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Is there any non-trivial algebra for which any non-constant algebraic expression has a root in that algebra?

For example the complex numbers have a solution for any basic polynomial, but do not have a solution to:

$$a (a (a a)) - (a a) (a a) + 2 = 0$$

So we could "extend" the complex numbers with a root to this. But there are likely still other algebraic expressions that don't have a root. So we add a solution to that ... will this process be able to terminate?

For the purposes of this question let's define a "$k$-closed" algebra to mean a root exists for any non-constant algebraic expression represented as a tree of no more than $k$ operations in that algebra. For example, the complex numbers would at least be 3-closed. If there is no non-trivial $\infty$-closed algebra, is there at least a $k$-closed algebra for each finite $k$?

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  • $\begingroup$ What is $\times$ here (between two complexes?) $\endgroup$ – coffeemath Aug 29 '18 at 1:11
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    $\begingroup$ What counts as an algebra? $\endgroup$ – patrik Aug 29 '18 at 8:09
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    $\begingroup$ @patrick, coffeemath -- algebras need not be associative (e.g. the Octonions). An algebra is a vector space $A$ equipped with a bilinear multiplication map $A\times A \rightarrow A$. It is not necessarily unital or associative. Because multiplication is bilinear, this means multiplication will distribute over addition. Here is an example of a division algebra which has a solution to the algebraic expression in the question: math.stackexchange.com/questions/2894403/… $\endgroup$ – PPenguin Aug 29 '18 at 16:12
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    $\begingroup$ @PPenguin I tried that question and it doesn't exist (big backward E etc.) $\endgroup$ – coffeemath Aug 30 '18 at 7:07
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    $\begingroup$ I think it should be straightforward to prove that any algebra $A_1$ is a sub-algebra of another algebra $A_2$ where all non-constant expressions with coefficients in $A_1$ have roots in $A_2$. Similarly there is an algebra $A_3$ for expressions with coefficients in $A_2$ and so on... Taking $A=\cup_{i=1}^\infty A_i$ should have the property you look for. I might get around to actually do the "straightforward" part and write a proper answer, if someone sees some obvious trouble with my approach please share and spare me the trouble :) $\endgroup$ – patrik Sep 1 '18 at 7:07

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