Describe $\Bbb{R}[x]/(x^2 + 1)$

Question

First by previous knowledge, I do know that $\Bbb{R}[x]/(x^2 + 1) \cong \Bbb{C}$, so this might seem trivial. But I am not here to ask about that and I don't want to use this fact

I am reading Dummit and Foote on Chapter 13 and this example (pg 515) describes the field as "an extension of degree $2$ of $\Bbb{R}$ in which $x^2 + 1$ has a root". I am not sure what this sentence means.

First there was a theorem that stated the basis elements of $K = \Bbb{R}[x]/(p(x))$ as described by $\theta \equiv x\pmod{p(x)}$ and preceding paragraphs describe multiplication in the quotient ring.

Okay here is the actual question, they described the elements as $a + b\theta$ with $a,b\in \Bbb{R}[x]$. How do we arrive at that form? There is no talk about the isomorphism to $\Bbb{C}$ until much later.

I do not not see how setting $x^2 + 1 \equiv 0$ turns $a_0 + a_1x + a_2x^2 + \dots a_nx^n\to a_0 + a_1x$ .

Answer 1

In $\mathbb{R}[x]/(1+x^2)$, $[x]$ (that is, $x+(x^2+1)$) is a root of the equation $x^2+1=0$, since $[x]^2+1=[x^2+1]=0$.

And if $p(x)\in\mathbb{R}[x]$ there are polynomials $q(x),r(x)\in\mathbb{R}[x]$ such that $p(x)=q(x)\times(x^2+1)+r(x)$ and $\deg r(x)\leqslant 1$. So, $r(x)=a+bx$ for some real numbers $a$ and $b$ and$$\bigl[p(x)\bigr]=\bigl[r(x)\bigr]=[a+bx]=a+b[x].$$

Answer 2

Well, if $x^2+1\equiv 0$ then $x^2\equiv -1$. That means that whenever you see $x^2$, you can replace it with $-1$. So for instance, $x^3\equiv -x$, $x^4\equiv -x^2\equiv 1$, $x^5\equiv -x^3\equiv x$, and so on. In this way a polynomial of any degree can be reduced to a linear polynomial.

(This "reduction" process on powers of $x$ is in fact the same as the polynomial division algorithm used in the other answer, but it is often very helpful to think through the process inside the quotient ring instead of in the original polynomial ring.)