# Sum and Product of two transcendental numbers cannot be simultaneously algebraic

If $\alpha$ and $\beta$ are real number and $\alpha$ and $\beta$ are transcendental over $\mathbb Q$, show that $\alpha \beta$ or $\alpha +\beta$ is also transcendental over $\mathbb Q$

Attempt: Our strategy should be that we will assume both $\alpha \beta$ and $\alpha +\beta$ are algebraic over $\mathbb Q$ and then arrive at a contradiction that $\alpha$ or $\beta$ is algebraic.

$\alpha$ and $\beta$ are transcendental over $\mathbb Q \implies ~\nexists f(x), g(x) \in \mathbb Q[x]$ such that $f(\alpha)=0, ~g(\beta)=0$.

Lets assume both $\alpha \beta$ and $\alpha +\beta$ are algebraic over $\mathbb Q$. So, I must construct an equation which if, has roots $\alpha \beta$ and $\alpha +\beta$, must have root $\alpha$ as well

I am not able to understand how the equation $~~x^2-( \alpha + \beta)x + \alpha \beta = (x-\alpha)(x-\beta)$ will help us in this regard.

$2.$ The splitting field for $x^4-x^2-2$ over $\mathbb Z_3$

Attempt: The roots of the given equation are $\pm i, \pm \sqrt 2$. Hence, splitting field is $\mathbb Z_3(i, \sqrt 2)$ . Am I correct?

$3$. Let $a$ be a complex zero of $x^2+x+1$ over $\mathbb Q$. prove that $Q(\sqrt a)=Q(a).$

Attempt: Suppose $a = c+di~|~c,d \in \mathbb Q$, then : $c+di-c \in Q(a) \implies d^{-1}di \in \mathbb Q(a) \implies i \in \mathbb Q(a) \implies \alpha + \beta i \in \mathbb Q(a)~~\forall~~\alpha,\beta \in \mathbb Q \implies Q(a) = \mathbb C$

Since, $\sqrt a$ is also a complex number, $\implies Q(a) = Q(\sqrt a) = \mathbb C$.

Is my proof correct?

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Those are really three separate questions. So you should ask them separately. – tomasz Aug 16 '14 at 11:59
Please, post only one question in one post. Posting several questions in the same post is discouraged and such questions may be put on hold, see meta. – Martin Sleziak Aug 16 '14 at 13:21
Yes, I understand. I will take care of this in the future. – Wanderer Aug 16 '14 at 13:22
The first question was also posted here: math.stackexchange.com/questions/779162/… – Martin Sleziak Aug 16 '14 at 13:22
I did have a look at this question. But, my query was actually how the given equation $x^2 - (\alpha + \beta)x + \alpha \beta$ satisfies our requirements. However, I understand your concern :-) – Wanderer Aug 16 '14 at 13:27

1. If $K$ is the field of numbers (in $\Bbb{C}$) that are algebraic over $\Bbb{Q}$, then by your contrapositive assumption the polynomial $$(x-\alpha)(x-\beta)=x^2-(\alpha+\beta)x+\alpha\beta$$ has coefficients in $K$. Hence its zeros are algebraic over $K$. Therefore...
2. Your anwser is correct. However, your teacher may want you to observe that in $\Bbb{Z}_3$ we have $2=-1$. Hence ... BTW can you factor this polynomial in $\Bbb{Z}_3[x]$.
3. I'm afraid you made a mistake at the starting gate. Why do you claim that the real and imaginary parts of $a$ would both be rational? What does the formula for the solutions of a quadratic tell you? (Also what Tomasz says. +1)
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Re: Your comment to 1. Have you covered a result stating that $K$ is algebraically closed? The phrasing of the end game depends on this. Your comment to 2. Correct, but you do realize that $\Bbb{Z}_3[\sqrt2]$ and $\Bbb{Z}_3[i]$ are the same field. For extra credit: It looks like the zeros of the polynomial are doubled. Can you write the polynomial as a square of a lower degree polynomial? BTW: I loosely denoted an element $u$ of some extension field of $\Bbb{Z}_3$ that sastisfies the equation $u^2=2$ as $\sqrt2$. This is common abuse, but a bit inaccurate in that there two such elements. – Jyrki Lahtonen Aug 16 '14 at 12:26
$1.$ I know this result: Every field $F$ has a unique algebraic extension that is algebraically close. Also, that a field with no proper algebraic extension is a splitting field. $2.$ Yes,I understand that $\mathbb Z_3[i]= \mathbb Z_3[ \sqrt 2]$ .. $x^4-x^2-2 = (x^2+1)(x^2-2)=(x^2-2)^2$ – Wanderer Aug 16 '14 at 12:37
Good. On with part three. Have you noticed that $x^2+x+1\mid x^3-1=(x-1)(x^2+x+1)$? Can you show that if $a$ is a zero of this, then so is $a^2$. Thus so is also $a^4$. But $a^3=1$, so... – Jyrki Lahtonen Aug 16 '14 at 12:40
Better and better! Can you also show that $a^4=a$? And that $\sqrt a=-\pm a^2$? – Jyrki Lahtonen Aug 16 '14 at 13:06
Also, $Q(a) \subseteq Q(\sqrt a) \implies Q(a) \subseteq Q(a^2) ....(2) \implies$ from $(1),(2) : Q(a) = Q(a^2) \implies Q(a) = Q(\sqrt a)$. Hence, Proved. Thank you for your help. Though, It was embarrassing for me the way I started approaching this question. – Wanderer Aug 16 '14 at 13:36

Those are really separate questions and should be asked separately.

I think the general fact you're supposed to use is that algebraic numbers are algebraically closed, i.e. a polynomial with algebraic coefficients has algebraic roots (so the result follows immediately from considering $(x-\alpha)(x-\beta)$).

For the second one, $\sqrt 2$ and $i$ are complex numbers. While $2$ and $-1$ certainly do have roots in an extension of ${\bf Z}_3$, no good will come out of calling them the same as you call their complex counterparts. Otherwise, I guess that is as good a description as any.

For the third, you're way off: ${\bf Q}(a)$ and ${\bf Q}(\sqrt a)$ are both countable, while ${\bf C}$ is certainly not.

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Thank you for your answer :-) – Wanderer Aug 16 '14 at 13:38