identifying a polynomial ring with the underlying field Let $F$ be any field. Consider the ring $F[x]_x$. We know that elements in this ring are the polynomials $f(x) \in F[x] $ such that $f(x) = g(x) + xp(x) $ for any $g,p \in F[x]$. So we can write 
$$ F[x]_x = \{ g(x) + x p(x) : g(x),p(x) \in F[x] \} $$
How can we identify this with the field $F$??
 A: You misunderstood the definition : $F[x]/(x)$ is not the set of elements $f\in F[x]$ such that $f = g+x\cdot p$ ; any $f$ has this form, taking $g=f$ and $p=0$.
Rather, each element of $F[x]/(x)$ is a subset $[p] = \{ g + x\cdot p\,|\, g\in F[x]\}\subset F[x]$ for some $p\in F[x]$. And each such set contains a unique constant polynomial : $p(0)\in F$ is the unique constant in $[p]$, which gives you a bijection between $F[x]/(x)$ and $F$.
Try to see that this is a ring isomorphism.
A: If $S$ is a ring, and $J$ is an ideal of $S$ (in your case, $S = F[x]$, and $J = (x) = \{ f \in S : f(0) = 0\}$), then $S/J$ is defined to be the set of cosets of $J$ in $S$.  A coset of $J$ in $S$ is a subset of $S$ of the form $$s + J = \{ s + j : j \in J\}$$ Any two cosets $s_1 + J, s_2 + J$ are either equal, or they are disjoint (their intersection is empty), and two cosets are equal if and only if $s_1 - s_2 \in J$.  You make $S/J$ into a ring by defining addition by the formula $$(s_1 + J) + (s_2 +J) = (s_1 + s_2) + J$$ and multiplication by $$(s_1+J) \cdot (s_2 + J) = s_1s_2 +J$$ These operations are well defined.  The zero element of $S/J$ is the coset $0 +J$, and the identity is the coset $1+J$.
Now, for $f(x) \in F[x]$, write the coset as $\overline{f(x)}$ instead of $f(x) + (x)$.  By definition of coset representatives, you have that $\overline{f(x)} = \overline{g(x)}$ if and only if $f(0) = g(0)$ (check this!).  If you write any polynomial out as $$f(x) = a_0 + a_1x + \cdots + x^n$$ then you have $\overline{f(x)} = \overline{a_0}$.  Also, if $f(x)$ and $g(x)$ have constant terms $a_0$ and $b_0$ respectively, then $\overline{f(x)} + \overline{g(x)} = \overline{a_0} + \overline{b_0}$, and $\overline{f(x)} \cdot \overline{g(x)} = \overline{a_0} \cdot \overline{b_0}$.  Thus, every element of $F[x]/(x)$ can be identified with an element in $F$, and this identification is compatible with addition and multiplication.  
