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The theorem of existence of the algebraic closure of a field can be written as the following assertion:

Let $Q$ be a field, then there exists a field $C$ (having the same characteristic than $Q$) that is an algebraic extension of $Q$ and such that if the field $H$ is an algebraic extension of $C$ we must have $H = C$.

If we replace every occurrence of the word "field" by an occurrence of the words "division ring", will the substituted assertion remain true?

J. W. Tanner
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Gérard Lang
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    I have no answer to give except for the following observation: the notion of algebraic element should be replaced by the notion of integral element over a ring. But we already encounter problems if we naively transpose the situation we have on fields. For example, for $A \to B$, an integral morphism between division rings, the integral closure of $A$ in $B$ may not be a division ring, which is already unfortunate. I don't have an example of such a situation at hand, but I would be delighted to see that my thinking is wrong. – Francis H. Nov 19 '25 at 10:46
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    I think it would be relevant to the question what exactly the definition of algebraic extension is – Elia Immanuel Auer Nov 19 '25 at 10:51
  • @EliaImmanuel Auer Let Q be a division ring, and C a division ring containing Q, and say that an element c in C is algebraic over Q iff c is a root in C of a (monic) polynomial with coefficients in Q. We say that the division ring C is an algebraic extension of the division ring Q iff every element c in C is algebraic over Q. – Gérard Lang Nov 19 '25 at 13:38
  • Note that if $R$ is a ring containing $Q$ and $r \in R$, then the set $Q[r]={p(r) : p \in Q[X] }$ is not necessarily closed under multiplication, unless $Q$ is in the center in $R$. – lhf Nov 19 '25 at 14:27
  • See also https://math.stackexchange.com/questions/806119/do-polynomials-make-sense-over-non-commutative-rings – lhf Nov 19 '25 at 14:30

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There is nothing like what you are asking about because the notion of “polynomial” is much more complicated over a noncommutative division ring, e.g., $ax^2$ and $xax$ and $x^2a$ are different polynomials of degree $2$ when $a$ is not in the center and, moreover, evaluating a polynomial on a noncommutative ring is not a homomorphism when the coefficients are not all in the center. The concept of integral closure in commutative algebra gets more complicated in noncommutative rings because the sum and product of two integral elements need not be integral.

The simplest polynomials have all monomials of the form $a_ix^i$ (“left polynomials”) or all monomials of the form $x^ia_i$ (“right polynomials”). Over Hamilton’s quaternions $\mathbf H$, Niven proved all nonconstant left and right polynomials over $\mathbf H$ have a zero. Baer proved a converse: if a noncommutative division ring $D$ is finite-dimensional over its center and every left (or every right) polynomial over $D$ has a root in $D$, then $D$ is the quaternions over a real-closed field.

Later Eilenberg and Niven showed a nonconstant polynomial over $\mathbf H$ has a zero if it has a unique monomial of largest degree (recall the three different types of monomials of degree $2$ that I mentioned earlier). Without a unique highest-degree monomial there can be problems: $f(x) = ix - xi + 1$ has no zero in $\mathbf H$.

A nice reference for some of this material is Section $1.8$ of Chapter $7$ of the book Numbers by Ebbinghaus et al.

If you focus on left (or just on right) polynomials then you can read a discussion of such polynomials over a division ring in Section $16$ of Chapter $5$ Lam’s A First Course in Noncommutative Rings, which has the nice result that a polynomial (again, left or right) of degree $n$ over a division ring has at most $n$ conjugacy classes of roots. But watch out: $x^2 + 1 = 0$ has infinitely many roots in $\mathbf H$ but its number of conjugacy classes of roots in $\mathbf H$ is just one, not two!

J. W. Tanner
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KCd
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  • But your answer doesn't mean that $\bf H$ has no "algebraic closure", just that it can't be a finite degree extension, right? – Kimball Nov 19 '25 at 23:16
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    @Kimball when I wrote at the start that there is no such thing as an algebraic closure of a noncommutative division ring, I meant that as a practical matter nobody has ever defined it. Have you ever heard of such a workable concept in some setting? – KCd Nov 20 '25 at 03:04
  • No, I haven't either, but there's a lot about division rings I don't know about. I see - I interpreted your first sentence as saying one cannot construct such an object, but that's not what you meant. – Kimball Nov 20 '25 at 14:02