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Let $R$ be a ring. Recall that $R$ is said to satisfy the strong rank condition if, whenever $R^m \to R^n$ is a monomorphism of right $R$-modules (with $m,n \in \mathbb N$), we have $m \leq n$.

It is known that every nonzero commutative ring satisfies the strong rank condition (see Lam's Lectures on Modules and Rings, Corollary 1.38).

Question: Does every nonzero graded-commutative ring satisfy the strong rank condition?

I'd be happy with a graded version of the strong rank condition, saying that if $\oplus_i R^{m_i}[i] \to \oplus_i R^{n_i}[i]$ is a graded monomorphism, we have $\sum m_i \leq \sum n_i$.

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  • $\begingroup$ Isn't this trivial? If $R^m \to R^n$ is a monomorphism of graded $R$-modules, in degree $0$ it yields a monomorphism of $R_0$-modules $R_0^m \to R_0^n$. Since $R_0$ is a non-trivial commutative ring, we get $m \leq n$. $\endgroup$
    – Martin Brandenburg
    Commented Nov 29, 2023 at 15:22
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    $\begingroup$ @MartinBrandenburg I agree that this argument works if $R^m \to R^n$ is a graded monomorphism. But if we restrict to graded monomorphisms, then I want to allow a bit more flexibility in what is meant by "free $R$-module" -- a "free graded $R$-module" for me means a direct sum of shifts of copies of $R$, so $\oplus_i R^{m_i}[i]$. And the trivial argument doesn't cut it in this case. $\endgroup$
    – Tim Campion
    Commented Nov 29, 2023 at 15:32
  • $\begingroup$ What happens if you just mod out the nil radical and then refer to the commutative case? $\endgroup$
    – Dave Benson
    Commented Dec 9, 2023 at 10:39
  • $\begingroup$ @DaveBenson Does modding out the nilradical preserve monomorphisms? $\endgroup$
    – Tim Campion
    Commented Dec 9, 2023 at 16:23

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If you look at the proof of the proposition, you can notice that proof goes through for a graded-commutative ring for free.

Let's use the following characterisation of SRC: every system of $m$ homogeneous linear equations in $m+k$ variables has a solution over $R$. Let $a_{ij}$ be coefficients of such linear system, with decomposition into even and odd parts $a_{ij} = e_{ij} + o_{ij}$.

Consider a subring $S_0 \subset R$, generated by $e_{ij}$ over integers. It's noetherian by Hilbert basis theorem. We can extend it to a ring $S_0[o_{ij} o_{kl}] =: S_1 \subset R$, which is still noetherian for same reason. But then $S_1[o_{ij}] =: S_2 \subset R$ is a finitely generated (by those $o$'s) module over $S_1$, therefore noetherian as an $S_1$-module, and therefore (both right and left) noetherian as a ring: every ideal in $S_2$, i. e. $S_2$-submodule, is definitely an $S_1$-submodule, so ACC on latter ones implies ACC on former.

But in (right) noetherian rings (right) strong rank condition holds, as remarked in Lam's LMR just before in 1.35; therefore, initial linear system had a solution already in $S_2$.

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  • $\begingroup$ I want to put you in the acknowledgements to a paper. I’ll put it as “Denis T” unless you let me know your full name so I can acknowledge you properly — either here or via email. $\endgroup$
    – Tim Campion
    Commented Apr 22 at 13:30

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