Verifying whether a quotient ring is indeed a ring. Take $$\frac{\Bbb{R[x]}}{\langle x^2+1\rangle}$$ This is a ring. In this quotient ring, the product of equivalence classes $[a+bx]$ and $[c+dx]$ is another equivalence class, as a ring is closed under multiplication. 
$[a+bx]=\{a+bx,a+bx+(x^2+1),a+bx+2(x^2+1),\dots\}$, and $[c+dx]=\{c+dx,c+dx+(x^2+1),c+dx+2(x^2+1),\dots\}$. 
Also, we should have $[a+bx][c+dx]=[(ac+bdx^2)+(ad+bc)x]$. 
Which two elements in $[a+bx]$ and $[c+dx]$ multiply with each other to give $(ac+bdx^2)+(ad+bc)x+(x^2+1)$, which is a member of $[(ac+bdx^2)+(ad+bc)x]$?
Note: I know that $\{a+bx,a+bx+(x^2+1),a+bx+2(x^2+1),\dots\}\neq[a+bx]$; it is only the inverse of $[a+bx]$. I only write all this for a clear exposition: $[m][n]=[mn]$ only because the inverse images of $[m]$ and $[n]$, under the canonical mapping, multiply together to give the inverse image of $[mn]$.
Thanks!   

To be honest, if you are trying to show this is indeed a ring, there are indirect methods that are far easier.  For example:
Consider the evaluation homomorphism $ev_i:\mathbb{R}[x] \rightarrow \mathbb{C}$ defined as follows:
$$f(x) \mapsto f(i)$$
This homomorphism is indeed surjective since, given any $(a + bi) \in \mathbb{C}$, simply take the polynomial $f(x) = a + bx$.  Further, $ker(\phi) = \langle x^2 + 1 \rangle$.  Now we apply the isomorphism theorem:
$$\mathbb{R}[x]/ \langle x^2 + 1 \rangle \cong \mathbb{C}$$
So it is certainly a ring, and even a field.
