# Product of transcendental numbers is not transcendental, or is it?

The transcendental numbers form a field, or so I thought. I'm familiar with the fact that the algebraic numbers form a field which implies that reciprocals of transcendental numbers must be again transcendental (if reciprocal is not transcendental, then the reciprocal of the reciprocal, the transcendental element itself, must be algebraic...). But I was wondering about sums and products of transcendental numbers which are covered in numerous threads here on MSE. However, I came across an awful contradiction after combining certain proofs from here.

Let's begin clear. Let $$L/K$$ be a field extension with $$\alpha,\beta\in L$$. Then obviously, it is true that $$\alpha$$ and $$\beta$$ are algebraic iff $$\alpha+\beta$$ and $$\alpha\beta$$ are algebraic; a simple proof of this is given using the polynomial $$f=x^2-(\alpha+\beta)x+\alpha\beta=(x-\alpha)(x-\beta)$$ in combination with the tower rule.

I want to prove and disprove that $$\alpha\beta$$ is transcendental when $$\alpha$$ and $$\beta$$ are both transcendental.

Let's assume that $$\alpha$$ and $$\beta$$ are transcendental.

First for the proof: if $$\alpha\beta$$ is not transcendental, then it must be algebraic and hence $$\alpha$$ and $$\beta$$ must be algebraic, but they were assumed to be transcendental. Hence, a contradiction and $$\alpha\beta$$ must be transcendental.

The "result" above is easily disproven: we know by the reasoning from earlier that $$\frac{1}{\alpha}$$ must also be transcendental; we take this reciprocal as our transcendental $$\beta$$. Now $$\alpha\beta=1$$ which is algebraic.

Where did I go wrong? Thanks for the time.

In addition: if we take the case $$\beta\neq\frac{1}{\gamma\alpha}$$ where $$\gamma$$ is algebraic, is it then the case that $$\alpha\beta$$ is always transcendental given $$\alpha$$ and $$\beta$$ transcendental.

EDIT: Thanks to the people from the comment section below, I now know what went wrong in my (wrong) argumentation. The answer here tells the story quite well and the part that is wrong in my text is that I also assumed that $$\alpha$$ and $$\beta$$ are algebraic iff $$\alpha\beta$$ algebraic, which is false. I'm going to leave this open so that anyone having the same issue in the future will find more (summarised) info here.

• Your error is here: “if $\alpha\beta$ is not transcendental, then it must be algebraic and hence $\alpha$ and $\beta$ must be algebraic” – Martin R May 30 '19 at 20:03
• A reciprocal of a transcendental is transcendental, yes? So $\pi\cdot\frac 1\pi = 1$ is an algebraic product od two transcendental numbers. Similarly for addition $e + (4-e) = 4.$ – CiaPan May 30 '19 at 20:06
• Are $0$ and $1$ transcendental? They are certainly elements of any field. – Mark Bennet May 30 '19 at 20:09
• @Algebear: That answer says that if both $\alpha+\beta$ and $\alpha \beta$ are algebraic then $\alpha$ and $\beta$ are algebraic. – Martin R May 30 '19 at 20:14
• Likewise, rational numbers form a field, but irrational numbers do not (sum or product of irrational numbers could be rational) – J. W. Tanner May 30 '19 at 20:57

## 1 Answer

The main question was already answered in several comments: as an example of $$T\cdot\frac 1T=1$$ shows, a product of two transcendental numbers needn't be transcendental.

To answer an additional question by OP from the comment, let's consider: $$\begin{cases}\alpha\beta=K \\ \alpha+\beta=L \end{cases}$$ Plugging $$\beta=L-\alpha$$ from the second equation to the first one yields: $$\alpha^2-L\alpha+K=0.$$ This results in $$\alpha=\frac{L \pm \sqrt{L^2 - 4K}}2$$ which is algebraic for algebraic $$K,L.$$
So, yes: if both the sum and the product of two real numbers $$\alpha,\beta$$ are algebraic, then also both $$\alpha$$ and $$\beta$$ are algebraic.

• Also a nice alternative! Can you also tell me more about my additional question? I wonder if we exclude this simple counterexample case, whether $\alpha\beta$ might be transcendental when $\alpha$ and $\beta\neq\frac{1}{\gamma\alpha}$ are transcendental, where $\gamma$ is algebraic. Shall I open a different question for this? – Algebear May 31 '19 at 12:21
• @Algebear 1. I'm not sure I can answer it (not right now, at least). 2. Definitely you should. This is a Question-and-Answer site – post each question separately, so that each one can get its own answer. You can describe a common context when posting a new question, or even link a previous one, but keeping each question on its own allows different people to answer different questions (no one has to answer all questions as a batch). Additionally, you can accept (green-tick) the best answer for each question, which needn't be posted by the same user. – CiaPan May 31 '19 at 12:51
• Thanks anyways! – Algebear Jun 1 '19 at 8:40