Constructing Factor Ring $\mathbb C / I$ where $I=_{\mathbb C[X]}\langle X^2+1 \rangle$

Question

I'm trying to get some intuition behind the Factor Ring $\mathbb C[X] / I$ where $I=_{\mathbb C[X]}\langle X^2+1 \rangle$.

1)What does it look like in the set notation?

2)Is the equivalence relation always $x\sim y\iff x-y\in I$?

3)Is it isomorphic/homomorphic to anything?

4)Is $\mathbb C[X]/I$ a field?

I know that $\mathbb C[X]$ is the set of all polynomials. I'm guessing $a_0x_0+...+a_nx^n$ where $a_i\in \mathbb C$. I also know that $I$ is the principal ideal.

The definition in my book says it should be the set $(\mathbb C[X]/ \sim)$ of cosets of $I$ in $\mathbb C$ which doesn't explain much to me.

Answer 1

1. It is a set of cosets, namely $\mathbb{C}[X]/I = \{ f + I \mid f \in \mathbb{C}[X]\}$ where by definition the coset $f + I = \{ f + i \mid i \in \langle X^2 + 1\rangle \} = \{g \in \mathbb{C}[X] \mid g \sim f \}$ is exactly the equivalence class of $f$, under the equivalence relation $g \sim f \iff g - f \in I$.

2. Yes, this is the default equivalence relation when constructing factor rings.

3. Since in the factor ring the relation $X^2 + 1 = 0$ holds, we can reduce every polynomial of degree $\geq 2$ to a polynomial of degree $\leq 1$. For example, in $\mathbb{C}[X]/I$, we have $X^3 + X^2 = X^2(X + 1) = (-1)(X+1) = -X - 1$. Note that $-X-1$ is exactly the remainder of $X^3 + X^2$ divided by $X^2 + 1$. This holds in general, and so every coset $f + I$ in $\mathbb{C}[X]/I$ can be uniquely written in the form $(aX + b) + I$, with $a,b \in \mathbb{C}$, obtained by considering the remainder after division by $X^2 + 1$ of $f$. Thus as a $\mathbb{C}$-vector space $\mathbb{C}[X]/\langle X^2 + 1 \rangle$ is isomorphic to $2$-dimensional vector space of polynomials with coefficients in $\mathbb{C}$ of degree at most $1$.

4. No, $\mathbb{C}[X]/I$ is not a field. A general result states that for any field $K$ and polynomial $f \in K[X]$, the ring $K[X]/\langle f \rangle$ is a field iff $f$ is irreducible. Here $X^2 + 1 = (X+i)(X-i)$ isn't irreducible over $\mathbb{C}$. Note however that $\mathbb{R}[X]/\langle X^2 + 1 \rangle \cong \mathbb{C}$ is a field!