Let A be a commutative ring. Let $I \subsetneq A$. Show that there exists a maximal ideal $m \subseteq A$ such that $I \subseteq m$ I have this exercise, but I feel like something is wrong. As far as I know, if $m$ is a maximal ideal, then $m \subsetneq A$. But with this hypothesis, I think to take $I = \lbrace 1_A \rbrace$, so the only ideal that contains $I$ is $A$, so $I$ is not maximal. 
EDIT:
I wrote something wrong. I didn't meant that I was the ideal, but that if $m$ is an ideal such that $I \subseteq m$, then $m = A$, so m is not maximal
 A: The set $I=\{1\}$ is usually not an ideal, unless $1$ is the only element of $A$. That is because the ideal axioms require that for any $a$, if $1\in I$, we also have $a=a\cdot 1\in I$. Hence, $A\subseteq I$ and consequently, $A=I$. So this is not a contradiction because only proper ideals are contained in a maximal one.
A: The ideal $(1)$ generates the whole ring, so we usually do not consider it an ideal. A different approach is via Zorn's lemma. If you take the collection of ideals that contain $I$, and any chain in it, we just have to show that the chain has an upper bound (as in an ideal that contains $I$.)
This amounts to checking that
$$M:=\bigcup_{i \in \mathcal{I}} I_i$$
is a proper ideal since it clearly contains $I$. It is proper since $1 \notin I_i$ for any $i \in \mathcal{I}$.
To see that it is an ideal ,note that it is closed under multiplication by any $r \in A$, since for any $a\in M$, we have that $a \in I_i$ for some $i$, so $ra \in I_i$. This method also shows the existence of additive inverses by taking $r=-1$.
Closure under addition is the part that uses the ordering. WLOG, suppose that $I_i \subset I_j$. Then $i+j \subset I_j$, since $i,j \in I_j$.
Hence, $M$ is an ideal that bounds the chain, so Zorn's lemma applies.  
