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Let $X$ be a complex normal projective variety, let $|L|$ be a non empty linear series on $X$ and let $b(|L|)$ be its base ideal. Suppose $f:X'\rightarrow X$ is a log resolution of the ideal $b(|L|)$.

Is $f$ a log resolution of the linear series $|L|$ (even if $X$ is not smooth)?

If it is do you have a proof or a reference for this?

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I don't think so. Suppose for example that $X$ is smooth, and the base locus of $|L|$ is set-theoretically a divisor with normal crossing, but it has an embedded component. In this case $X$ itself will be a log-resolution of the base locus, but not of the linear system $|L|$.

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Maybe we are using two different definitions. When I speak about a log resolution of the base ideal, I mean that I want the inverse image of the ideal that defines the scheme-theoretical base locus to become a line bundle (defining a divisor with snc support). In particular I'm using the definitions of Lazarsfeld's "Positivity in algebraic geometry", chap 9.1. Using this definition, in the same chapter, Lazarsfeld remarks that for smooth varieties the claim holds. – Gianni Bello Jun 17 '10 at 19:07
I guess we are using different definitions. With the definitions I seem to understand you have in mind, the result seems trivial to me. The point is that when you subtract the inverse image of the base ideal, which is supported on a divisor with normal crossing, the linear system becames base point free, so this is a log-resolution, by definition. Is there something wrong with this argument? If so, can you spell out your defitions exactly. – Angelo Jun 17 '10 at 20:38
Ok, this is what I had in mind. The point I lack is why the inverse image of the base ideal is the base ideal of the pullback of the linear series. I suppose this is trivial but it is not clear to me. – Gianni Bello Jun 17 '10 at 21:11
This seems pretty obvious to me. The linear system corresponds to a line bundle $L$ with a number of independent sections $s_i$, whose zero scheme is the base locus. When you pullback the $s_i$ the base scheme pulls back. – Angelo Jun 18 '10 at 5:25
No problem; you are learning. – Angelo Jun 18 '10 at 15:52

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