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!
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).