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Let $R=k[x_1,\ldots,x_6]$ be a polynomial ring and $I=(x_1x_5-x_2x_4,x_2x_6-x_3x_5)$ be an ideal.

How to show that, $(I^2:x_1x_5-x_2x_4)=I$ ?

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  • $\begingroup$ These are determinantal ideals. Any generalization? $\endgroup$ – darij grinberg Sep 4 '16 at 3:12
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Put $P=x_1x_5-x_2x_4$, $Q=x_2x_6-x_3x_5$. Assume a polynomial $A$ satisfies $AP\in I^2$. Then we have a relation $$(*)\quad AP=UP^2+VQ^2+WPQ.$$ In particular, $P$ divides $VQ^2$. Since $P$ is prime and does not divide $Q$, it must divide $V$. Simplifying $(*)$ by $P$ gives the result.

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If you are open to using a computer, this can be computed in Macaulay2. Below is the computation.

i1 : R = QQ[x1,x2,x3,x4,x5,x6]

o1 = R

o1 : PolynomialRing

i2 : I = ideal(x1*x5 - x2*x4, x2*x6 - x3*x5)

o2 = ideal (- x2*x4 + x1*x5, - x3*x5 + x2*x6)

o2 : Ideal of R

i3 : I == I^2 : ideal(x1*x5 - x2*x4)

o3 = true

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  • $\begingroup$ Thanks. Can you show without M2 ? $\endgroup$ – user177523 Sep 4 '16 at 1:17
  • $\begingroup$ Thinking more about it. I believe $r$ is in the colon ideal iff $r(x_1 x_5 - x_2 x_4)$ is in $I^2$ iff $r$ is in $I$. $\endgroup$ – John Machacek Sep 4 '16 at 2:01
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    $\begingroup$ The first iff is the definition of colon ideal, the second iff takes a quick argument now in another answer. $\endgroup$ – John Machacek Sep 4 '16 at 15:28

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