How can I prove that $\mathbb{Q}(\sqrt[3]{2},i)=\mathbb{Q}(\sqrt[3]{2}+i)$?

Question

It is easy to show that $\mathbb{Q}(\sqrt[3]{2}+i)\subseteq \mathbb{Q}(\sqrt[3]{2},i)$.

But how I can show that $\mathbb{Q}(\sqrt[3]{2},i)\subseteq\mathbb{Q}(\sqrt[3]{2}+i)$?

I can't find a way to express $\sqrt[3]{2}$ in terms of $\sqrt[3]{2}+i$.

Answer 1

Follows a sequence of maneuvers which shows how to express $i$ and $\sqrt[3]2$ in terms of $\sqrt[3]2 + i$, and hence that $\Bbb Q(\sqrt[3]2, i) = \Bbb Q(\sqrt[3]2 + i)$:

$\alpha = \sqrt[3]2 + i; \tag 1$

$\alpha - i = \sqrt[3]2; \tag 2$

$(\alpha - i)^3 = 2; \tag 3$

$\alpha^3 - 3 \alpha^2 i - 3 \alpha + i = 2; \tag 4$

$i(1 - 3\alpha^2) = 2 + 3\alpha - \alpha^3; \tag 5$

$i = \dfrac{\alpha^3 - 3\alpha - 2}{3\alpha^2 - 1} = \dfrac{2 + 3\alpha - \alpha^3}{1 - 3\alpha^2} \in \Bbb Q(\alpha) = \Bbb Q(\sqrt[3]2 + i); \tag 6$

$\sqrt[3]2 = \alpha - i \in \Bbb Q(\alpha) = \Bbb Q(\sqrt[3]2 + i); \tag 7$

from (6), (7) and the fact that $\Bbb Q(\alpha) = \Bbb Q(\sqrt[3]2 + i) \subset \Bbb Q(\sqrt[3]2, i)$ we affirm

$\Bbb Q(\sqrt[3]2, i) \subset \Bbb Q(\alpha) \Longrightarrow \Bbb Q(\sqrt[3]2, i) = \Bbb Q(\alpha) = \Bbb Q(\sqrt[3]2 + i). \tag 8$

N.B. We may in fact derive from (6) a sixth degree polynomial satisfied by $\alpha$:

$\alpha^3 - 3\alpha - 2 = i(3\alpha^2 - 1), \tag 9$

$(\alpha^3 - 3\alpha - 2)^2 = -(3\alpha^2 - 1)^2, \tag{10}$

$\alpha^6 + 9\alpha^2 + 4 - 6\alpha^4 - 4\alpha^3 + 6\alpha = -9\alpha^4 + 6\alpha^2 - 1, \tag{11}$

$\alpha^6 + 3\alpha^4 - 4\alpha^3 + 3\alpha^2 + 6\alpha + 1 = 0, \tag{12}$

consistent with the results of Parcly Taxel.

Answer 2

Consider numbers of the form $$x_0+x_1a+x_2a^2+x_3i+x_4ai+x_5a^2i$$ where $\sqrt[3]2=a$ and $x_i\in\mathbb Q$. It is easily shown that these numbers form a vector space $V$ under addition and are closed under multiplication.

Now consider the powers of $\sqrt[3]2+i=a+i=z$ from $z^0=1$ to $z^5$. Certainly all these numbers are of the form above, and the six powers of $z$ are linearly independent in $V$, so they are a basis. By inverting this basis, it is possible to derive an expression for $\sqrt[3]2$ in terms of powers of $z$, showing that it is in $\mathbb Q(\sqrt[3]2,i)$, and similarly for $i$.

Specifically, $$\sqrt[3]2=\frac1{22}(91+100z-78z^2+40z^3-9z^4+12z^5)$$

Answer 3

If you are lazy and do not want to compute by hand, or you want to check a greater number of similar examples and be sure there is no slip of the pen in the solution, you can do it with Macaulay2. Note:

1) iJ is generated by the irreducible polynomial, that $z = i + 2^{1/3}$ has to fulfill.

2) The primaryDecomposition calculations below give the representations of $v = p(z)$ for $v = i$ ($v^2 + 1 = 0$) and $v = 2^{1/3}$ ($v^3 - 2 = 0$). You can read them off from the factors of the primary decomposition. If you have several possibilities (several factors contain an $p_i(z) = v$, here the case at $v^2+1=0$) then you must embed everything into (a floating point approximation) of the complex (or real) numbers and calculate which $p_i(z) = v$ is fulfilled. Increase precision, until only one $p_{i_0}(z) = v$ is floating point zero.

Macaulay2, version 1.12
with packages: ConwayPolynomials, Elimination, IntegralClosure, InverseSystems, LLLBases,
               PrimaryDecomposition, ReesAlgebra, TangentCone

i1 : R=QQ[x,y]

o1 = R

o1 : PolynomialRing

i2 : iI = ideal(x^2+1, y^3-2)

             2       3
o2 = ideal (x  + 1, y  - 2)

o2 : Ideal of R

i3 : R1=R/iI

o3 = R1

o3 : QuotientRing

i4 : S=QQ[z]

o4 = S

o4 : PolynomialRing

i5 : phi=map(R1,S,{x+y})

o5 = map(R1,S,{x + y})

o5 : RingMap R1 <--- S

i6 : iJ = ker phi

            6     4     3     2
o6 = ideal(z  + 3z  - 4z  + 3z  + 12z + 5)

o6 : Ideal of S

i7 : S1 = S/iJ

o7 = S1

o7 : QuotientRing

i8 : S11 = S1[v]

o8 = S11

o8 : PolynomialRing

i9 : primaryDecomposition ideal (S11/ideal(v^2 + 1))

              2               5     4      3      2                      2               5     4  
o9 = {ideal (v  + 1, 22v + 12z  - 9z  + 40z  - 78z  + 78z + 91), ideal (v  + 1, 22v - 12z  + 9z  -
     -----------------------------------------------------------------------------------------------
        3      2
     40z  + 78z  - 78z - 91)}

o9 : List

i10 : primaryDecomposition ideal (S11/ideal(v^3-2))

               3               5     4      3      2                       3         2       5     4
o10 = {ideal (v  - 2, 22v - 12z  + 9z  - 40z  + 78z  - 100z - 91), ideal (v  - 2, 22v  + (12z  - 9z 
      ----------------------------------------------------------------------------------------------
           3      2                    5     4      3       2
      + 40z  - 78z  + 100z + 91)v - 18z  + 8z  - 60z  + 106z  - 106z - 142)}

o10 : List