Let $S=K[x_0,...,x_3]$. Eisenbud claimed in his book "The Geometry of Syzygies" that the dimension of the space of cubics in the ideal generated by the three quadrics is at most $3\times 4=12$, because $S$ has only $4$ linear forms (I assume he meant $x_0,...,x_3$). However I failed to see how the number of linear forms related to the dimension of the said space, can someone the explain calculation in more detail?
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2$\begingroup$ Hint: how do you get a cubic from a quadric? You multiply by a linear form, right? $\endgroup$– KReiserMay 12, 2020 at 1:21
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$\begingroup$ @KReiser Ah you are right... $\endgroup$– MathChopperMay 12, 2020 at 1:46
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1$\begingroup$ If your issue's been resolved, I suggest writing an answer to your own question. $\endgroup$– KReiserMay 12, 2020 at 2:07
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1 Answer
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As the comment suggested, we know that the space is generated by three quadrics, and the basis of the space of cubics generated by this quadrics comes from multiplying linear forms to the them, each quadrics has four choices of linear forms $x_0,...,x_3$. So the dimension of the space of cubics is $3\times 4=12$.