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A rational elliptic fibration can be obtained from the total space of a pencil of cubics (i.e. $\mathbb{P}^2$), by Blp the 9 basepoints of the pencil. If the basepoints are distinct, the resulting elliptic fibration has 12 singular, nodal fibers. Most proofs of this I've seen rely on, e.g. using the Euler characteristic of the total space. Is there a way to understand this using just the geometry of the pencil (open to seeing different interpretations of what that means)?

Caliper
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    The space of all plane cubic forms a 9-dimensional projective space. The locus where the plane cubics are singular form a degree 12 hypersurface, which is called the discriminant. The pencil $aF+bG=0$ of cubic that you considered corresponds to a general line in $\mathbb P^9$, which intersects the discriminant at 12 smooth points, so correspond to 12 nodal cubics. – AG learner Jan 31 '21 at 01:48
  • Thank you - this makes complete sense. – Caliper Jan 31 '21 at 01:51

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