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If we define $\mathbb{R}[x] = \{ c_0 + c_1 x + \cdots + c_n x^n : n\in\mathbb{N}, \forall c_k \in \mathbb{R} \}$ as the set of polynomials with coefficients in $\mathbb{R}$ then I'm familiar with the result that $\mathbb{R}[i] = \mathbb{C}$, where $i$ is a root of the irreducible (over $\mathbb{R}$) polynomial $x^2 + 1$.

Obviously all polynomials of degree 3 or higher are reducible, since all odd powers have at least 1 real root, and every even degree can be factored into irreducible quadratics by the fundamental theorem of algebra — like how $$x^4 + 1 = (x^2 + \sqrt{2}x+1)(x^2 - \sqrt{2}x + 1).$$

But what if we try adjoining roots of other irreducible quadratics to $\mathbb{R}$, different from $i$?

For example, what if $\alpha$ is a root of the irreducible quadratic $x^2 - x + 1$, meaning that $\alpha^2 = \alpha - 1$? Obviously this number can be expressed as a traditional complex number involving $i$, but what can we say directly about the set $\mathbb{R}[\alpha]$?

  • Is $\mathbb{R}[\alpha]$ isomorphic to $\mathbb{C}$ since both $\alpha$ and $i$ are roots of a degree 2 polynomial?
Bill Dubuque
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Patch
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    Yes, it's isomorphic to $\mathbb{C}$, you don't get anything new this way. – Qiaochu Yuan Jun 27 '24 at 19:56
  • Ah, well then. I guess there are no "interesting" extensions of $\mathbb{R}$ then. Too bad. – Patch Jun 27 '24 at 20:02
  • But you can start with $\Bbb Q$ and take the completion $\Bbb Q_p$ with respect to the $p$-adic norm (instead of obtaining $\Bbb R$), and then take the algebraic closure of $\Bbb Q_p$, instead of $\Bbb R$. – Dietrich Burde Jun 27 '24 at 20:04
  • @Patch: well, some would consider the quaternions or the octonions to be quite interesting, or Clifford algebras, or... it all depends on what you mean by "extension"! – Qiaochu Yuan Jun 27 '24 at 20:11
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    Otoh, there are non-complex rings (vs. fields) that can be obtained by adjoining roots of polynomials, e.g. the real dual numbers $,\Bbb R[x]/x^n,,$ which prove handy as algebraic models of tangent and jet-spaces, and are applied in deformation theory, numerical analysis (along with Levi-Civita fields), Synthetic Differential Geometry, etc, see here. – Bill Dubuque Jun 27 '24 at 20:12
  • If you are into other extensions of the real numbers "similar" to the complex numbers. Perhaps hypercomplex numbers (like dual numbers), which can contain other and more imaginary units, would be something for you. If you want a different direction there would also be something like surreal numbers (real, infinitesimal and infinite number). – The Art Of Repetition Jun 27 '24 at 20:18
  • Please ask only one question per post. I removed the 2nd question. Ask it separately if need be. – Bill Dubuque Jun 27 '24 at 20:32
  • Please also search for answers before posting questions. It is very easy to find many prior posts on this and related topics. – Bill Dubuque Jun 27 '24 at 20:36
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    Please unaccept the answer (click on the green checkmark) so that the post can be more efficiently processed. – Bill Dubuque Jun 27 '24 at 22:16

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