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Based on the above comment of Holger Partsch, you can't factor a lot of stuff. You can do silly things like:

A necessary condition for factoring $f$ through a given $S'$ is that you can embed: $$K(S) \subseteq K(S') \subseteq K(X).$$

If you are willing to blow-up $X$ to resolve indeterminacies, then this might be the only obstruction in the geometric setting anyway (assuming $S$ and $S'$ are normal).

On the other hand THISTHIS question, and the numerous excellent answers, seems like it might be very relevant depending on your context.

Based on the above comment of Holger Partsch, you can't factor a lot of stuff. You can do silly things like:

A necessary condition for factoring $f$ through a given $S'$ is that you can embed: $$K(S) \subseteq K(S') \subseteq K(X).$$

If you are willing to blow-up $X$ to resolve indeterminacies, then this might be the only obstruction in the geometric setting anyway (assuming $S$ and $S'$ are normal).

On the other hand THIS question, and the numerous excellent answers, seems like it might be very relevant depending on your context.

Based on the above comment of Holger Partsch, you can't factor a lot of stuff. You can do silly things like:

A necessary condition for factoring $f$ through a given $S'$ is that you can embed: $$K(S) \subseteq K(S') \subseteq K(X).$$

If you are willing to blow-up $X$ to resolve indeterminacies, then this might be the only obstruction in the geometric setting anyway (assuming $S$ and $S'$ are normal).

On the other hand THIS question, and the numerous excellent answers, seems like it might be very relevant depending on your context.

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Karl Schwede
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Based on the above comment of Holger Partsch, you can't factor a lot of stuff. You can do silly things like:

A necessary condition for factoring $f$ through a given $S'$ is that you can embed: $$K(S) \subseteq K(S') \subseteq K(X).$$

If you are willing to blow-up $X$ to resolve indeterminacies, then this shouldmight be the only obstruction in the geometric setting anyway (assuming $S$ and $S'$ are normal).

On the other hand THIS question, and the numerous excellent answers, seems like it might be very relevant depending on your context.

Based on the above comment of Holger Partsch, you can't factor a lot of stuff. You can do silly things like:

A necessary condition for factoring $f$ through a given $S'$ is that you can embed: $$K(S) \subseteq K(S') \subseteq K(X).$$

If you are willing to blow-up $X$ to resolve indeterminacies, then this should be the only obstruction in the geometric setting anyway (assuming $S$ and $S'$ are normal).

On the other hand THIS question, and the numerous excellent answers, seems like it might be very relevant depending on your context.

Based on the above comment of Holger Partsch, you can't factor a lot of stuff. You can do silly things like:

A necessary condition for factoring $f$ through a given $S'$ is that you can embed: $$K(S) \subseteq K(S') \subseteq K(X).$$

If you are willing to blow-up $X$ to resolve indeterminacies, then this might be the only obstruction in the geometric setting anyway (assuming $S$ and $S'$ are normal).

On the other hand THIS question, and the numerous excellent answers, seems like it might be very relevant depending on your context.

Source Link
Karl Schwede
  • 20.5k
  • 3
  • 53
  • 98

Based on the above comment of Holger Partsch, you can't factor a lot of stuff. You can do silly things like:

A necessary condition for factoring $f$ through a given $S'$ is that you can embed: $$K(S) \subseteq K(S') \subseteq K(X).$$

If you are willing to blow-up $X$ to resolve indeterminacies, then this should be the only obstruction in the geometric setting anyway (assuming $S$ and $S'$ are normal).

On the other hand THIS question, and the numerous excellent answers, seems like it might be very relevant depending on your context.