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of course the problem is how to prove if a and b are both algebraic real numbers then a+b and ab is also an algebraic number .

would you explain it without using vector spaces or extensions or etc. things we know are :

  1. there exists a polynomial which its root is a

  2. there exists a polynomial which its root is b

please just keep it simple!

user26857
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KFkf
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  • You may find this useful. – Marcin Łoś Sep 27 '14 at 12:22
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    It's so much easier to use vector spaces. It's the simple approach. The complicated approach is to insist on finding polynomials for $a+b$ and $ab$. – lhf Sep 27 '14 at 13:00
  • Because the set of algebraic numbers is the union of all algebraic extensions of $\Bbb{Q}$: for any algebraic $a$ and $b$ there is a finitely generated extention that contains them both, and the union of fields is again a field. Is this correct? – Marc Bogaerts Sep 27 '14 at 15:44

2 Answers2

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If you like Galois theory let $a_1, \ldots a_n$ be the conjugates of $a$ and $b_1, \ldots b_m$ be the conjugates of $b$ then the polynomial

$$\prod_{i,j} (x-(a_i+b_j))$$ is invariant under all automorphisms and so its coefficients lie in the ground field.

Rene Schipperus
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I think the simplest way to show it is to see first that any element generates a finite extension if and only if it is algebraic, and that the degree of field extension is monotone and multiplicative. Both facts are rather straightforward.

Then just pick any $a,b$ algebraic, and notice that $[{\bf Q}(a,b):{\bf Q}(a)],[{\bf Q}(a),{\bf Q}]$ are both finite, therefore $[{\bf Q}(a,b):{\bf Q}]$ is finite as well, and by monotonicity so is $[{\bf Q}(a+b):{\bf Q}]$ as well as $[{\bf Q}(ab):{\bf Q}]$.

tomasz
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