# If $\alpha$ is algebraic over $\mathbb{Q}$, then is it always true that $\mathbb{Q}[\alpha] = \mathbb{Q}(\alpha)$? [duplicate]

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?

• 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.
– lulu
Sep 13, 2023 at 11:25
• 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! Sep 14, 2023 at 11:30
• 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 $\alpha$." Sep 15, 2023 at 17:51
• @FShrike: this answer is really helpful and presents your idea applied to any member of $k[\alpha ]$. Sep 16, 2023 at 2:20
• @ParamanandSingh Thanks. I forgot the essential fact it’s not just $\alpha$ but actually any element that is a root of a polynomial Sep 16, 2023 at 10:39

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 $$\mathbb{Q}[\alpha]=\mathbb{Q}(\alpha)$$.