec 6 at 11:23










  • Over $Bbb C$??
    – ajotatxe
    Dec 6 at 11:33








3




3




The roots of $P$ are roots of unity. They are the zeroes of $X^8-1$ which aren't roots of $X^2+1$, $X+1$ or $X-1$. In other words, they are the primitive 8th roots of unity.
– Arthur
Dec 6 at 11:02






The roots of $P$ are roots of unity. They are the zeroes of $X^8-1$ which aren't roots of $X^2+1$, $X+1$ or $X-1$. In other words, they are the primitive 8th roots of unity.
– Arthur
Dec 6 at 11:02






2




2




You may want to use that $$x^4+1=(x^2+sqrt2,x+1)(x^2-sqrt2,x+1)$$
– DonAntonio
Dec 6 at 11:04




You may want to use that $$x^4+1=(x^2+sqrt2,x+1)(x^2-sqrt2,x+1)$$
– DonAntonio
Dec 6 at 11:04




1




1




What is $sqrt i$? The number $i$ has two square roots.
– José Carlos Santos
Dec 6 at 11:11




What is $sqrt i$? The number $i$ has two square roots.
– José Carlos Santos
Dec 6 at 11:11












Hint to connect above two comments: calculate $(1+i)^2$ and find (both values of) $sqrt i$ explicitly.
– Berci
Dec 6 at 11:23




Hint to connect above two comments: calculate $(1+i)^2$ and find (both values of) $sqrt i$ explicitly.
– Berci
Dec 6 at 11:23












Over $Bbb C$??
– ajotatxe
Dec 6 at 11:33




Over $Bbb C$??
– ajotatxe
Dec 6 at 11:33










2 Answers
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Let $alpha=dfrac1{sqrt 2}(1+i)$. The roots of $P$ are $alpha$, $alpha^3$, $alpha^5$ and $alpha^7$. So the splitting field over $Bbb Q$ is $Bbb Q(alpha)$.






share|cite|improve this answer





























    1














    I'd use the decomposition $$X^4+1 = (X^2-i)(X^2+i).$$
    Then
    $$X^2-i = (X-alpha)(X+alpha)quadmbox{and}quad X^2+i = (X+beta)(X-beta),$$ where $i=alpha^2$ and $-i=beta^2$. So $alpha$ and $beta$ are 8-th roots of unity.
    Take the primitive 8-th root of unity $xi=e^{2pi i/8}$. Then $alpha=xi$ and $beta=xi^3$.



    Added: There is a simpler way to arrive at the decomposition. For this, consider
    $$X^8-1 = (X^4+1)(X^4-1).$$
    The decomposition of $X^4-1$ are the 4th roots of unity:
    $$X^4-1 = (X-1)(X+1)(X-i)(X+i).$$
    So in view of the above notation, $xi^0=1$, $xi^2=i$, $xi^4=-1$, $xi^6=-i$, and so
    $$X^4+1 = (X-xi)(X-xi^3)(X-xi^5)(X-xi^7).$$
    The zeros are all the primitive 8th roots of unity. Cheers!






    share|cite|improve this answer























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      2 Answers
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      1














      Let $alpha=dfrac1{sqrt 2}(1+i)$. The roots of $P$ are $alpha$, $alpha^3$, $alpha^5$ and $alpha^7$. So the splitting field over $Bbb Q$ is $Bbb Q(alpha)$.






      share|cite|improve this answer


























        1














        Let $alpha=dfrac1{sqrt 2}(1+i)$. The roots of $P$ are $alpha$, $alpha^3$, $alpha^5$ and $alpha^7$. So the splitting field over $Bbb Q$ is $Bbb Q(alpha)$.






        share|cite|improve this answer
























          1












          1








          1






          Let $alpha=dfrac1{sqrt 2}(1+i)$. The roots of $P$ are $alpha$, $alpha^3$, $alpha^5$ and $alpha^7$. So the splitting field over $Bbb Q$ is $Bbb Q(alpha)$.






          share|cite|improve this answer












          Let $alpha=dfrac1{sqrt 2}(1+i)$. The roots of $P$ are $alpha$, $alpha^3$, $alpha^5$ and $alpha^7$. So the splitting field over $Bbb Q$ is $Bbb Q(alpha)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 6 at 11:34









          ajotatxe

          53.1k23890




          53.1k23890























              1














              I'd use the decomposition $$X^4+1 = (X^2-i)(X^2+i).$$
              Then
              $$X^2-i = (X-alpha)(X+alpha)quadmbox{and}quad X^2+i = (X+beta)(X-beta),$$ where $i=alpha^2$ and $-i=beta^2$. So $alpha$ and $beta$ are 8-th roots of unity.
              Take the primitive 8-th root of unity $xi=e^{2pi i/8}$. Then $alpha=xi$ and $beta=xi^3$.



              Added: There is a simpler way to arrive at the decomposition. For this, consider
              $$X^8-1 = (X^4+1)(X^4-1).$$
              The decomposition of $X^4-1$ are the 4th roots of unity:
              $$X^4-1 = (X-1)(X+1)(X-i)(X+i).$$
              So in view of the above notation, $xi^0=1$, $xi^2=i$, $xi^4=-1$, $xi^6=-i$, and so
              $$X^4+1 = (X-xi)(X-xi^3)(X-xi^5)(X-xi^7).$$
              The zeros are all the primitive 8th roots of unity. Cheers!






              share|cite|improve this answer




























                1














                I'd use the decomposition $$X^4+1 = (X^2-i)(X^2+i).$$
                Then
                $$X^2-i = (X-alpha)(X+alpha)quadmbox{and}quad X^2+i = (X+beta)(X-beta),$$ where $i=alpha^2$ and $-i=beta^2$. So $alpha$ and $beta$ are 8-th roots of unity.
                Take the primitive 8-th root of unity $xi=e^{2pi i/8}$. Then $alpha=xi$ and $beta=xi^3$.



                Added: There is a simpler way to arrive at the decomposition. For this, consider
                $$X^8-1 = (X^4+1)(X^4-1).$$
                The decomposition of $X^4-1$ are the 4th roots of unity:
                $$X^4-1 = (X-1)(X+1)(X-i)(X+i).$$
                So in view of the above notation, $xi^0=1$, $xi^2=i$, $xi^4=-1$, $xi^6=-i$, and so
                $$X^4+1 = (X-xi)(X-xi^3)(X-xi^5)(X-xi^7).$$
                The zeros are all the primitive 8th roots of unity. Cheers!






                share|cite|improve this answer


























                  1












                  1








                  1






                  I'd use the decomposition $$X^4+1 = (X^2-i)(X^2+i).$$
                  Then
                  $$X^2-i = (X-alpha)(X+alpha)quadmbox{and}quad X^2+i = (X+beta)(X-beta),$$ where $i=alpha^2$ and $-i=beta^2$. So $alpha$ and $beta$ are 8-th roots of unity.
                  Take the primitive 8-th root of unity $xi=e^{2pi i/8}$. Then $alpha=xi$ and $beta=xi^3$.



                  Added: There is a simpler way to arrive at the decomposition. For this, consider
                  $$X^8-1 = (X^4+1)(X^4-1).$$
                  The decomposition of $X^4-1$ are the 4th roots of unity:
                  $$X^4-1 = (X-1)(X+1)(X-i)(X+i).$$
                  So in view of the above notation, $xi^0=1$, $xi^2=i$, $xi^4=-1$, $xi^6=-i$, and so
                  $$X^4+1 = (X-xi)(X-xi^3)(X-xi^5)(X-xi^7).$$
                  The zeros are all the primitive 8th roots of unity. Cheers!






                  share|cite|improve this answer














                  I'd use the decomposition $$X^4+1 = (X^2-i)(X^2+i).$$
                  Then
                  $$X^2-i = (X-alpha)(X+alpha)quadmbox{and}quad X^2+i = (X+beta)(X-beta),$$ where $i=alpha^2$ and $-i=beta^2$. So $alpha$ and $beta$ are 8-th roots of unity.
                  Take the primitive 8-th root of unity $xi=e^{2pi i/8}$. Then $alpha=xi$ and $beta=xi^3$.



                  Added: There is a simpler way to arrive at the decomposition. For this, consider
                  $$X^8-1 = (X^4+1)(X^4-1).$$
                  The decomposition of $X^4-1$ are the 4th roots of unity:
                  $$X^4-1 = (X-1)(X+1)(X-i)(X+i).$$
                  So in view of the above notation, $xi^0=1$, $xi^2=i$, $xi^4=-1$, $xi^6=-i$, and so
                  $$X^4+1 = (X-xi)(X-xi^3)(X-xi^5)(X-xi^7).$$
                  The zeros are all the primitive 8th roots of unity. Cheers!







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 8 at 16:26

























                  answered Dec 6 at 12:09









                  Wuestenfux

                  3,1281410




                  3,1281410






























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