Noetherian Ring under a Homomorphism / homomorphic function / map 
Assume R is noetherian and $f:R \to S$ is a ring homomorphism. Is $f(R)$ noetherian? 



*

*Reading Dylan Moreland commentary: "If $\mathfrak{b}$ is an ideal of $f(R)$, then $f^{-1}(\mathfrak{b})$ is an ideal of R and $f(f^{-1}(\mathfrak{b}))=\mathfrak{b}$ ." Then he writes: "if you have an  increasing chain of ideals in $f(R)$" . So because there are only finitely many ideals in $f^{-1}(\mathfrak{b})$, which is noetherian, there can also only be finitely many ideals in $f(R)$. So we can say from this that $f(R)$ is noetherian? 

*Reading Arturo Magidins commentary: I don't know the lattice isomorphism theorem. I can't find it on google either. Do you maybe have a link for me so I can read it? For the second definition of a noetherian ring: every ideal is finitely generated because: $f(a_{0},a_{1},....,a_{n}) = f(a_{0}),f(a_{1}),f(a_{2}),...,f(a_{n})= f(R)$ follows directly because f is a ring homomorphism. Is this a correct proof already? 
Thank you. 
 A: This is the same as asking whether a quotient of a Noetherian ring is also Noetherian. How might you show that the answer is yes? If $\mathfrak b$ is an ideal of $f(R)$, then $f^{-1}(\mathfrak b)$ is an ideal of $R$ and $f(f^{-1}(\mathfrak b)) = \mathfrak b$. If you have an increasing chain of ideals in $f(R)$,
\[
\mathfrak b_1 \subset \mathfrak b_2 \subset \cdots
\]
can you use these facts to show that the chain must stabilize, i.e. that there exists an $n$ such that $\mathfrak b_n = \mathfrak b_{n+1} = \cdots$?
I think it would be instructive to also prove this from the viewpoint that "Noetherian" means "every ideal is finitely generated".
A: Here's three equivalent definitions of "Noetherian ring" (equivalent in ZFC, at any rate):


*

*Every ascending chain of ideals stabilizes.

*Every ideal is finitely generated.

*Every nonempty collection of ideals has a maximal element.


For the first and third definitions, the proof follows easily from the Lattice Isomorphism Theorem, which tells you that there is an order-preserving bijection between ideals of $f(R)$ and ideals of $R$ that contains $\mathrm{ker}(f)$. 
For the second definition, it follows easily from the general fact that finitely generated subobjects remain finitely generated in the image: if $I=(a_0,\ldots,a_n)$, show that $f(I)=(f(a_0),\ldots,f(a_n))$ in $f(R)$. So again, using the correspondence above, you can use information about ideals in $R$ to get information about ideals in $f(R)$. 

Comments on edited version of the question:


*

*"So because there are only finitely many ideals in $f^{−1}(\mathfrak{b})$, which is noetherian" is nonsensical. $f^{-1}(\mathfrak{b})$ is an ideal of $R$, not a noetherian anything. (Okay, technically, it is a noetherian module over $R$, since $R$ is noetherian and $f^{-1}(\mathfrak{b})$ is finitely generated; but the point is that there is an obvious categorical confusion here,  between ideals, rings, preimages, etc).

*"...there can also only be finitely many ideals in $f(R)$". That's false. $\mathbb{Z}$ is noetherian, and its images may have infinitely many ideals. Specifically, if $f$ is an embedding, then $f(\mathbb{Z})$ would have infinitely many ideals. 

*The Lattice Isomorphism Theorem states what I said it states: there is an order-preserving bijection between the ideals of $f(R)$ and the ideals of $R$ that contain $\mathrm{ker}(f)$.

*The equations $f(a_{0},a_{1},....,a_{n}) = f(a_{0}),f(a_{1}),f(a_{2}),...,f(a_{n})= f(R)$ are nonsensical as written.
My observation is that you are just not very careful with notation, definitions, or notions; that is likely to be at least half your problem with these proofs. You are not clear on what you are assuming, or on what you want to conclude.


*

*We want to show that any ascending chain of ideals in $f(R)$ stabilizes. Let
$$\mathfrak{b}_1\subseteq \mathfrak{b}_2\subseteq\cdots$$
be an ascending chain of ideals in $f(R)$. We want to show that there exists $n$ such that for all $k\geq n$, $\mathfrak{b}_n=\mathfrak{b}_k$.
Consider the chain of ideals in $R$ obtained via de Lattice Isomorphism Theorem: 
$$f^{-1}(\mathfrak{b}_1)\subseteq f^{-1}(\mathfrak{b}_2)\subseteq\cdots.$$
Since $R$ is noetherian, there exists $n$ such that for all $k\geq n$, $f^{-1}(\mathfrak{b}_n) = f^{-1}(\mathfrak{b}_k)$. Applying $f$ to both sides of the equation, we get that for all $k\geq n$, 
$$\mathfrak{b}_n = f\left(f^{-1}(\mathfrak{b}_n)\right) = f\left(f^{-1}(\mathfrak{b}_k)\right) = \mathfrak{b}_k.$$
This is exactly what we wanted to prove, so the chain stabilizes. Thus, $f(R)$ is noetherian.

*Let $\mathfrak{b}$ be an ideal of $f(R)$. We want to show that it is finitely generated. Consider $f^{-1}(\mathfrak{b})$, which is an ideal of $R$ by the Lattice Isomorphism Theorem. Since $R$ is noetherian, every ideal of $R$ is finitely generated, so there exist $a_1,\ldots,a_n\in R$ such that $f^{-1}(\mathfrak{b})=(a_1,\ldots,a_n)$. I claim that $\mathfrak{b} = (f(a_1),\ldots,f(a_n))$, where the ideal on the left is the ideal in $f(R)$. Indeed, since $a_1,\ldots,a_n\in f^{-1}(\mathfrak{b})$, then $f(a_i)\in\mathfrak{b}$ for $i=1,\ldots,n$; so $(f(a_1),\ldots,f(a_n))\subseteq \mathfrak{b}$. Now let $b\in\mathfrak{b}$. Then there exists $a\in A$ such that $f(a)=b$ (since $f$ is into $f(R)$); thus, $a\in f^{-1}(\mathfrak{b})$, so $a\in (a_1,\ldots,a_n)$. Therefore, there exist $\alpha_1,\ldots,\alpha_n)\in R$ such that 
$$a=\alpha_1a_1+\cdots+\alpha_na_n.$$
Therefore,
$$b = f(a) = f(\alpha_1a_1+\cdots+\alpha_na_n) = f(\alpha_1)f(a_1)+\cdots + f(\alpha_n)f(a_n) \in (f(a_1),\ldots,f(a_n)).$$
This proves that $\mathfrak{b}\subseteq (f(a_1),\ldots,f(a_n))$. Together with the other inclusion, we conclude that $\mathfrak{b}=(f(a_1),\ldots,f(a_n))$, hence $\mathfrak{b}$ is finitely generated.
Thus, every ideal of $f(R)$ is finitely generated, so $f(R)$ is noetherian.

*Let $M$ be a nonempty collection of ideals of $f(R)$. We want to show that $M$ has maximal elements. Let $N = \{f^{-1}(\mathfrak{b})\mid \mathfrak{b}\in M\}$. Then $N$ is a nonempty collection of ideals of $R$. Since $R$ is noetherian, $N$ has a maximal element, $f^{-1}(\mathfrak{b_0})$. I claim that $\mathfrak{b_0}$ is a maximal element of $M$. Indeed, let $\mathfrak{b}\in M$ be such that $\mathfrak{b}_0\subseteq \mathfrak{b}$. Then $f^{-1}(\mathfrak{b}_0)\subseteq f^{-1}(\mathfrak{b})$. By the maximality of $f^{-1}(\mathfrak{b}_0)$ in $N$, since $f^{-1}(\mathfrak{b})\in N$ we conclude that $f^{-1}(\mathfrak{b}_0) = f^{-1}(\mathfrak{b})$. Applying $f$ to both we get 
$$\mathfrak{b}_0 = f(f^{-1}(\mathfrak{b})) = f(f^{-1}(\mathfrak{b})) = \mathfrak{b}.$$
That is: we have shown that if $\mathfrak{b}\in M$ and $\mathfrak{b}_0\subseteq \mathfrak{b}$, then $\mathfrak{b}_0 = \mathfrak{b}$. This proves that $\mathfrak{b}_0$ is maixmal in $M$.
Thus, any nonempty collection of ideals in $f(R)$ has a maximal element. Therefore, $f(R)$ is noetherian.
