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Proof/Counterexample: If $z$ is a complex number and $z\notin \mathbb Q$, then $\mathbb Q(z)=\mathbb Q(z^3,z^5)$.

First, $\mathbb Q(z)\subseteq \mathbb Q(z^3,z^5)$ would be trivial, right? Then we are left with showing $\mathbb Q(z)\supseteq \mathbb Q(z^3,z^5)$.

If $z\notin \mathbb Q$, then for $z=a+bi$ where $a,b\in \mathbb Q$, $b\ne 0$.

$a+bi=(a+bi)^3$
$a+bi=a^3+3a^2bi-3b^2a-b^3i$
$a+bi=(a^3-3b^2a)+(3a^2b-b^3)i$

Then because $a,b\in \mathbb Q$, $(a^3-3b^2a)+(3a^2b-b^3)i$ is contained in $\mathbb Q(z)$. Thus, $\mathbb Q(z)\supseteq \mathbb Q(z^3,z^5)$?

Does it work like this, or am I way off base with this attempt?

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  • $\begingroup$ You are way off base... I don't understand half of your reasoning. Why assume $z \not\in \mathbb{Q}$ when you know it is actually rational by hypothesis? How can you write $a+bi = (a+bi)^3$ without knowing more about $z$? $\endgroup$
    – Najib Idrissi
    Commented Oct 4, 2014 at 9:02
  • $\begingroup$ I wrote it wrong. $z\notin Q$. Let me fix. Though, that isn't to say it makes the rest of it any better. $\endgroup$
    – Vincent
    Commented Oct 4, 2014 at 9:03
  • $\begingroup$ That still doesn't explain why you write $a+bi = (a+bi)^3$. The answer is much simpler than that, by the way. $\endgroup$
    – Najib Idrissi
    Commented Oct 4, 2014 at 9:04
  • $\begingroup$ Hmm. I was trying to find a way to correlate $z$ and $z^3$/$z^5$. $\endgroup$
    – Vincent
    Commented Oct 4, 2014 at 9:05
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    $\begingroup$ non trivial part is not $\mathbb Q(z)\supseteq \mathbb Q(z^3,z^5)$ but $\mathbb{Q}(z)\subset \mathbb{Q}(z^3,z^5)$ $\endgroup$
    – user87543
    Commented Oct 4, 2014 at 9:08

1 Answer 1

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No, you are way off base. Your reasoning doesn't make much sense (see my comments above).

In fact you start from the wrong foot. The trivial inclusion is $\mathbb{Q}(z^3,z^5) \subseteq \mathbb{Q}(z)$. Indeed, it's clear that both $z^3$ and $z^5$ can be written as rational fractions of $z$: they're both polynomial in $z$.

To prove the other inclusion $\mathbb{Q}(z) \subseteq \mathbb{Q}(z^3,z^5)$, you need to prove that $z$ can be written in terms of $z^3$ and $z^5$ using only addition, multiplication, division, and multiplication by a rational number. You can do that without using the representation $z=a+bi$, it's purely formal: since $5 \times 2 - 3 \times 3 = 1$, you find that: $$z = z^1 = z^{5 \times 2 - 3 \times 3} = \frac{(z^5)^2}{(z^3)^3}.$$

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  • $\begingroup$ From another example I'm looking at: $\mathbb Q(-3+i\sqrt{2},2-\sqrt{9})=\mathbb Q(i,\sqrt{2})$. I eventually showed: $\sqrt{2}=(2-(2-\sqrt{8}))(2)^{-1}\in \mathbb Q(-3+i\sqrt{2},2-\sqrt{8})$. Is this the procedure to which you're referring? $\endgroup$
    – Vincent
    Commented Oct 4, 2014 at 9:15
  • $\begingroup$ No, that's really not the point. What you need to do is write $z = P(z^3,z^5)/Q(z^3,z^5)$ where $P$, $Q$ are polynomials with rational coefficients. If you don't see why that is, you need to reread the definitions of the objects you're studying. $\endgroup$
    – Najib Idrissi
    Commented Oct 4, 2014 at 9:17
  • $\begingroup$ Not that it matters since it isn't the right procedure but the $\sqrt{9}$ was suppose to be $\sqrt{8}$. I'm definitely missing something fundamental to this material. I can understand writing $z$ in terms of $z^5$ and $z^3$ but I'm not quite understanding the $\mathbb P(z^3,z^5)/\mathbb Q(z^3,z^5)$ part unfortunately. The only nonsense I mustered was $z=\large\frac{z^5}{z^3}\cdot z^{-1}$. $\endgroup$
    – Vincent
    Commented Oct 4, 2014 at 9:45
  • $\begingroup$ If $K$ is a field the notazion $K(z)$ means the field of rational functions in $z$ with coefficients in $K$, i.e. the field of all expressions $A(z)/B(z)$ where $A$ and $B$ are polynomials such that $B(z)\neq0$. $\endgroup$
    – Andrea Mori
    Commented Oct 4, 2014 at 9:48
  • $\begingroup$ Is there a way to provide an example that isn't a solution or that gives it away? $\endgroup$
    – Vincent
    Commented Oct 4, 2014 at 10:02

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