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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 $k$?

PPenguin
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  • What is $\times$ here (between two complexes?) – coffeemath Aug 29 '18 at 01:11
  • @coffeemath I was explicitly representing the multiplication operation. I was torn whether to make it implicit, but since the multiplication would have to be non-associative to have a solution I made it explicit. Implicit does look nicer. – PPenguin Aug 29 '18 at 01:20
  • Multiplication of complexes is associative, so if that's what your $\times$ means, your equation becomes $2=0,$ which doesn't have a solution. So you must mean some nonassociative binary operation on complexes, and I don't know of any. – coffeemath Aug 29 '18 at 07:41
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    What counts as an algebra? – patrik Aug 29 '18 at 08:09
  • @coffeemath "Multiplication of complexes is associative, so ... doesn't have a solution" Correct, it doesn't have a solution in the complex numbers. Therefore to be algebraically closed for more complicated algebraic expressions we need to consider a different algebra, which as you've noticed must be non-associative. – PPenguin Aug 29 '18 at 16:12
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    @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: https://math.stackexchange.com/questions/2894403/what-is-this-2d-division-algebrahttps://math.stackexchange.com/questions/2894403/what-is-this-2d-division-algebra – PPenguin Aug 29 '18 at 16:12
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    @PPenguin I tried that question and it doesn't exist (big backward E etc.) – coffeemath Aug 30 '18 at 07:07
  • Non-commutativity already introduces problems like this. I do know that quaternions are associative, and that example needs associativity, but I suspect this to be asking too much in general. – Jyrki Lahtonen Aug 30 '18 at 18:16
  • @coffeemath I must have accidentally hit paste twice as the url is doubled. Trying again https://math.stackexchange.com/questions/2894403/what-is-this-2d-division-algebra – PPenguin Aug 31 '18 at 01:02
  • @PPenguin Thanks, that link worked. Good example. – coffeemath Aug 31 '18 at 17:10
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    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 :)

     – patrik Sep 01 '18 at 07:07
  • @patrik Please show that if you find it straightforward! Do observe that the discussion in the linked thread shows that the quaternions have no (associative) extension ring that contains a solution $x$ of the quaternion equation $ix+xi=j$. You can use some universal ring extension, and then force such a relation to hold, but the resulting quotient ring may not be an extension of the one you started with, and in that example case the quotient became the zero ring. I have no intuition about what changes, if we drop associativity. Over to you. – Jyrki Lahtonen Oct 25 '18 at 05:17

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