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.
Best Answer
Here's three equivalent definitions of "Noetherian ring" (equivalent in ZFC, at any rate):
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.