Let $F[[X]]$ be the ring of formal power series over the field $F$. Show that $(X)$ is a maximal ideal. I need to prove the following:
Let $F[[X]]$ be the ring of formal power series over the field $F$. Show that $(X)$ is a maximal ideal.
I think I can prove this in three different ways
(1) $(X$) is a maximal ideal if and only if $F[[X]]/(X)$ is a field
(2) We can show that $(X)$ is a prime ideal since prime ideals are maximal.
(3) We can assume that $(X)$ is not maximal and get a contradiction.
This is my attempt for (1), which I think is not complete:
Let $(X) = \{\sum_{i = 1}^{\infty}a_iX^i: a_i \in F \}$ (all elements whose constant term is zero).
We can show that $F[[X]]/(X) \cong F$ and that would imply that $F[[X]]/(X)$ is also a field.
We have that
$$F[[X]]/(X) = \{ f + (X): f \in F[[X]] \mbox { and the constant term of } f  \mbox{ is not zero } \} $$
Hence, for any element $p \in F[[X]]/(X)$, 
$$p = a_0 + a_1X^1 + a_2X^2 + ... + b_1X^1 + b_2X^2 + ...  $$
$$p = a_0 + (a_1 + b_1)X^1 + (a_2 + b_2)X^2 + ...$$
Then required isomorphism $\phi: F[[X]]/(X) \to F$, is given by
$$ \phi(p) = a_0 \in F$$
I must show that for any $p,q \in F[[X]]/(X)$,
(1) $\phi(pq) = \phi(p)\phi(q)$
(2) $\phi(p + q) = \phi(a) + \phi(b)$
(3) $\phi(1_{F[[X]]/(X)}) = 1_F = 0$
Now since $p,q \in F[[x]]/(X)$, 
$$p = a_0 + (a_1 + b_1)X^1 + (a_2 + b_2)X^2 + ... = p_0 + p_1X^1 + p_2X^2 + ...$$
$$q = c_0 + (c_1 + d_1)X^1 + (c_2 + d_2)X^2 + ... = q_0 + q_1X^1 + q_2X^2 + ...$$
Now by the Cauchy product, we have that
$$pq = \sum_{n = 0}^{\infty}c_nX^n$$
where $c_n = \sum_{i = 0}^{n}p_iq_{n-i}$, therefore
$$ \phi(pq) = c_0 = p_0q_0 = a_0c_0 = \phi(p)\phi(q) $$
Also,
$$ \phi(p + q) = p_0 + a_0 = a_0 + c_0 = \phi(p) + \phi(q) $$
Now $1_{F[[X]]/(X)} = 0 + (X)$, hence
$$\phi(0 + (X)) = 0 = 1_F$$ 
$$\therefore \phi \mbox{ is a ring (field) isomorphism.} $$
$$\therefore F[[X]]/(X) \cong F, \therefore F[[X]]/(X) \mbox{ is a field since F is.}$$
$$\therefore(X) \mbox{ is a maximal ideal. } \square $$
Can anyone verify my proof for part (1), and help me with part (2) or (3) ?
I think my approach is very long and uses an important result. I want to prove it in a more direct way.
 A: Once one sees that
$(X) = \displaystyle \left \{ \sum_1^\infty a_i X^i \mid \forall i \; a_i \in F  \right \}, \tag 1$
that is, the principal ideal $(X)$ consists of all those formal power series in $X$ such that the constant term $a_0 = 0 \in F$, then showing $(X)$ is maximal is basically a one-liner.  
To see that (1) binds, we first observe that
$\displaystyle \left \{ \sum_1^\infty a_i X^i \mid \forall i \; a_i \in F  \right \} \subset (X), \tag 2$
since any element of $\{ \sum_1^\infty a_i X^i \mid \forall i \; a_i \in F \}$ satisfies
$\displaystyle  \sum_1^\infty a_i X^i = X  \sum_1^\infty a_iX^{i - 1} \in (X), \tag 3$
and we also have, for
$b(X) = \displaystyle \sum_0^\infty b_i X^i \in F[[X]], \tag 4$
$Xb(X) = \displaystyle X \sum_0^\infty b_i X^i =  \sum_0^\infty b_i X^{i + 1} = \sum_1^\infty b_{i - 1} X^i \in  \left \{ \sum_1^\infty a_i X^i \mid \forall i \; a_i \in F \right \}, \tag 5$
which shows that
$(X) \subset \displaystyle \left \{ \sum_1^\infty a_i X^i \mid \forall i \; a_i \in F  \right \}; \tag 6$
(2) and (6) together imply (1).
With this little result in hand, the promised one-liner goes as follows:
Suppose
$M \supsetneq (X) \tag 7$
is an ideal properly containing $(X)$; since it is clear from (1) that $(X)$ consists precisely of those elements of $F[[X]]$ whose zero-degree term is $0 \in F$, $M$ must contain at least one formal power series of the form
$c(X) =  \displaystyle \sum_0^\infty c_i X^i, \; c_0 \ne 0; \tag 8$
but it is clear from (1) that
$ \displaystyle \sum_1^\infty c_i X^i \in (X) \subset M \tag 9$
and thus, since $M$ is an ideal,
$c_0 = c(X) - \displaystyle \sum_1^\infty c_i X^i \in M; \tag{10}$
but $c_0 \ne 0 \in F$ is a unit; thus
$1 = c_0^{-1} c_0 \in M, \tag{11}$
which of course implies for any $a(X) \in F[[X]]$,
$a(X) = 1a(X) \in M; \tag{12}$
thus
$M = F[[X]], \tag{13}$
and so $(X)$ must be maximal.
