How should $\mathbb{C}\triangleq\mathbb{R}[X]/(X^2+1)$ be read?

Question

I have been told that the set of all complex numbers $\mathbb{C}$ is technically defined as the field extension: $$\mathbb{C}\triangleq\mathbb{R}[X]/(X^2+1)$$ How should this be read? Should is be:

1. "The set of all polynomials in $\mathbb{R}$ of variable $X$ extended by the polynomial $X^2+1=0$" or
2. "The set of solutions to all polynomials in $\mathbb{R}$ of variable $X$ extended by the set of solutions to the polynomial $X^2+1=0$

I guess by question is really-- Does this field extension extend a set of polynomials, or rather extend the set of solutions to said polynomials?

Answer 1

It's neither of your options. I'm guessing you were thinking this was an instance of the notation "$K/L$" meaning "$K$ is an extension field of $L$", but it's not: instead it's just the notation for a quotient ring, which is totally unrelated despite having the same notation.

Specifically, it's the quotient of the ring $\mathbb{R}[X]$ of polynomials in a variable $X$ with coefficients in $\mathbb{R}$ by the ideal $(X^2+1)$ of all polynomials that are multiples of $X^2+1$. So an element of $\mathbb{R}[X]/(X^2+1)$ is really a coset $f(X)+(X^2+1)$, for some $f(X)\in \mathbb{R}[X]$.

How is this an extension of $\mathbb{R}$? Well, by definition an extension of $\mathbb{R}$ is a field $K$ together with a homomorphism $j:\mathbb{R}\to K$. In this case it turns out that the quotient ring $\mathbb{R}[X]/(X^2+1)$ is indeed a field, and we can define a homomorphism $j:\mathbb{R}\to \mathbb{R}[X]/(X^2+1)$ by $j(r)=r+(X^2+1)$.

However, it doesn't really make sense to say that $\mathbb{R}[X]/(X^2+1)$ "extends polynomials"--the ring $\mathbb{R}[X]$ of polynomials is not a field, and we normally only talk about extensions of fields, not more general rings. And if we did talk about extensions of more general rings, we would require the homomorphism between the rings to be injective. But the canonial homomorphism $\mathbb{R}[X]\to \mathbb{R}[X]/(X^2+1)$ is not injective (its kernel is $(X^2+1)$).

Answer 2

There seems to be confusion about the concept of "field extension". If $K$ is a field that contains the subfield $F$ ("contains" means one has an inclusion map $F\hookrightarrow K$), we say that $K$ is an extension field of $F$, which is denoted as $K/F$. Note that the slash here has nothing to do with the slash in the quotient ring notation.

$\mathbb{R}[x]/(x^2+1)$ is an extension field of $\mathbb{R}$. Now $\mathbb{C}$ is defined as $\mathbb{R}[x]/(x^2+1)$.

If you have question about what $\mathbb{R}[x]/(x^2+1)$ really is, then you might probably want to clarify first

• what is the ring $\mathbb{R}[x]$;
• what is the ideal $(x^2+1)$;
• and what is a quotient ring.

If you are wondering how $k:=\mathbb{R}[x]/(x^2+1)$ coincides with the usual "$a+bi$" definition, you would need the study the ring homomorphism $\varphi:k\to\mathbb{C}$ defined by $$ a+bx\mapsto a+bi. $$

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Now to wrap things up, the notation $$ \mathbb{C}\overset{\Delta}{=}\mathbb{R}[x]/(x^2+1) $$ can be understood as the following

$\mathbb{C}$ is defined as the quotient ring $\mathbb{R}[x]/(x^2+1)$, which happens to be a field. And the field $\mathbb{R}[x]/(x^2+1)$ is a field extension of $\mathbb{R}$. That's why people say that $\mathbb{C}$ is defined as a field extension (of $\mathbb{R}$).