How do I go between generators and elements of an ideal in $_2[x_1,...,x_n]$?

Question

Part 1: Elements of ideals defined by generators (answered)

Consider the ideal $I=⟨f_1,f_2,f_3⟩⊂_2 [x_1,x_2,x_3,x_4 ]$. $f_1,f_2$ and $f_3$ are generators of the ideal $I$.

$f_1=(1−x_1)x_2 \\ f_2=x_1 (1−x_3 ) \\f_3=x_3 x_4$

I can see how the ideal $I$ contains polynomials like $h=x_1 f_3= x_1 x_3 x_4$ or $w=x_3^2 f_3=x_3^3 x_4$. But I have been told that $g=1 f_1 + x_2 f_2=x_1 x_2 x_3 - x_2$, and I have a problem understanding why. I came up with the following derivation although I'm not sure if it's correct:

$g=1 f_1 + x_2 f_2 \\ g=1(x_1 - 1) x_2 + x_1 x_2 (x_3 - 1) \\ g=x_1 x_2 - x_2 + x_1 x_2 x_3 - x_1 x_2 \\ g=x_1 x_2 x_3 - x_2$

Perhaps this is correct. But why do I have to switch the bracketed terms' order? For example, why does $f_1$ become $(x_1 - 1) x_2$ in the derivation when it was originally defined as $(1 - x_1) x_2$?

Part 2: Generators can be contained as elements in ideals generated by other generators

Consider the ideal $I=⟨k_1,k_2,k_3⟩⊂_2 [x_1,x_2,x_3]$.

$k_1=x_1 (1−x_2 ) \\ k_2=x_2 (1−x_3 ) \\ k_3=x_1 (1−x_3 )$

How can I show that $k_3 \in ⟨k_1, k_2⟩$? (An earlier version of this question had a typo, where I asked: How can I show that $k_1 \in ⟨k_2, k_3⟩$)

An ideal is closed under addition with itself, so I can try $k_1 + k_2$ (using the same derivation method above where I swap the in-bracket terms), but this doesn't seem to get me anywhere:

$y = k_1 + k_2 \\ y = x_1 (x_2−1 ) + x_2 (x_3 - 1 ) \\ y = x_1x_2 - x_1 + x_2 x_3 - x_2$

I know I can also multiply $k_2$ or $k_3$ by anything of the form $x_1^a x_2^b x_3^b$, where $a, b$ and $c$ can be any integer (I think?). But how do I find the correct $a, b$ and $c$? Or is there no systematic way and I just have to try different combinations until I find $k_1$?

Answer 1

Part 1

Thanks to @plop for answering this. They pointed out in their comment below that:

$1−x_1=−(x_1−1)=x_1−1 $, where the latter equation is because $(x_1−1)+(x_1−1)=(1+1)x_1−(1+1)=0$, since $1+1=0$ in $_2$. The switch in order wasn't such much a "have to" as it was a "can do".

When I saw $-(x_1 - 1)$, I was seeing $-x$ and $+1$. But then I also remembered that in $_2$, $-1=1$. So, it's also true that this equation shows $+x$ and $-1$. Maybe the main way of remembering this $_2$ strangeness is: ignore sign, is it $0$ or $1$? I also found this PDF of notes titled The finite field with 2 elements helpful.

Part 2

In my original question, I mistakenly asked how to show $k_1 \in ⟨k_2,k_3⟩$. Actually, I meant to ask how we can show $k_3 \in ⟨k_1,k_2⟩$. Apologies to all for that.

A friend of mine gave this answer via e-mail based on trying to show $k_1 \in ⟨k_2,k_3⟩$, which is false:

--- e-mail from friend starts ---

For the sake of completeness, I suggest the book "Ideals, Variety, and Algorithm" for the details. There are at least 3 ways to solve this problem:

1. Using Division Algorithm as in Theorem 3 page 64 in the book. I tried this one, but then I got the non-zero remainder. This means $k_1$ is not divided by $k_2$ and $k_3$.

Note: Before doing the division algorithm, please read about monomial ordering (lexicographic order is enough I suppose), multidegree, leading monomial, leading term, etc in the previous part of the book (pages 54-60 in the book)

2. By checking variety (Affine Variety) inclusion as in Proposition 8 (i) page 34 Using similar way as we did before, I found that:

$V(k_1)= \{ 001, 000, 010, 011, 110, 111 \}$

and

$V(k_2,k_3) = \{ 000, 001, 011, 101, 111 \}$.

We can see that, 101 is in $V(k_2,k_3)$ but not in $V(k_1)$. This means, the ideal $⟨k_1⟩$ is not contained in the ideals $⟨k_2,k_3⟩$ (based on the contraposition of Proposition 8 (i)). This is not what we hope actually.

3. Checking ideal membership by first computing the Gröbner bases (GB) of $I =⟨k_2,k_3⟩$. I tried to calculate reduced GB of $I$ using sympy in Python as follows (I use the code from: https://docs.sympy.org/latest/modules/polys/basics.html#groebner-bases):

from sympy import *

k= GF(2)

x = Symbol('x')

y = Symbol('y')

z = Symbol('z')

groebner([y*(1+z), x*(1+z)], x, y, z, domain=k, order='lex')

Here I use $x = x_1$, $y = x_2$, and $z = x_3$. It turns out I got the following output: ((+, +),(, , ))

which means ${x(1+z), y(1+z)}$ is already the reduced GB for $I$. Using the division algorithm as before, we can see that the remainder of $k_1$ divided by all the elements of GB ($k_2$ and $k_3$) is not zero. This confirms the previous two observations.

For the details in Ideal Membership checking, please check pages 97-98 in the book.

--- e-mail from friend ends ---

Based on this, I tried method 2 and found that:

$V(k_3)= \{ 101, 001, 000, 111, 011, 010 \}$

$V(k_1,k_2)= \{ 101, 001, 000, 111, 011, 010 \}$

Fact: $V ⊆ W$ if and only if $I(V) ⊇ I(W)$, where $V$ and $W$ are varieties (Proposition 8 (i), page 34 of "Ideals, Variety, and Algorithm" book).

Since $V(k_1,k_2) ⊆ V(k_3)$, then $I(V(k_1,k_2)) ⊇ I(V(k_3))$ and $k_3 \in ⟨k_1,k_2⟩$.