Let $A$ be a noetherian, graded ring and $M$ be a projective, graded $A$-module. Denote by $M_{\ge d}:=\oplus_{\ell \ge d} M_\ell$ the sub-module of $M$. Is $M_{\ge d}$ again a projective $A$-module?
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1$\begingroup$ $A=k$, $M=k[x,y]$. $\endgroup$– Steven LandsburgCommented Nov 17, 2019 at 22:47
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$\begingroup$ A module is projective iff it is a direct summand of a free module. Therefore the answer to your question is yes. $\endgroup$– ChrisCommented Nov 17, 2019 at 23:05
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1$\begingroup$ @Chris The direct sum $M = M_{\geq d} \oplus M_{< d}$ is not a direct sum of $A$-modules. $\endgroup$– WhatsUpCommented Nov 17, 2019 at 23:44
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$\begingroup$ @WhatsUp You're right, for some reason I missed the part where also $A$ is assumed to be graded. If the grading is the trivial one (i.e. $A_0=A$) then the result is valid, but it's quite uninteresting of course. $\endgroup$– ChrisCommented Nov 17, 2019 at 23:47
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A counter example would be $A = k[x, y]$, graded by the total degree, and $M = A$, $d = 1$.
The module $M_1$ is then the ideal of $A$ generated by $x$ and $y$. It is not projective.
Because:
Consider the surjection $A^2 \rightarrow M_1$, sending $(u, v)$ to $xu + yv$. It's an exercise to show that this map doesn't have a section.
