# $\mathbb{C}$ 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 $$x^2 + 1$$?

Tags :

• 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 $$\pm \bar{x} \mapsto \pm i$$.