4.3 Points in almost general position
4.3.2 The game for almost general points
Without forgetting the ultimate goal of this section (to prove Stan (I) = Sm (I) for certain ideals) we show two statements about leading monomials of some product ideals, which actually will be crucial in the proof of the main theorem.
Proposition 4.3.4. Suppose that I1 and I2 are zero dimensional splitting ideals such that their points can be distinguished by their last coordinate (that is if y is a point of I1 and z is a point of I2 then yn6=zn).
If xwxmn1 ∈Lm (I1) and xwxmn2 ∈Lm (I2) then xwxmn1+m2 ∈Lm (I1I2).
Proof. For every y∈Fn associated to I1, the primary component of I1 cor- responding to y is hx1 −y1, . . . , xn−yni-primary and thus contains a poly- nomial (xn−yn)cy for some cy ∈ N. Multiplying these together, we get a polynomial f1(xn) ∈ I1. Similarly we have f2(xn) ∈ I2. As the irreducible components of f1(xn) are of the form xn −yn, where yn is a possible last coordinate of a point of I1, it follows that f1 and f2 are relatively prime polynomials. Therefore there exist g1(xn), g2(xn)∈F[xn] such that
1−f1(xn)g1(xn)−f2(xn)g2(xn) = 0. (4.4) We also have two polynomialsxwpj(xn)−hj(x)∈Ij (j = 1,2) such that deg(pj) = mj and hj(x) ≺ xw. Such polynomials do exist since xwxmnj ∈ Lm (Ij), hence in any polynomial showing this fact we can collect together the terms divisible with xw in a polynomial xwp(xn), and put every other term in −h(x).
Multiplying (4.4) with xwp1(xn)p2(xn) we get xwp1(xn)p2(xn)−xwp2(xn)f1(xn)p1(xn)g1(xn)−
xwp1(xn)f2(xn)p2(xn)g2(xn) = 0 Note that¡
xwp2(xn)−h2(x)¢
f1(xn)∈I1I2and¡
xwp1(xn)−h1(x)¢
f2(xn)∈ I1I2, and so
f(x) = xwp1(xn)p2(xn)−h2(x)f1(xn)p1(xn)g1(xn)−h1(x)f2(xn)p2(xn)g2(xn) is in I1I2. The leading monomial of f is lm (f) = xwxdeg(pn 1p2) =xwxmn1+m2 using that h1(x) ≺ xw (and h2(x) ≺ xw), so the same is true for h1f2p2g2
(and h2f1p1g1) by the properties of the lexicographic order. This proves the statement.
CHAPTER 4. THE LEX GAME 46 The next proposition (and its proof as well) looks similar, except that we have to apply Corollary 4.3.3 in the proof, which was not at all trivial to verify. In fact, the reason why Stan monomials are in general not equal to standard monomials is that the straightforward generalization of these propositions is not true (see Example 4.4.1).
Proposition 4.3.5. Suppose that I1 and I2 are zero dimensional splitting ideals such that their points have all the same last coordinates but they can be distinguished by their second to the last coordinate (that is if y is a point of I1 and z is a point of I2 then yn =zn and yn−1 6=zn−1).
If exwexmn−11 xwnn ∈ Lm (I1) and exwexmn−12 xwnn ∈ Lm (I2) then exwexmn−11+m2xwnn ∈ Lm (I1I2).
Proof. Without loss of generality, we may suppose that the common last coordinate of the points of I1 and I2 is 0, since a change of variables x0n = xn−yn does not affect the standard monomials.
The ideal I1 has dimension zero, thus a power of xn is in I1, that is it satisfies the conditions of Corollary 4.3.3. Applying our usual trick, any polynomial which shows xewexmn−11 xwnn ∈ Lm (I1) can be written in the form xewep(xn−1, xn)−h(x) from which by Corollary 4.3.3 we get that there exist polynomials such that xewep1(xn−1, xn)xwnn −h1(x) ∈ I1, lm (p1) = xmn−11 and h1 ≺xewe. Similarly we have xewep2(xn−1, xn)xwnn −h2(x)∈I2, lm (p2) =xmn−12 and h2 ≺exwe.
Exactly as in the proof of Proposition 4.3.4, there are relatively prime polynomialsf1(xn−1)∈I1andf2(xn−1)∈I2andg1(xn−1), g1(xn−1)∈F[xn−1] such that
1−f1(xn−1)g1(xn−1)−f2(xn−1)g2(xn−1) = 0 Multiplying this with xewep1(xn−1)p2(xn−1)xwnn we get
xewep1(xn−1, xn)p2(xn−1, xn)xwnn−
xewep2(xn−1, xn)xwnnf1(xn−1)p1(xn−1, xn)g1(xn−1)− e
xwep1(xn−1, xn)xwnnf2(xn−1)p2(xn−1, xn)g2(xn−1) = 0.
As we see that the polynomials ¡ e
xwep2(xn−1, xn)xwnn−h2(x)¢
f1(xn−1) and
¡exwep1(xn−1, xn)xwnn −h1(x)¢
f2(xn−1) are both inI1I2, we have that f(x) :=exwep1(xn−1, xn)p2(xn−1, xn)xwnn −
h2(x)f1(xn−1)p1(xn−1, xn)g1(xn−1)−h1(x)f2(xn−1)p2(xn−1, xn)g2(xn−1) is in I1I2. The leading monomial of f is lm (f) = exwe ·lm (p1p2)· xwnn = e
xwexmn−11+m2xwnn since byh1(x)≺exwe (andh2(x)≺xewe) we have h1f2p2g2 ≺exwe (and h2f1p1g1 ≺exwe). The proposition follows.
CHAPTER 4. THE LEX GAME 47 Theorem 4.3.6. If I is a zero dimensional splitting ideal, and for all differ- ent points y,z of I either yn6=zn or yn−1 6=zn−1 then
Stan (I) = Sm (I).
In particular if n = 2, then the standard and the Stan monomials are the same for every zero dimensional splitting ideal.
Proof. Suppose that n ≥ 2. Let the primary decomposition be I = Q
y∈Fn
Qy
and for y, y0 ∈Fput
Iy = Y
y∈Fn yn=y
Qy and Iy,y0 = Y
y∈Fn yn=y,yn−1=y0
Qy.
In fact, the assumption on I implies that Iy,y0 is either F[x] or a primary component of I.
A game Lex (Iy,y0;w) is quite simple. If Iy,y0 is primary, then there is only one point y of Iy,y0, so if Lea guesses for the coordinates of y in each round respectively, then the result vector r is either w (if Stan’s point was y) or 0 (if it was any other). But in the latter case Lea wins the game as x0 = 1 ∈ Lm (F[x]). Therefore we see that Stan (Iy,y0) = Sm (Iy,y0) (and obviously Stan (F[x]) = ∅= Sm (F[x])).
We now prove that Stan (Iy) = Sm (Iy). As |Stan (Iy)| = |Sm (Iy)| (see Proposition 4.4.2 in the next section), it is enough to show that the left hand side contains the right.
Fix a w ∈ Nn such that Lea can win the game Lex (Iy;w). We have to see that xw ∈Lm (Iy). Set
my,y0 = min©
m ∈N : xewexmn−1xwnn ∈Lm (Iy,y0)ª .
In the first round Lea guessesy as this is the only possible last coordinate of the points of Iy. If wn−1 < P
y0∈F
my,y0, then in the next round there exists ay0 for which Lea can guess only m < my,y0 times. We claim that if Stan picks such a point ofIythen Lea cannot win the game. This contradiction will yield wn−1 ≥ P
y0∈F
my,y0. The result vector is now (∗, . . . ,∗, m, wn), which means that Lea can win this game if and only if she can win Lex (Iy,y0; (w, m, we n)) We have just seen that this latter is equivalent to exwexmn−1xwnn ∈ Lm (Iy,y0), which is not the case as m < my,y0.
CHAPTER 4. THE LEX GAME 48 Therefore we have wn−1 ≥ P
y0∈F
my,y0. We now apply Proposition 4.3.5 to get from exwexmn−1y,y0xwnn ∈Lm (Iy,y0) that
exwex
P
y0 ∈Fmy,y0
n−1 xwnn ∈Lm ÃY
y0∈F
Iy,y0
!
= Lm (Iy).
Therefore alsoxw =exwexwn−1n−1xwnn ∈Lm (Iy). This proves Stan (Iy) = Sm (Iy).
To show Stan (I) = Sm (I) we will do essentially the same thing. Again fix a w∈Nn such that Lea wins Lex (I;w). Put
my = min©
m ∈N : xwxmn ∈Lm (Iy)ª . If wn < P
y∈F
my then there has to be a y that is guessed by Lea in the first round only m < my times. Lea can still win the game, thus she can win Lex (Iy; (w, m)) as well which implies xwxmn ∈ Lm (Iy), a contradiction to the minimality of my.
Using Proposition 4.3.4 we have xwx
P
y∈Fmy
n ∈Lm
ÃY
y∈F
Iy
!
= Lm (I)
and by wn ≥ P
y∈F
my also xw=xwxwnn ∈Lm (I).
We conclude that Stan (I)⊇Sm (I) which yields our statement.
CHAPTER 4. THE LEX GAME 49