Let $\mathbb{C}[u]$ be the algebra of polynomials with the indeterminate u.
What is the quotient $\frac{\mathbb{C}[u]}{u\mathbb{C}[u]}$ ?
My attempt is that it's the set with two classes $\lbrace [a], [u] \rbrace $, where $a \in \mathbb{C} - \lbrace 0\rbrace$. Am I right ?
This is not true, as not every two nonzero complex numbers are quotiented together in $\mathbb C[u]/u\mathbb C[u]$. The equivalence relation defining this is $f \sim g$ if $u \mid (f - g)$. If $a, b \in \mathbb C$ are distinct, then their difference $a - b$ is not a multiple of $u$.
So what is this quotient? Well another way to think of the equivalence relation is to recall that $f(0)=0$ iff $u \mid f$ in $\mathbb C[u]$. Hence, $f \sim g$ iff $(f-g)(0) = 0$ iff $f(0) = g(0)$. Now, if you know the first isomorphism theorem, that'd immediately tell you that the map $\mathbb C[u] \longrightarrow \mathbb C$ via $f \mapsto f(0)$ has kernel $u \mathbb C[u]$ so $\mathbb C[u]/u\mathbb C[u] \cong \mathbb C$.
If you don't know the first isomorphism theorem, then let's work directly with equivalence classes. Take some $f \in \mathbb C[u]$ and let $\overline f$ denote its equivalence class in $\mathbb C[u]/u\mathbb C[u]$. Now, $\overline f$ is the set of polynomials which agree with $f$ at $0$. A great example of such a polynomial is the constant polynomial $f(0)$ itself! That is, $f \sim f(0)$, where elements of $\mathbb C$ are thought of as constant polynomials. Hence, we can rewrite $\overline f = \overline{f(0)}$, so every equivalence class is represented by a constant. Furthermore, we already said above that if $a, b \in \mathbb C$ are distinct then $u \nmid (a-b)$, i.e. $a \not\sim b$. Thus, the constant polynomials form a system of representatives of $\mathbb C[u]/u\mathbb C[u]$. This will give us an isomorphism $\mathbb C \cong \mathbb C[u]/u\mathbb C[u]$ of rings.