Non-integer domain which every ideal is a principal ideal Let $F$ be field and $A=F[t]\setminus (t^2)$, where $(t^2)$ is the ideal of $F[t]$
(a) Show that every ideal of $A$ is principal ideal
(b) Find all prime ideals of $A$
I know $A$ is not integer domain because $t^2$ is reducible, So it is just commutative ring with unity. Thus, it shows that there exists non-integer domain which every ideal is a principal ideal.
To prove it, let $I$ be an ideal of $A$, I need to find one generator of $I$. But I couldn't. I don't think there is special theorem to solve it. I guess I just need to use the definition of ideal and the structure of the factor ring. Could anyone help me to solve it..? I just need a few hints. Thanks!
 A: (1) $F[t]$ is a PID and thus an ideal in $F[t]/(t^2)$ is of the form $(P(t))/(t^2)$ where $P(t)$ is an element of $F(t)$ such that $(t^2)\subseteq (P(t))\iff P(t)|t^2$.
Show that this ideal is generated by the coset $P(t) + (t^2)$.
(2) The correspondence between ideals of the quotient ring and ideals of the ring containing the ideal you divide out preserves prime ideals. Thus you can reduce the question of giving prime ideals in the quotient ring to prime ideals of the ring $F[t]$ which contain $(t^2)$. What are these?
A: Result 1 : A quotient ring of a PID $R$ will be a principal ideal ring: Every ideal of $R/I$ is principal, where $I$ is an ideal in $R$.
Proof: 
Indeed, let $K$ be an ideal of $R/I$. By the Correspondence Theorem $K$ corresponds to an ideal $J$ of $R$ that contains $I$. Since $R$ is assumed to be a PID, then $J=(j)$ for some $j\in R$. The claim is that $K = (j+I)(R/I)$: let $k+I\in K$. Then $k+I \in J+I$, so there exists $a\in J$ such that  $k+I = a+I$, which means $a-k\in I$; since $I\subseteq J$, we conclude that $a-(a-k) = k\in J$. Therefore, $k=jx$ for some $x\in R$, so $k+I = jx+I = (j+I)(x+I)\in (j+I)(R/I)$. Thus, $K\subseteq (j+I)(R/I)$. And since $j+I\in K$ and $K$ is an ideal, then $(j+I)(R/I)\subseteq K$, giving equality.
Result 2: $F$ is a field iff $F[t]$ is a PID.
Proof : Exercise.
Now can you use these results to complete (a).
