# 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?

• Those are really three separate questions. So you should ask them separately. 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. Aug 16 '14 at 13:21
• Yes, I understand. I will take care of this in the future. Aug 16 '14 at 13:22
• The first question was also posted here: math.stackexchange.com/questions/779162/… 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 :-) 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)
• 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. 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 $\mathbb Z_3[i]= \mathbb Z_3[ \sqrt 2]$ .. $x^4-x^2-2 = (x^2+1)(x^2-2)=(x^2-2)^2$ 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... Aug 16 '14 at 12:40
• Better and better! Can you also show that $a^4=a$? And that $\sqrt a=-\pm a^2$? 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. 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.