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I need help proving this stunning result.

Let me state it this way.

Let $\mathbb L$ be a field and $\mathbb K$ be a field with $\mathbb K \subset \mathbb L$

Let $A$ be the set of algebraic numbers ie $A=\{a \in \mathbb L, \exists P \in \mathbb K [X], P(a)=0 \}$

I managed to prove that $A$ is a ring with the following reasoning:

If $a,b \in A$ then $ \mathbb K[a]$ is a ring.

And $ \mathbb K \subset \mathbb K[a] \subset \mathbb K[a][b] $

Some finite-dimension argument(ask details if needed) proves that $\mathbb K[a][b]$ has finite dimension as a vector space over the field $\mathbb K$

Reminding that $\mathbb K[a+b] \subset \mathbb K[a][b]$ and $\mathbb K[ab] \subset \mathbb K[a][b]$ proves that they both have finite dimension, hence $a$ and $b$ algebraic.

How should I proceed with inverses ?

Gabriel Romon
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  • I might be lost here, but it sounds like you already prove that $A$ is a field...I meant, all the operation are inherited from the big field $L$ so the only thing you need to check is closure. – Gina Jan 03 '14 at 23:53
  • @Gina, I believe that is what he is checking, not the existence of $a^{-1}$, but a proof that $a^{-1}$ is algebraic. – Alex Wertheim Jan 03 '14 at 23:54
  • Huh? But isn't $K[a]=K[a^{-1}]$ so both of them are finite extension? – Gina Jan 03 '14 at 23:56
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    If $p(x)=0$ and $q(y)=0$ where $p$ and $q$ are known polynomial functions, how does one find a polynomial of which $x+y$ is a root? From proofs of this kind, one can actually construct such a polynomial. I did it once. I've forgotten the details. – Michael Hardy Jan 04 '14 at 01:28
  • @MichaelHardy:I believe you can find it in a standard proof of algebraic INTEGER. I struggled for days on that in Herstein and end up have to look up the details. For just algebraic number the finite dimension argument is good enough though. – Gina Jan 04 '14 at 02:53

2 Answers2

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Hint: Let $r\ne 0$ be a root of the equation $a_0x^n+a_1x^{n-1}+\cdots +a_n=0$. Then $\frac{1}{r}$ is a root of the equation $a_nx^n +a_{n-1}x^{n-1}+\cdots +a_1x+a_0=0$.

André Nicolas
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Hints:

For any $\;0\neq a\in A\;$ :

$$a\in\Bbb K[a]=\Bbb K(a)\implies a^{-1}\in \Bbb K(a)\le A\ldots$$

DonAntonio
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  • Hmm I don't know much about $\mathbb K(a)$. Is it the field of fractions ? – Gabriel Romon Jan 03 '14 at 23:48
  • Yes, of the ring $;\Bbb K[a];$ , but this is one of the first things that are usually studied in fields theory: if $;L/K;$ is a fields extension, then an element $;a\in L;$ is algebraic over $;K;$ iff $;K[a]=K(a);$ . A simple, very important, characterization of algebraic elements. – DonAntonio Jan 04 '14 at 00:00