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In the theory of Cohen-Macaulay rings, a basic theorem is that if $(R,\mathfrak{m})$ is a Cohen-Macaulay local noetherian ring, and $x_1,\dots,x_r\in\mathfrak{m}$ is a sequence satisfying $\dim R/(x_1,\dots,x_r)R = \dim R - r$, then $x_1,\dots,x_r$ is a regular sequence.

This theorem fails without the local assumption. For example, let $R = k[x,y,z]$ for $k$ a field (clearly a Cohen-Macaulay ring), and consider the sequence $xy,xz,1-x$. (This is a standard example of a sequence that is not regular in this order but it is regular in a different order.) Then the ideal $(xy,xz,1-x) = (y,z,1-x)$ is maximal and we have $R/(y,z,1-x)$ is dimension zero, but the sequence isn't regular since $xz$ is not regular in $R/(xy)$.

But there is often an analogy between the local case and the graded case, so I'm wondering if there is a graded analogy. In particular, I want to know:

If $R$ is a Cohen-Macaulay, nonnegatively graded ring, and $x_1,\dots,x_r$ are homogeneous elements of positive degree such that $\dim R/(x_1,\dots,x_r)R = \dim R - r$, is $x_1,\dots,x_r$ a regular sequence?

Note that I am not assuming that the degree zero component of $R$ is a field. (In the case of interest to me, $R_0=\mathbb{Z}$.) But if that assumption would change the status from false to true, then I'd like to know that.

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    For the case over a field, Mel Hochster in [these notes](http://www.math.lsa.umich.edu/~hochster/615W07/L01.29.pdf) shows on p. 5 that a sequence of your form can be completed to a system of parameters. The theorem on p. 6 shows that this system of parameters is a regular sequence, hence the original sequence you had is a system of parameters. I haven't looked at the proofs too carefully, but I don't recall where the assumption that $R_0$ is a field is used.2017-02-19

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