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This is of course true, for any semi-separated scheme (i.e. the diagonal is affine), or maybe you assume $X$ is separated if you like, and you can take any point (not necessarily closed). The reason that the sheaf is flat is that ${\rm Spec}(O_{X,x})\to X$ is affine and flat. In general if $f: X\to Y$ is flat and affine then $f_*O_X$ is $O_Y-$flat. This is obvious.

But I don't think it is true that $f: X\to Y$ is flat implies $f_*O_X$ is $O_Y-$flat. See link text

In the answer of Jason Starr, the map is flat, while the 0-th direct image is not flat at all.

This is of course true, for any semi-separated scheme (i.e. the diagonal is affine), or maybe you assume $X$ is separated if you like, and you can take any point (not necessarily closed). The reason that the sheaf is flat is that ${\rm Spec}(O_{X,x})\to X$ is affine and flat. In general if $f: X\to Y$ is flat and affine then $f_*O_X$ is $O_Y-$flat. This is obvious.

But I don't think it is true that $f: X\to Y$ is flat implies $f_*O_X$ is $O_Y-$flat. See link text

In the answer of Jason Starr, the map is flat, while the 0-th direct image is not flat at all.

This is of course true, for any semi-separated scheme (i.e. the diagonal is affine), or maybe you assume $X$ is separated if you like, and you can take any point (not necessarily closed). The reason that the sheaf is flat is that ${\rm Spec}(O_{X,x})\to X$ is affine and flat. In general if $f: X\to Y$ is flat and affine then $f_*O_X$ is $O_Y-$flat. This is obvious.

But I don't think it is true that $f: X\to Y$ is flat implies $f_*O_X$ is $O_Y-$flat. I would be happy if someone can back me up with a counter example here.

This is of course true, for any semi-separated scheme (i.e. the diagonal is affine), or maybe you assume $X$ is separated if you like, and you can take any point (not necessarily closed). The reason that the sheaf is flat is that ${\rm Spec}(O_{X,x})\to X$ is affine and flat. In general if $f: X\to Y$ is flat and affine then $f_*O_X$ is $O_Y-$flat. This is obvious.

But I don't think it is true that $f: X\to Y$ is flat implies $f_*O_X$ is $O_Y-$flat. See link text

In the answer of Jason Starr, the map is flat, while the 0-th direct image is not flat at all.

This is of course true, for any semi-separated scheme (i.e. the diagonal is affine), or maybe you assume $X$ is separated if you like. The reason that the sheaf is flat is that ${\rm Spec}(O_{X,x})\to X$ is affine and flat. In general if $f: X\to Y$ is flat and affine then $f_*O_X$ is $O_Y-$flat. This is obvious.

But I don't think it is true that $f: X\to Y$ is flat implies $f_*O_X$ is $O_Y-$flat. I would be happy if someone can back me up with a counter example here.

This is of course true, for any semi-separated scheme (i.e. the diagonal is affine), or maybe you assume $X$ is separated if you like, and you can take any point (not necessarily closed). The reason that the sheaf is flat is that ${\rm Spec}(O_{X,x})\to X$ is affine and flat. In general if $f: X\to Y$ is flat and affine then $f_*O_X$ is $O_Y-$flat. This is obvious.

But I don't think it is true that $f: X\to Y$ is flat implies $f_*O_X$ is $O_Y-$flat. I would be happy if someone can back me up with a counter example here.