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 $R$?

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

2 Answers 2


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

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


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


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

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

$r \in R \tag{19}$


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

and now

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


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

(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

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


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


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

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.

  • $\begingroup$ 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 $(4)$? $\endgroup$
    – user838843
    Dec 8, 2020 at 11:56
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    $\begingroup$ 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 $\varphi(2 \mathbb Z).$ $\endgroup$ Dec 8, 2020 at 15:49
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    $\begingroup$ 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 $S.$ $\endgroup$ Dec 8, 2020 at 15:51
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    $\begingroup$ I think in eq.22 should be J not I. $\endgroup$
    – user838843
    Dec 8, 2020 at 20:58
  • $\begingroup$ @Confusion: agreed. Will fix. Thanks. $\endgroup$ 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.

  • $\begingroup$ How do you prove $(3)$? $\endgroup$
    – user838843
    Dec 8, 2020 at 11:26
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    $\begingroup$ 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 $\varphi^{-1}(J).$ $\endgroup$ Dec 8, 2020 at 15:43
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    $\begingroup$ For the same proof (so that you can go upvote it), check here. $\endgroup$ Dec 8, 2020 at 15:45
  • $\begingroup$ I think I misstated my question but thank you! $\endgroup$
    – user838843
    Dec 8, 2020 at 16:58

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