# 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!

• F[t] is a P.I.D and thus inverse image of ideal through quotient map would be principal ideal. – Akash Yadav Mar 26 at 6:45

(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?

• Thanks for your comment. I am trying to follow your comment !! – fivestar Mar 26 at 7:27
• Basically for (2) the question becomes: what are the prime ideals containing $(t^2)$ in $F[t]$. – ε-δ Mar 26 at 7:30
• Then, the only possible case is that $P(t)=t$ because $F[t]/(t)$ is filed, it means that $(t)$ is also prime ideal of $F[t]$. And if $P(t)=t^2$, it is not prime because the factor ring is not integer domain. thus $t + (t^2)$ is sole prime ideal of $A$. Right? – fivestar Mar 26 at 8:11
• Sounds good! But you should notice that $(0)$ is a prime ideal in $F[T]$. Why not in the quotient? – ε-δ Mar 26 at 9:20
• That's right, I had to consider it! Anyway, it was really helpful with your posting. Thanks! – fivestar Mar 26 at 11:05

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).

• In the way for you to find the ideal $J$, Do you just inverse canonical (natural homomorphism from $R$ to its factor ring) map which sends K to the ideal of $J$ you said. – fivestar Mar 26 at 7:22
• Yes, J is just the inverse image of ideal K via the quotient homomorphism. – Akash Yadav Mar 26 at 7:25
• Thanks for your help. I fully understood what you explained! – fivestar Mar 26 at 7:26