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Theorem : If $x$ is algebraic over $\mathbb{Q} $, then $\mathbb{Q}[x]= \mathbb{Q}(x)$

My doubt : Here $\mathbb{Q}(x)$ is field and $\mathbb{Q}[x]$ is not field since $\frac{1}{x} \notin \mathbb{Q} $ but $\mathbb{Q}[x]$ is ring

we know that field $\neq$ ring

Then why $\mathbb{Q}[x]= \mathbb{Q}(x)?$

jasmine
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    Try here! https://math.stackexchange.com/q/1098032/914078 – Pavan C. May 02 '21 at 06:39
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    I think you're confused about ambiguous notation: in the context of the theorem $\Bbb Q[x]$ is not the polynomial ring, but the subring generated by $\Bbb Q$ and $x$ (in some field extension) – Lukas Heger May 02 '21 at 06:39
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    You can first show that $\mathbb{Q}[x]$ is a field. See here: https://math.stackexchange.com/questions/397733/ring-inside-an-algebraic-field-extension . Then note that $\mathbb{Q}(x)$ is the field of fraction of $\mathbb{Q}[x]$. – Soby May 02 '21 at 06:52
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    $\frac 1x\notin\mathbb Q$ does not show $\mathbb Q[x]$ is not a field. You have to show it's not in $\mathbb Q[x]$. And the thing is, the inverse is actually in it when $x$ is algebraic, and eventually it is a field. – Tesla Daybreak May 02 '21 at 07:29
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    Consider, say, $\Bbb Q(\sqrt[3]2)$. What does a typical element in that field look like? Then consider $\Bbb Q[\sqrt[3]2]$. What does a typical element in that ring look like? See any similarities that might hint at an isomorphism? (And "field $\neq$ ring" is ambiguous. It's true that the requirements of a field and the requirements of a ring aren't the same, and the collection of all fields and the collection of all rings are different collections. But any field is a ring, with additional requirements.) – Arthur May 02 '21 at 07:49
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    This is standard theorem that if $\alpha$ is algebraic over any field $F$ then $F[\alpha] =F(\alpha) $ ie rational functions in $\alpha$ with coefficients in $F$ can also be expressed as polynomials in $\alpha$ with coefficients in $F$. This is very straightforward for simple examples when $\alpha=\sqrt{2}$, but the proof for general algebraic $\alpha$ needs the idea of GCD of polynomials. – Paramanand Singh May 02 '21 at 08:23
  • I hope your textbook includes a proof of the theorem and you should try to study it carefully. – Paramanand Singh May 02 '21 at 08:26
  • @Arthur can you check my answer in answer box whether my solution is correct or not ? – jasmine May 03 '21 at 06:50

1 Answers1

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Take $f(x)=x^2 +1 \in \mathbb{Q}[x]$.

$x^2+1=0 \implies x= i$

Here $$\frac{\mathbb{Q}[x]}{f(x)} \cong \mathbb{Q}[i]$$

since $\langle f(x)\rangle$ is maximal $\implies \mathbb{Q}[i] \subseteq \mathbb{Q}(i) \tag1$

Also $x^{-1} \in \mathbb{Q}[x]$ this $\implies \mathbb{Q}(i) \subseteq \mathbb{Q}[i]\tag2$

From $(1)$ and $(2)$ we have $\mathbb{Q}[i] = \mathbb{Q}(i)$

jasmine
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    This is not enough. What immediately jumps out to me is that you shouldn't just claim that $i^{-1}\in \Bbb Q[i]$. You should prove it. For instance, you could point out that we have $i^{-1} = -i\in \Bbb Q[i]$. More fundamentally, though, you haven't proven that, say, $(i+1)^{-1}\in \Bbb Q[i]$. Or any other non-zero complex number. They all need to be there. And once you've done that, you've proven that $\Bbb Q[i] = \Bbb Q(i)$. But that's just one algebraic number. It should be proven for all of them. – Arthur May 03 '21 at 17:12