Equivalence classes with modulo $m(x)$ 
Show that $W = \{f(x)m(x) | f(x)\in\mathbb{F}_2[x]\}$ is a vector subspace of $\mathbb{F}_2[x]$, whose cosets are exactly the equivalence classes of $\mathbb{F}_2[x]$ with respect to congruence modulo $m(x)$.


I am confusing this problem, since $w=f(x)m(x)$, for any elements $g(x)$ in $\mathbb{F}_2[x]$, $g(x)f(x)m(x)\mod{m(x)}$ and $f(x)m(x)g(x)\mod{m(x)}$ give zero. I don't think that is right. Could someone help me to clear out my confusion? 

EDIT:
I think I get the idea. This is what I have so far, but it seems not right.
Define the left coset be $g(x)+W$ and right coset be $W+g(x)$ with $g(x)\in\mathbb{F}_2[x]$. For any non-zero $m(x)\in\mathbb{F}_2[x]$, we have $$g(x)+f(x)m(x)=g(x)\pmod{m(x)}\qquad\text{and}\qquad f(x)m(x)+g(x)=g(x)\pmod{m(x)}$$
 A: If I've understood your question correctly, your question applies to a more general setting where $K$ is any field, $R$ is any commutative ring with identity containing $K$, and $m \in R$ is any element.  In your notation, you can take $K = \mathbb{F}_2, R = \mathbb{F}_2[X]$, and $m = m(x)$.  Actually you can generalize much further, but it doesn't matter.
Now $(m) := \{ f m : f \in R\}$ (in your notation, this is what you call $W$) is a subgroup of $R$ with respect to addition, and it is also closed under scalar multiplication by $K$, making it a vector subspace of $R$ (I think as part of your problem they want you to prove this!).  It is actually closed under multiplication by any element of $R$, making it an ideal, but you don't seem to need that.
Since the group operation here (addition) is commutative, right and left cosets are the same thing.  If $g \in R$, the coset $g + (m)$ is by definition the set of all sums $g + h$, where $h \in (m)$, or in other words $g+ (m)$ is the set of all sums $g + fm$, where $f \in R$.  So of course $g + (m) = (m) + g$.  I didn't really understand what you were getting at in your edit.
If $g_1, g_2 \in R$, one says that $g_1, g_2$ are congruent modulo $m$ if there exists an $f \in R$ such that $g_1 - g_2 = f m$.  This definition is symmetric in $g_1$ and $g_2$: I can say $g_1$ is congruent to $g_2$, or $g_2$ is congruent to $g_1$, or $g_1$ and $g_2$ are congruent, it doesn't matter how you like to phrase it.  In other words, $g_1$ and $g_2$ are congruent modulo $m$ if $g_1 - g_2$ is a member of $(m)$.
As to your question, I think what you want to show is the following:

If $g_1 \in R$, the equivalence class of $g_1$ modulo $m$ is equal to the coset $g_1 + (m)$.

Let $S$ be the equivalence class of $g_1$ modulo $m$.  By definition $S$ is the set of all elements $g_2 \in R$ with the property that $g_2$ is congruent to $g_1$ modulo $m$.  But $g_2$ is congruent to $g_1$ modulo $m$ if and only if $g_1 - g_2 \in (m)$, if and only if $g_1 \in g_2 + (m)$.  So there is really nothing to this.
