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I have an ideal $I$ generated by quadratic and cubic homogeneous polynomials in $10$ variables.

Macaulay2 tells me that $I$ defines an irreducible variety $X$ of dimension $5$ and degree $10$ in $\mathbb{P}^9$, and that $I$ is not radical.

When I ask for the primary decomposition of $I$ Macaulay2 gives me two ideals $I_0,I_1$. He says that $I_0$ is radical and defines an irreducible variety of dimension $5$ and degree $10$, while $I_1$ is not radical and defines a variety of dimension $-1$. Indeed, when I ask for the radical of $I_1$ Macaulay2 gives me the ideal generated by all the homogeneous coordinates of $\mathbb{P}^9$ thus corresponding to the empty set. This last fact is really confusing me.

What is then the meaning of $I_1$ in the primary decomposition of $I$?

Thank you very much in advance.

Here are the ideals I am considering

P9 = QQ[a00,a01,a02,a03,a11,a12,a13,a22,a23,a33]

F0 = a00*a12-a01*a02+a01*a13-a03*a11

F1 = a00*a23+a01*a33-a02*a03-a03*a13

F2 = a01*a22-a02*a12+a11*a23-a12*a13

F3 = a02*a23-a03*a22+a12*a33-a13*a23

F4 = a00*a22-a02^2-a11*a33+a13^2

J = ideal(F0,F1,F2,F3,F4)

M = matrix{{a00,a01,a02,a03},{a01,a11,a12,a13},{a02,a12,a22,a23},{a03,a13,a23,a33}}

I = J + minors(3,M)

L = primaryDecomposition(I)

X = variety L_0

Y = variety L_1

dim(X)

dim(Y)
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  • $\begingroup$ Just to have it available: faculty.math.illinois.edu/Macaulay2/doc/Macaulay2-1.12/share/… . $\endgroup$ – LSpice Oct 31 '20 at 19:38
  • $\begingroup$ In terms of the definition, it seems to me that we would expect $I = I_0 \cap I_1$ (which I think we can't judge from the information given) and $\sqrt{I_0} \ne \sqrt{I_1}$ (which is the case in your computation). I think that there is no expectation that $I_1$ itself be radical. $\endgroup$ – LSpice Oct 31 '20 at 19:42
  • $\begingroup$ I posted the ideals I am considering. What is really unclear to me is the difference between the variety defined by $I$ and the variety defined by $L_0$. $\endgroup$ – user125056 Oct 31 '20 at 19:59
  • $\begingroup$ When you post code, please make sure to format it as code (either by manually adding 4 spaces at the beginning of each line, or by using the format-as-code button in the GUI). Otherwise, Markdown gets at it and makes a mess. I have edited accordingly. $\endgroup$ – LSpice Oct 31 '20 at 20:07
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    $\begingroup$ I am sorry. Next time I will do it. $\endgroup$ – user125056 Oct 31 '20 at 20:58
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Commutative algebra is NOT the same as algebraic geometry, especially projective algebraic geometry.

The variety in $\mathbb{P}^9$ defined by $I$ and the variety in $\mathbb{P}^9$ defined by $I_0$ are the same variety.

If you were to work in $\mathbb{A}^{10}$, then $I$ and $I_0$ would define different affine schemes; the scheme defined by $I$ has some extra fatness at the origin, but since the origin is not part of projective space, you don't see this difference in $\mathbb{P}^9$.

EDIT: The relevant bit of Hartshorne is Exercise II.5.10

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  • $\begingroup$ Thank you very much for the answer. I suspected that the varieties defined by $I$ and $I_0$ had to coincide. Anyway I am still confused about $I$ not being radical. Can we have a non radical ideal $I$ such that $I$ and $r(I)$ define the same scheme? This seems to be the case. Could you elaborate a bit more on the facr that $I$ and $I_0$ define the same scheme in $\mathbb{P}^9$? $\endgroup$ – user125056 Oct 31 '20 at 21:00
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    $\begingroup$ Your question is, I think, exactly @AlexanderWoo’s point: a non-radical ideal and its radical will always define the same variety, but never the same scheme. It may help to simplify your picture to $I = (x^2)$; then the variety determined by $I$ is the single point $\{0\}$ in affine space, but the scheme determined by $I$ is a doubled point $\{0\}$. You can see the difference, for example, by finding that the tangent space to the scheme at its sole point is 1-dimensional. $\endgroup$ – LSpice Nov 1 '20 at 1:04
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    $\begingroup$ No I think that Alexander Woo is claiming that the ideal $I$ and its radical define the same scheme in $\mathbb{P}^9$ but different schemes in $\mathbb{A}^{10}$. $\endgroup$ – user125056 Nov 1 '20 at 9:36
  • $\begingroup$ @user125056, you are right. Sorry! $\endgroup$ – LSpice Nov 1 '20 at 14:34
  • $\begingroup$ @user125056: You should look at Exercise II.5.10 in Hartshorne for more details. $\endgroup$ – Alexander Woo Nov 3 '20 at 4:49

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