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, $(I^2:x_1x_5-x_2x_4)=I$ ?
MathOverflow is a question and answer site for professional mathematicians. It only takes a minute to sign up.Sign up to join this community
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.
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