Is this proof of a ring being Noetherian correct or we should prove that it is left Noetherian and then right Noetherian and why?

Here is the question I want to answer:

Let $$\varphi : R \rightarrow S$$ be a surjective homomorphism of commutative rings. Show that, if $$R$$ is Noetherian, then $$S$$ is Noetherian.

Here is a trial to the solution:

Let $$\varphi : R \rightarrow S$$ be a surjective homomorphism of commutative rings and assume that $$R$$ is Noetherian. We want to show that $$S$$ is Noetherian.

Now, since we know the following definition of a Noetherian ring : a commutative ring $$S$$ is Noetherian iff it satisfies the $$ACC$$ condition. And since we know that a commutative ring $$S$$ has the ascending chain condition $$(ACC),$$ if, for every ascending chain of ideals $$I_1 \subset I_2 \subset I_3 \subset \dots$$ in $$S,$$ there is an $$N > 0,$$ so that $$I_n = I_{n+1}$$ for $$n \geq N.$$

So, let $$I_1 \subset I_2 \subset I_3 \subset \dots$$ be an ascending chain of ideals in $$S,$$ Now since we have a surjective ring homomorphism, then the preimage of the ideal $$I_k$$ in $$S$$ (i.e. $$f^{-1}(I_k)$$) is an ideal of $$R$$(to be proved at the end). And similarly for all other ideals. So we get the following ascending (we will prove later why it is ascending) chain of ideals in $$R,$$ $$f^{-1}(I_1) \subset f^{-1}(I_2) \subset f^{-1}(I_3) \subset \dots \subset f^{-1}(I_k) \subset \dots$$

Now, since by assumption, $$R$$ is Noetherian, so its ascending chain of ideals terminate. That is $$\exists N \in \mathbb Z$$ such that $$f^{-1}(I_n) = f^{-1}(I_{n+1})$$ for all $$n \geq N.$$

Now since $$f$$ is surjective, we have $$f (f^{-1}(I_k)) = I_k, \quad \forall k > 0.$$

Therefore, it follows that we have $$I_n = I_{n+1}, \quad \forall n \geq N.$$ So, each ascending chain of ideals of $$S$$ terminates, as we have taken an arbitrary ascending chain of $$S$$ and so $$S$$ is Noetherian as required.

My questions are:

1-Is this proof correct? or we should prove that $$S$$ is left Noetherian and then right Noetherian? If so, why?

2- why we are sure that we get an ascending chain of ideals in $$R$$?

• If you can convert your image to $\LaTeX$, it would be readable. Commented Dec 7, 2020 at 22:20
• $S$ is trivially commutative so there's no reason to talk about sides if you don't have to. Commented Dec 7, 2020 at 22:30
• @RobertLewis ok I will sorry about that .... just give me sometime to typeset my answer.
– user838843
Commented Dec 8, 2020 at 0:05
• Thanks or converting your image to $\LaTeX$! Cheers!!! Commented Dec 8, 2020 at 3:22
• @RobertLewis its my pleasure .... sorry I was just in a hurry of posting to know if the proof is correct or no Cheers!!!
– user838843
Commented Dec 8, 2020 at 3:26

As has been pointed out in the comments to the question itself, and is in fact easily seen, the hypothesis that $$R$$ and $$S$$ be commutative implies that every ideal in either ring is two-sided, so we needn't separately consider left or right ideals.

It strikes me that the essential fact here is the

Assertion that for surjective homomorphisms

$$\varphi:R \to S \tag 1$$

the image $$\varphi(I)$$ of an ideal

$$I \subset R \tag 2$$

is in fact an ideal in $$S$$.

It should be noted that if we lift the hypothesis that $$\varphi$$ is surjective, we still have that

$$I = \varphi^{-1}(J) \subset R \tag 3$$

is an ideal for any ideal

$$J \subset S; \tag 4$$

for if

$$i_1, i_2 \in I, \tag 5$$

we have

$$\varphi(i_1), \varphi(i_2) \in J, \tag 7$$

whence

$$\varphi(i_1 - i_2) = \varphi(i_1) - \varphi(i_2) \in J; \tag 8$$

thus

$$i_1 - i_2 \in I; \tag 9$$

also, for

$$r \in R, \tag{10}$$

$$\varphi(ri_1) = \varphi(r) \varphi(i_1) \in J, \tag{11}$$

since $$J$$ is an ideal; from this,

$$r i_1 \in I; \tag{12}$$

(9) and (12) together show that $$I$$ is an ideal in $$R$$.

The present Assertion is thus in a sense a logical complement of the above result, for it allows the affirmation that $$I$$ is a ideal if and only if $$J$$ is in the event of surjective $$\varphi$$.

Now once again assuming that $$\varphi$$ is surjective, we let $$I$$ be an ideal in $$R$$ and let

$$J = \varphi(I) \tag{13}$$

as a set. Then for

$$j_1, j_2 \in J \tag{14}$$

we may find

$$i_1, i_2 \in I \tag{15}$$

such that

$$\varphi(i_1) = j_1, \; \varphi(i_2) = j_2; \tag{16}$$

$$j_1 - j_2 = \varphi(i_1) - \varphi(i_2) = \varphi(i_1 - i_2) \in J, \tag{17}$$

and if

$$s \in S, \tag{18}$$

using the surjectivity of $$\varphi$$ we have some

$$r \in R \tag{19}$$

with

$$\varphi(r) = s; \tag{20}$$

and now

$$sj_1 = \varphi(r) \varphi(i_1) = \varphi(ri_1) \in J, \tag{21}$$

since

$$ri_1 \in I; \tag{22}$$

(17) and (21) in concert show that $$J$$ is in fact an ideal in $$S$$, proving the Assertion.

Thus we see that, in the event $$\varphi$$ is surjective, $$I$$ is an ideal in $$R$$ if and only if $$J = \varphi(I)$$ is an ideal in $$S$$.

It is a short step from this to the requested result, for if $$R$$ is Noetherian and $$J_k$$ is an ascending sequence of ideals in $$S$$, that is

$$J_i \subset J_{i + 1}, \tag{23}$$

we can form the corresponding sequence of ideals

$$I_i = \varphi^{-1}(J_i) \subset R; \tag{24}$$

we know the $$I_i$$ are ideals in $$R$$ by virtue of what has just been proven in the above. Now since $$R$$ is Noetherian, the sequence $$I_i$$ stabilizes at some point, whence

$$\exists n \in \Bbb N, \; I_j = I_n \; \text{for} \; j \ge n; \tag{25}$$

from elementary set theory it follows that

$$J_j = \varphi(\varphi^{-1}(J_j)) = \varphi(I_j) = \varphi(I_n) = J_n \; \text{for} \; j \ge n, \tag{26}$$

which shows the sequence of ideals $$J_j$$ stabilizes in $$S$$ and thus that $$S$$, like $$R$$, is a Noetherian ring.

Note Added in Edit, Friday 11 December 2020 10:22 PM PST: A few final remarks meant to address our OP Confusion's two closing questions. Yes, the proof given in the text of the question itself appears to be quite correct; indeed, it is for all intents and purposes the same as mine, and I'm pretty damn sure that is OK, so . . . I'm pretty damn sure Confusion's proof is OK as well. There is no need to consider right and left Noetherian rings separately: indeed, since the rings in question are commutative, there is not distinction 'twixt right, left, and two-sided ideals. As for Confusion's second question, that is, how do we know the sequence of ideals (24) (in the notation of my answer) is in fact ascending, note that

$$r \in I_i = \varphi^{-1}(J_i) \tag{27}$$

implies

$$\varphi(r) \in J_i \subset J_{i + 1}, \tag{28}$$

whence

$$r \in \varphi^{-1}(J_{i + 1}) = I_{i + 1}, \tag{29}$$

which shows that

$$I_i \subset I_{i + 1}, \tag{30}$$

that is, the ideals $$I_i$$ comprise an ascending chain in $$R$$. End of Note.

• Are you saying that the surjectivity of $\varphi$ is required in the proof of the assertion you mentioned in $(1)$ and $(2)$ it is not required in the prove of the assertion you mentioned in $(3)$ and $(4)$?
– user838843
Commented Dec 8, 2020 at 11:56
• It is necessary that $\varphi : R \to S$ be surjective in order to conclude that $\varphi(I)$ is an ideal of $S$ for all ideals $I$ of $R.$ Observe that $\varphi : \mathbb Z \to \mathbb Z[x]$ defined by $\varphi(n) = n$ is an injective ring homomorphism; however, for the ideal $2 \mathbb Z$ of $\mathbb Z,$ we have that $\varphi(2 \mathbb Z)$ is not an ideal of $\mathbb Z[x]$ since it is not closed under multiplication by ring elements. Particularly, the polynomial $2x$ is not in $\varphi(2 \mathbb Z).$ Commented Dec 8, 2020 at 15:49
• On the other hand, it is always true that the pre-image of an ideal under a ring homomorphism is an ideal, i.e., if $\varphi : R \to S$ is a ring homomorphism, then $\varphi^{-1}(I)$ is an ideal of $R$ for all ideals $I$ of $S.$ Commented Dec 8, 2020 at 15:51
• I think in eq.22 should be J not I.
– user838843
Commented Dec 8, 2020 at 20:58
• @Confusion: agreed. Will fix. Thanks. Commented Dec 8, 2020 at 21:01

Considering that $$R$$ and $$S$$ are commutative, there is no need to talk about left- or right-Noetherian because all ideals of $$R$$ and $$S$$ are two-sided. Consequently, $$S$$ is left-Noetherian if and only if it is right-Noetherian if and only if it is Noetherian.

For the rest of the proof, we need a few key ingredients. Consider a homomorphism $$\varphi : R \to S$$ of commutative rings.

1.) For any ideal $$I$$ of $$S,$$ we have that $$\varphi^{-1}(I)$$ is an ideal of $$R$$ (called the contraction of $$I$$ in $$R$$).

2.) If $$\varphi$$ is surjective, then for any ideal $$J$$ of $$R,$$ we have that $$\varphi(J)$$ is an ideal of $$S$$ (called the extension of $$J$$ in $$S$$).

3.) If $$I \subseteq J$$ are ideal of $$S,$$ then $$\varphi^{-1}(I) \subseteq \varphi^{-1}(J).$$

4.) If $$I \subseteq J$$ are ideal of $$R,$$ then $$\varphi(I) \subseteq \varphi(J).$$

5.) If $$\varphi$$ is surjective, then for any ideal $$I$$ of $$S,$$ we have that $$\varphi(\varphi^{-1}(I)) = I.$$

Putting these all together (as you have done) gives the proof: for any ascending chain $$I_1 \subseteq I_2 \subseteq \cdots$$ of ideals of $$S,$$ we have that $$\varphi^{-1}(I_1) \subseteq \varphi^{-1}(I_2) \subseteq \cdots$$ is an ascending chain of ideals of $$R.$$ By hypothesis that $$R$$ is Noetherian, there exists an integer $$N \gg 0$$ such that $$\varphi^{-1}(I_n) = \varphi^{-1}(I_{n + 1})$$ for each integer $$n \geq N.$$ Considering that $$\varphi$$ is surjective, it follows that $$\varphi(\varphi^{-1}(I_k)) = I_k$$ for all integers $$k \geq 1.$$ By viewing the original chain $$I_1 \subseteq I_2 \subseteq \cdots$$ of ideals in $$S$$ as the ascending chain of ideals $$\varphi(\varphi^{-1}(I_k))$$ and using the fact that $$\varphi^{-1}(I_n) = \varphi^{-1}(I_{n + 1})$$ for all integers $$n \geq N,$$ we conclude that $$I_1 \subseteq I_2 \subseteq \cdots$$ stabilizes. QED.

By the way, if you know the equivalent condition that a commutative ring $$R$$ is Noetherian if and only if all of its (two-sided) ideals are finitely generated, then there is an alternate proof.

Proof. Consider an ideal $$I$$ of $$S.$$ By the first point above, we have that $$\varphi^{-1}(I)$$ is an ideal of $$R.$$ By hypothesis that $$R$$ is Noetherian, there exist elements $$i_1, \dots, i_n$$ of $$\varphi^{-1}(I)$$ such that every element of $$\varphi^{-1}(I)$$ can be written as $$r_1 i_1 + \cdots + r_n i_n$$ for some elements $$r_1, \dots, r_n$$ of $$R.$$ Considering that $$\varphi$$ is surjective, by the fifth point above, we have that $$I = \varphi(\varphi^{-1}(I)),$$ hence every element of $$I$$ can be written as $$\varphi(r_1 i_1 + \cdots + r_n i_n) = \varphi(r_1) \varphi(i_1) + \cdots + \varphi(r_n) \varphi(i_n)$$ for some elements $$\varphi(r_1), \dots, \varphi(r_n)$$ of $$S$$ and some elements $$\varphi(i_1), \dots, \varphi(i_n)$$ of $$I.$$ Consequently, the ideal $$I$$ of $$S$$ is finitely generated by the elements $$\varphi(i_1), \dots, \varphi(i_n)$$ of $$I.$$ QED.

• How do you prove $(3)$?
– user838843
Commented Dec 8, 2020 at 11:26
• Let $x$ be an element of $\varphi^{-1}(I).$ By definition, we have that $\varphi(x)$ is an element of $I$ and so an element of $J.$ But this implies that $x$ is in $\varphi^{-1}(J).$ Commented Dec 8, 2020 at 15:43
• For the same proof (so that you can go upvote it), check here. Commented Dec 8, 2020 at 15:45
• I think I misstated my question but thank you!
– user838843
Commented Dec 8, 2020 at 16:58