Show that every maximal ideal of $R$ contains the element $a$. 
Let $R$ be a commutative ring with identity and let $a\in R$ such that $a^n=0$, for some positive integer $n$. Suppose that $I$ is and ideal of $R$. Define $(I,a)=\{x+ra\ :\ x\in I\ \text{and}\ r\in R\}$. Let $a^n$ be zero, but $a^{n-1}$ is not zero for some positive integer $n$.
Show that every maximal ideal of $R$ contains the element $a$.

I tried: Let $M$ is maximal ideal of $R$. Then $M$ is not same $R$.
And if $M \subseteq I \subseteq R$ then $M=R$ or $I=R$.
$(I,a)$ is and ideal of $R$.
 A: Let $M$ be a maximal ideal of $R$ and let $n$ be the smallest positive integer such that $a^{n}=0$. Since $0\in M$ we see that $a^{n}\in M$ because $a^{n}=0$. Since $M$ is a maximal ideal it is also a prime ideal, so noting that $a^{n}=a^{n-1}a$ implies that either $a^{n-1}\in M$ or $a\in M$. If $a\in M$ we're done. If $a^{n-1}\in M$ we continue by noting that $a^{n-1}=a^{n-2}a$ implies that $a^{n-2}\in M$ or $a\in M$ and so on and so forth. Therefore we conclude that $a\in M$.
A: Here's another way:
Suppose it's not in a particular maximal ideal $M$.  Then $(a, M)=R$, and you can find $r\in R$ such that $ar+m=1$.
But then $1=1^n=(ar)^n+(\text{a sum of things in $M$)}=0+(\text{a sum of things in $M$)}\in M$.
This would imply that $M=R$, contradicting the definition of maximality of $M$.
A: More generally, if $1-bc$ is invertible for every $c$, then $b$ belongs to every maximal ideal.
Indeed, if $M$ is a maximal ideal and $b\notin M$, then $(b,M)=R$ and therefore $1=bx+y$, with $x\in R$ and $y\in M$. But then $y=1-bx$ would be invertible: a contradiction.
If $a$ is nilpotent, that is, $a^n=0$ for some $n>0$, then $1-ac$ is invertible for every $c$, because the inverse is
$$
1+ac+a^2c^2+\dots+a^{n-1}c^{n-1}
$$
as it can be seen by computation.

Note. The condition about $1-bc$ being invertible for every $c$ is actually equivalent to $b$ belonging to every maximal ideals.
