$\mathbb{C}$ as a quotient of $\mathbb{R}[x]$ by the principal ideal of $x^2 + 1$

by 0xd34df00d   Last Updated July 14, 2019 20:20 PM - source

Aluffi illustrates (circa III.4.2) the division by monic polynomials in a commutative polynomial ring $R$ by showing that $\frac{R[x]}{(f(x))} \cong R^{\oplus d}$ as groups, where $f(x) \in R[x]$ is monic, $d = \deg f(x)$, and $(a)$ is the principal ideal generated by $a$.

As a particular example, he shows that $\mathbb{C} \cong \frac{\mathbb{R}[x]}{(x^2+1)}$ as rings. He further explains the intuition behind this isomorphism:

The point is that the polynomial equation $x^2 + 1 = 0$ has no solutions in $\mathbb{R}$; the quotient $\frac{\mathbb{R}[x]}{(x^2 + 1)}$ produces a ring containing a copy of $\mathbb{R}$ and in which the polynomial does have roots (that is, $\pm$ the class of $x$ in the quotient).

I'm not sure I follow this. Indeed, $x^2 + 1$ does not have roots in $\mathbb{R}$, but

  • I only know that $\frac{\mathbb{R}[x]}{(x^2 + 1)}$ contains a copy of $\mathbb{R}$ because I know $\frac{\mathbb{R}[x]}{(x^2 + 1)} \cong \mathbb{R} \oplus \mathbb{R}$. Is there another nice way to see this, perhaps tied specifically to the structure of $\mathbb{R}$?

  • What would it mean for the polynomial (and I assume that's $x^2 + 1$ that's meant) to have roots in this quotient? This quotient is just a set of cosets of $(x^2 + 1)$ (that is, a set of sets $\{ g(x) + h(x)(x^2 + 1) \mid h(x) \in \mathbb{R}[x] \}$ for each $g(x) \in \mathbb{R}[x]$) that happens to be a ring. What would be a root here and how would it relate to $x^2 + 1$?



Answers 1


This is a common example to demonstrate quotient rings (and to see that polynomial rings are quite universal).

  • The quotient $\mathbb{R}[x]/(x^2+1)$ contains a copy of $\mathbb{R}$ because of the canonical injection $\mathbb{R} \hookrightarrow \mathbb{R}[x]/(x^2+1)$.
  • By modding out $(x^2+1)$, we introduce the relation $x^2+1 = 0$. This means that $x + (x^2+1)$ and $-x + (x^2+1)$ solve the polynomial equation $t^2 + 1 = 0$, i.e. $\pm \bar{x}$ fulfil the relation $x^2 + 1$. We wish to have this because solutions to that equation represent exactly $\pm i \in \mathbb{C}$! The isomorphism $\mathbb{R}[x]/(x^2+1) \xrightarrow{\sim} \mathbb{C}$ will hence map $\pm \bar{x} \mapsto \pm i$.
Kezer
Kezer
July 14, 2019 20:02 PM

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