There is this isomorphism in my notes but there is no explanation. So I tried to reason myself but still not convincing enough, or my reasoning may even be wrong. I will appreciate if anyone is willing to lend some helps.
$\mathbb{R}[X]/(X^2+1)\cong\mathbb{C}$.
I have heard that we can achieve that by putting $i$ into $X$ in the denominator, but if we do that then the denominator will be $0$, so $\mathbb{R}[X]/(0)=\mathbb{R}[X]$? Then how can it be isomorphic to $\mathbb{C}$?
Thanks so much!
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 $\mathbb{C}$.
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.
$\newcommand{\C}{\mathbb{C}}\newcommand{\R}{\mathbb{R}} \mathbb{C}$ 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.
I was surprised that nobody added an argument using the first isomorphism theorem and elementary means.
Define $$f: \mathbb{R}[X] \to \mathbb{C}: P \mapsto P(i)$$
It is easy to see that this is a ring epimorphism.
It is also easy to see that $(X^2 +1)\subseteq ker f$. For the other inclusion, assume $P\in ker f$ and consider $P$ as a complex polynomial. Then $P(i) =0$ means that $(X-i)|P$. Note that $P(-i) =0$, so $(X+i)|P$ as well.
It follows that $(X^2+1)|P$ in $\mathbb{C}[X]$. A moment of consideration tells us that then $(X^2+1)|P$ in $\mathbb{R}[X]$ as well which means $P \in (X^2 +1)$.
Everyone else has given very good answers, so let me just make a few hand-wavey remarks:
Consider $\mathbb{R}[X]/(x^2+1)\ni a+bX\mapsto a+i b\in \mathbb{C}$.
The polynomial $X^2+1$ is irreducible of degree 2.
Thus, the ideal $(X^2+1)$ is maximal in the ring $\Bbb R[X]$.
Thus, the quotient $\Bbb R[X]/(X^2+1)$ is a field extension of degree $2$ over $\Bbb R$.
How many are there?
$\mathbb{R}[x]/x^2+1$ has the elements like: {$g(x)\cdot(x^2+1)+h(x)$}, the degree of h(x) is 0 or 1.
We can get $h_1, h_2 \in \mathbb{R}[x]/x^2+1$, hence they can be written as ($a_i,b_i \in \mathbb{R}$): $$h_1 = a_1\cdot x+b_1$$ $$h_2 = a_2\cdot x+b_2$$
It's easy to verify that: $$h_1+h_2 = (a_1 + a_2)\cdot x + (b_1 + b_2)$$ $$h_1\cdot h_2 = a_1a_2\cdot (x^2+1) + (a_1b_2+b_1a_2)\cdot x + b_1b_2-a_1a_2$$ $$= (a_1b_2+b_1a_2)\cdot x + b_1b_2-a_1a_2$$
So we have the morphism $$f:\space\space \mathbb{R}[x]/(x^2+1) \to \mathbb{C}$$ $$f(a\cdot x+b) \to a\cdot i+b$$ which preserves $+ \space and \space \cdot$ and apparently is bijective, this is an isomorphism.
