MHB Statements about subrings and quotient

[mathmari]

Hey!

I want to check if the following statement are true.

Let $R$ be a ring, $S$ a subring and $I$ an ideal.

1. If $R$ is Noetherian then $S$ is also.
2. If $R$ is Noetherian then $R/I$ is also.
3. If $R$ is Artinian then $S$ is also.
4. If $R$ is Artinian then $R/I$ is also.
5. If $R$ is P.I.D. then $S$ is also.
6. If $R$ is P.I.D. then $R/I$ is also.
7. If $R$ is U.F.D. then $S$ is also.
8. If $R$ is U.F.D. then $R/I$ is also.
9. If $R$ is an euclidean domain then $S$ is also.

I have done the following:

1. $R$ is Noetherian iff each increasing sequence of ideal $I_1\subseteq I_2 \subseteq I_3 \subseteq \dots \subseteq I_k\subseteq \dots $ stops, i.e., $\exists k$ such that $I_k=I_{k+1}$, right?
   Then since $S$ is a subring of $R$, not all $I_i$ are contained in $S$. Therefore, the above condition isn't necessarily satisfied. So, $S$ is not necessarily Noetherian.
   is this correct? (Wondering)
2. What can we say in that case? Does the increasing sequence stop? (Wondering)
3. $R$ is Artinian iff each decreasing sequence of ideal $I_1\supseteq I_2 \supseteq I_3 \supseteq \dots \supseteq I_k\supseteq \dots $ stops, i.e., $\exists k$ such that $I_k=I_{k+1}$, right?
   Then since $S$ is a subring of $R$, not all $I_i$ are contained in $S$. Therefore, the above condition isn't necessarily satisfied. So, $S$ is not necessarily Artinian.
   is this correct? (Wondering)
4. What can we say in that case? Does the decreasing sequence stop? (Wondering)
5. If $R$ is P.I.D. then the ideals are prime, therefore $S$ contain also only prime ideals. So, $S$ is also P.I.D., right? (Wondering)
6. What can we say in this case? (Wondering)
7. If $R$ is U.F.D. then $\forall r\in R\setminus \{0\}$, $r\notin U(R)$: $r=a_1 \cdots a_k$ with $a_i$ irreducible, and if $r=a_1\cdots a_k=b_1\cdots b_t$ with $a_i, b_i$ irreducible then $k=t$ and $a_i=b_iu_i$ with $u_i\in U(R), \forall i=1, \dots , k$.
   Does the same hold also for $S$ ? (Wondering)
8. And also in this case? (Wondering)
9. How can we check that? (Wondering)

 

[Deveno]

Your "proofs" for 1 & 3 are not correct. It is not the case that an ideal of a subring $S$ is an ideal of its containing ring $R$.

For example, $\Bbb Z$ is a subring of $\Bbb Q$, but $2\Bbb Z$ is an ideal of $\Bbb Q$, but not of $\Bbb Q$.

Also, just because some of the ideals of $R$ lie outside of $S$, does not mean that the ideals of $S$ fail to satisfy the Noetherian ascending chain condition.

What you want is a *counter-example*. Here is something to get you started on #1:

Any field is Noetherian, and any integral domain can be extended to a field. Can you find a non-Noetherian integral domain?

For 2, consider the correspondence theorem for rings.

For number 3, consider $S = \Bbb Z$. This ring is non-Artinian, can you find an Artinian ring it is a sub-ring of?

We'll discuss the other questions later.

 

[mathmari]

Any field is Noetherian, and any integral domain can be extended to a field. Can you find a non-Noetherian integral domain?

Let $K$ be a field, then $K[x_1, x_2, \dots ]$ is not Noetherian, since the chain $(x_1) \subseteq (x_1, x_2) \subseteq \dots $ never stops, or not? (Wondering)

 

[mathmari]

What you want is a *counter-example*. Here is something to get you started on #1:

Any field is Noetherian, and any integral domain can be extended to a field. Can you find a non-Noetherian integral domain?

Do we maybe use the Hilbert basis theorem?

For 2, consider the correspondence theorem for rings.

How exactly do we use this theorem? I got stuck righ now? (Wondering)

For number 3, consider $S = \Bbb Z$. This ring is non-Artinian, can you find an Artinian ring it is a sub-ring of?

The only ideals of $\mathbb{R}$ are $0$ and $\mathbb{R}$.
So, the sequence $\mathbb{R}\supset 0$ is finite, therefore $\mathbb{R}$ is Artinian, right? (Wondering)