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First, $\mathbb{Q}[\sqrt{2}] = \mathbb{Q}(\sqrt{2})$ since $\frac{1}{\sqrt{2}} = \frac{1}{2}\sqrt{2}$.

and as for more complicated examples:

Let $\alpha = \sqrt{\sqrt{3}+\sqrt{2}}$, $A = \mathbb{Q}[\alpha]$ and $B = \mathbb{Q}(\alpha)$.

Then $\frac{1}{\alpha} = \sqrt{\sqrt{3}- \sqrt{2}} = 10\alpha^3- \alpha^7$, which shows that $A = B$.

My conjecture is that $\mathbb{Q}[\alpha] = \mathbb{Q}(\alpha)$ is always true as long as $\alpha$ is algebraic, but I do not have a proof of this claim, just trials of many difference cases, and I am wondering if someone has a counterexample? or maybe a known criterion for when this is true?

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    $\begingroup$ If $p(x)$ is the minimal polynomial for $\alpha$ over $\mathbb Q$ then $p(0)\neq 0$ (why?) so you can use $p(x)$ to write down the inverse for $\alpha$. In your case, we have $x^8-10x^4+1$ as the minimum polynomial, so $\alpha(\alpha^7-10\alpha^3)=-1\implies \frac 1{\alpha}=10\alpha^4-\alpha^7$, for example. $\endgroup$
    – lulu
    Sep 13, 2023 at 11:25
  • $\begingroup$ In your question you only focused on inverting $\alpha$. However that doesn't necessarily imply the whole thing is a field. Although by other, much cleaner arguments, $k[\alpha]$ is always a field for $k$ a field and $\alpha$ algebraic. So my answer isn't exactly wrong but it's misleading. I can't remove it though, since you've accepted it! $\endgroup$
    – FShrike
    Sep 14, 2023 at 11:30
  • $\begingroup$ Answer converted to comment: "Of course we can assume $n\ge1$ and $a_0\neq 0$. In the ring $k[\alpha]$, $a_n\alpha^n+\cdots+a_1\alpha=-a_0$ and because $k$ is a field, $-a_0$ is invertible; $(-a_0)^{-1}(a_n\alpha^n+\cdots+a_1\alpha)=1$. Note we can factor out $\alpha$: $$(-a_0)^{-1}(a_n\alpha^{n-1}+\cdots+a_1)\cdot\alpha=1$$Therefore $(-a_0)^{-1}(a_n\alpha^{n-1}+\cdots+a_1)\in k[\alpha]$ is an inverse to $\alpha$." $\endgroup$
    – FShrike
    Sep 15, 2023 at 17:51
  • $\begingroup$ @FShrike: this answer is really helpful and presents your idea applied to any member of $k[\alpha ]$. $\endgroup$
    – Paramanand Singh
    Sep 16, 2023 at 2:20
  • $\begingroup$ @ParamanandSingh Thanks. I forgot the essential fact it’s not just $\alpha$ but actually any element that is a root of a polynomial $\endgroup$
    – FShrike
    Sep 16, 2023 at 10:39

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Assuming $\alpha\neq 0$ is algebraic, so $\mathbb{Q}[\alpha]$ is a finite-dimensional $\mathbb{Q}$-vectorspace. Multiplication by any $q(\alpha)=q_0+q_1\alpha+\dots+q_k\alpha^k\neq 0$ gives a $\mathbb{Q}$-linear endomorphism of $\mathbb{Q}[\alpha]$ which is injective, hence also surjective by finite-dimensionality. So $\frac{1}{q(\alpha)}\in\mathbb{Q}[\alpha]$ and $\mathbb{Q}[\alpha]=\mathbb{Q}(\alpha)$.

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