Why $\mathbb{R}[X]/(X^2+1)\cong\mathbb{C}$?

Solution 1:

When you quotient a polynomial ring by (the ideal generated by) a polynomial, you automatically get a root to it. In this case, the element represented by $X$ satisfies $X^2 + 1$ (just by definition).

Using the polynomial division algorithm, you can write any element $f$ of $\mathbb{R}[X]$ as $f = q(X^2 + 1) + r$, where $r$ has degree $1$ or $0$. In the quotient, $f$ and $r$ represent the same element.

So an arbitrary element of your quotient has the form $aX + b$. If you write out what happens when you multiply and then reduce two such expressions, you recover the multiplication equation you get for $\mathbb{C}$.

Solution 2:

Hint: the evaluation map $\,\Bbb R[x] \to \Bbb C\,$ where $\,f(x)\mapsto f(i)\,$ is onto, and its kernel is generated by the minimal polynomial of $\,i,\,$ i.e. $\,x^2+1,\,$ since $\,\Bbb R[x]\,$ is Euclidean so a PID. Therefore $\ \Bbb R[x]/(x^2+1)\cong \Bbb C\ $ by the first isomorphism theorem.

Solution 3:

$\newcommand{\C}{\mathbb{C}}\newcommand{\R}{\mathbb{R}} \mathbb{C}$ is often described as being constructed from $\mathbb{R}$ by freely adjoining an element $i$ satisfying $i^2 = -1$ (equivalently, $i^2 + 1 = 0$). This isomorphism is one way of saying that formally, since the polynomial ring freely adjoins a new element, and then quotienting forces the equation $x^2 + 1 = 0$.

What does this mean, precisely? The following two facts are the core of it:

Proposition. Let $R$, $S$ be any rings. Then ring homomorphisms $R[x] \to S$ correspond to pairs of a homomorphism $R \to S$ (the image of the constant polynomials), together with an element of $S$ (the image of $x$).

This formalises the idea that $R[x]$ freely adjoins an element to $R$ — any time you have a ring with a map of $R$ and a chosen element, these extend uniquely to a map from $R[x]$, since the ring homomorphism conditions determine what the image of any polynomial must be once you know the images of the variable and the coefficients.

Proposition. Let $R$ be a ring, $a \in R$ any element. Then for any other ring $S$, homomorphisms $R/(a) \to S$ correspond to homomorphisms $R \to S$ that send $a$ to $0$.

This similarly formalises the idea that quotienting by $(a)$ is exactly forcing the equation $a = 0$.

Putting these together, $\R[x]/(x^2 + 1)$ is “freely adding an element with $x^2 = -1$ to $\R$”; so this explains intuitively why you should expect it to be isomorphic to $\C$.

To construct the isomorphism, put together the facts above. For any ring $S$, a homomorphism $\R[x]/(x^2+1) \to S$ is determined by specifying a homomorphism from $\R$, together with an element $a$, such that $a^2 + 1 = 0$; and this extension is unique, so to check that two homomorphisms out of $\R[x]/(x^2+1)$ agree, it’s enough to check what they do on $\R$ and $x$. This fact lets you easily construct a map $\R[x]/(x^2+1) \to \C$; then you can explicitly give a map $\C \to \R[x]/(x^2+1)$, and check it’s a homomorphism; and then you can use this fact again in checking that the maps are mutually inverse.