If $R$ is a Noetherian ring, then $R^n$ is Noetherian

Question

I'm working with a Noetherian ring $R$. As an $R$-module, $R^n = R \oplus ... \oplus R$. I want to show that $R^n$ is Noetherian in the sense that it obeys the ascending chain condition for its submodules (or equivalently that all submodules are finitely generated). I do not know how to do this. Because the ideals in $R^n$ don't necessarily need to "split" over the different coordinate components in the direct sum I don't know how to use the hypothesis that $R$ is Noetherian. I think this indicates perhaps we should try a contradictive proof but I don't know how to proceed with this.

Answer 1

Being "Noetherian" has two meanings here. (i) $R^n$ is Noetherian as a ring; (ii) $R^n$ is Noetherian as an $R$ module. If you want to show (i), you can use that any ideal $I$ of $R^n$ is split, namely there are ideals $I_1,\dots,I_n\subset R$ such that $I=I_1\oplus \cdots I_n$. You can see this by the idempotent decomposition $$1_{R^n}=(1,0,\dots,0)+\cdots+(0,0,\dots,1)$$ as you may already know. Then, since $R$ is Noetherian, each $I_i$ is finitely generated and so is $I$. To show (ii), we can use the fact that finitely generated $R$ modules are Noetherian if $R$ is a Noetherian ring. This is elementary, see Atiyah-MacDonald Cor. 6.4., for example. Your remark applies here, $R$-submodules of $R^n$ are not necessarily split.

Edit: Let us see (ii) by induction. The case $n=1$ is already assumed. For $n+1$, we denote by $f\colon R^{n+1}\rightarrow R^n$ the projection onto first $n$ factors. Take a chain of submodules $$S_1\subset S_2\subset \cdots$$ of $R^{n+1}$, then the chain $$f(S_1)\subset f(S_2)\subset \cdots$$ in $R^n$ terminates by induction at, say, $f(S_p)$ and similarly the chain $$S_{p}\cap \mathrm{ker}f\subset S_{p+1}\cap \mathrm{ker}f\subset \cdots$$ in $R=\mathrm{ker}f$ terminates at, say, $S_{p+q}\cap \mathrm{ker}f$. After all, each term of the portion $$S_{p+q}\subset S_{p+q+1}\subset \cdots$$ of the original chain has constant image and kernel with respect to $f$. We denote this common image by $I\subset R^{n}$ and kernel $K\subset R$. For $r\geq p+q$, take $a\in S_{r}$. Then, since $f(a)\in I$, we find $b\in S_{p+q}$ with $f(a)=f(b)$. Then $c:=a-b\in S_{r}\cap \mathrm{ker}f=K$, so we see $c\in S_{p+q}$ and $a=b+c\in S_{p+q}$. Therefore, we see $S_r=S_{p+q}$ and the original chain terminates at $S_{p+q}$.

Answer 2

Your observation that the ideals of $R^n$ do not split componentwise is not correct. I asked about it in this question.

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Here's a way to show it using the definition of a Noetherian ring.

A commutative ring is Noetherian if and only if every ascending chain of ideals eventually stabilizes.

This is a proof for the ring part only, not the module part.

Theorem 51: If $R$ is a Noetherian ring, then $R^n$ is a Noetherian ring.

First, let's note that multiplication in $R^n$ is componentwise, so $(a, b) \times_{\!R^2} (c, d) = (a \times_{\!R} c, b \times_{\!R} d)$.

Suppose by way of contrapositive that $R^n$ is not Noetherian, this means that it contains an infinite ascending chain of ideals:

$$ I(1) \subsetneq I(2) \subsetneq I(3) \subsetneq \cdots $$

Note that $I(i)$ it the $i$th ideal in the sequence; I'm doing this so I can select components using subscripts.

The above is equal to:

$$ \bigg(I_1(1), I_2(1) \cdots I_n(1)\bigg) \subsetneq \bigg(I_2(1), I_2(2), \cdots I_n(2)\bigg) \subsetneq \cdots $$

If every component $I_a(1) \subsetneq I_a(2) \subsetneq I_a(3) \subsetneq \cdots$ stabilized, then the whole sequence would stabilize.

Thus at least one of the components doesn't stabilize.

End of proof of theorem 51.

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There's an important subtlety in the $R$-module case.

Consider the submodule $((1,2))$ in the $\mathbb{Z}$-module $\mathbb{Z}^2$.

Each component of $((1,2))$ is a $\mathbb{Z}$-ideal $\mathbb{Z}$ and $\mathbb{2Z}$, however,

$((1, 2))$ is NOT equal to $\mathbb{Z} \times 2\mathbb{Z}$.

I don't have a proof at the moment that a chain of submodules with constant "component ideals" that nevertheless fails to stabilize is impossible.

Answer 3

If $R$ is noetherian, the polynomial ring $R[x_1,\dotsc,x_n]$ is also noetherian. Consider the ideals $$I_k = (x_1, x_2, \dotsc, x_k-1, \dotsc, x_n).$$ The quotients $R[x_1,\dotsc, x_n]/I_k$ are isomorphic to $R$, because the variables are identified with $0$ or $1$. Also note that the $I_k$ are pairwise relatively prime, i.e. $$I_j + I_k = R[x_1, \dotsc, x_n] \quad \text{for} \quad j \neq k.$$ By the Chinese remainder theorem you get an isomorphism $$R[x_1,\dotsc,x_n]/I_1 \cap \dotsm \cap I_n \cong R \times \dotsm \times R,$$ and the left hand side is noetherian as a quotient of a noetherian ring.