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I want to find an algebraic division algebra that is not finite dimensional, but i don't want to do it in terms of field extensions nor anything like that.

Instead of that, what i want to do is to give a particular division algebra, making sure that is algebraic and showing that the underlying vector space has an infinite dimension base.

I am trying to work it out with an algebra generated by a single element. For example $\mathbb{R}[(\sqrt[n]{2})]$ $(n\geq1)$ which is easy to see that $\left\lbrace \sqrt{2},\sqrt[2]{2},\cdots,\sqrt[n]{2} \right\rbrace $ is a base. And i know that $x^n-2$ is a polynomial that makes $0$ every $\sqrt[n]{2}$ but i can't find a polynomial that makes $0$ an element which is a linear combination of all elements in the base and i don't know how can i proceed now.

Javiervil
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  • I might be mistaken, but if $(A,+,)$ is a division algebra, isn't $(A^\Bbb{N},+',')$ - where $x+'y=z\iff x_i+y_i=z_i$ and $x'y=z\iff x_iy_i=z_i$ for all $i\in\Bbb{N}$ - trivially a division algebra? – R. Burton Jun 20 '20 at 16:58
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    I don't understand what you're trying to do in your example. The (an) n-th root of 2 is in $\mathbb R$. Are you trying to add a different one? In any case, if you have an element that satisfies an algebraic relation over a field, the extension generated by it will be finite dimensional. – Kimball Jun 21 '20 at 01:12
  • @JyrkiLahtonen I said I might be mistaken. Now I know that I was. – R. Burton Jun 22 '20 at 15:19
  • @R.Burton The fact is that I need an algebraic infinite dimensional algebra, so if we have an element that has no multiplication inverse that means no problem since every element is algebraic and the dimension is infinite. – Javiervil Jun 22 '20 at 16:28
  • @Kimball I don't recall any result as you have just said. If I have a finite dimensional division algebra, then every element will be algebraic, but I don't think that the reciprocal is true, in fact that is what I am trying to proof. And you have some examples in terms of field extensions with $\mathbb{Q}$ – Javiervil Jun 22 '20 at 16:33
  • Javiervil, (one of) the problem(s) Kimball pointed out is that for all $n$ $\Bbb{R}(\root n\of 2)=\Bbb{R}$. – Jyrki Lahtonen Jun 22 '20 at 17:08
  • Anyway, would (Hamilton) quaternions with the four coordinates restricted to real algebraic numbers fit the bill? Everything in sight is algebraic. It is infinite dimensional over $\Bbb{Q}$ (even though the dimension over the center is only four). It is, in a sense defined in terms of field extension, and that is a somewhat unclear requirement. Anyway, often in the case of division algebras the interesting dimension is that over the center. I think even that can be arranged with a suitable nested union, but I would need to think harder. May be Kimball can help (they know this better than I do) – Jyrki Lahtonen Jun 22 '20 at 17:14
  • @JyrkiLahtonen of course you are right, that's why on most examples the usual is to use $\mathbb{Q}$. The problem is, even if I use $\mathbb{Q}$ I don't know how to construct a polynomial so any linear combination of the elements in the base is root of that. That's why I am thinking maybe it's best to try to proof that is algebraic in some other way or maybe try another example. – Javiervil Jun 22 '20 at 17:16
  • @JyrkiLahtonen wow, I love the idea about $\mathbb{H}$. I will think about that, thank you very much. I will keep an eye on what you write. – Javiervil Jun 22 '20 at 17:23
  • If I have a finite dimensional division algebra, then every element will be algebraic, but I don't think that the reciprocal is true - Indeed the converse is false, but you need to adjoin more than just one element to get something infinite dimensional. As a commutative example, you can look at the field $\overline{\mathbb Q}$ of all algebraic numbers. Every element is algebraic over $\mathbb Q$ by definition, but $\overline{ \mathbb Q}/\mathbb Q$ is infinite dimensional. – Kimball Jun 22 '20 at 17:23
  • @Kimball sorry if this is a dumb question but although I see that $\overline{ \mathbb Q}/\mathbb Q$ is infinite dimensional, i can't see so easily that every element of $\overline{\mathbb{Q}}$ is algebraic over $\mathbb{Q}$. If i have a linear combination of elements in $\overline{\mathbb{Q}}$ is it trivial that such element is root of a polynomial with coefficients in $\mathbb{Q}$? Thank you very much. – Javiervil Jun 22 '20 at 18:59
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    It's a basic result in algebraic number theory. See here: https://math.stackexchange.com/q/155122/11323 – Kimball Jun 22 '20 at 20:20

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