Convex polytopes and Newton polytopes
Andrzej Nowicki 24.10.2016, version nt-07
In this note we recall some well known properties of convex polytopes. We prove several propositions on convex sets, convex hulls, convex polytopes, Minkowski sums, vertices and Newton polytopes. It is well known that if f, g are Laurent polynomials in n variables, then the Newton polytope of the product f g is the Minkowski sum of the Newton polytopes of f and g. We present an elementary proof of this fact.
1 Convex sets
Let n > 1 be a fixed natural number. If a, b ∈ Rn then the set D
a, bE
=n
αa + βb; α, β ∈ R, α > 0, β > 0, α + β = 1o
is the line segment in Rn connecting the points a and b. A set A in Rn is said to be convex if it contains the line segment connecting any two points in A. By definition , the empty set is convex. We say that (λ1, . . . , λm) is an m-sequence if λ1, . . . , λm are nonnegative real numbers such that λ1+ · · · + λm = 1. If a1, . . . , am are points in Rn then all points in Rn of the form
λ1a1+ · · · + λmam,
where (λ1, . . . , λm) is an m-sequence, are called convex combinations of a1, . . . , am. In particular, the line segment ha, bi is the set of all convex combinations of the points a, b.
Proposition 1.1. Let A be a convex set in Rn and let (λ1, . . . , λm) be an m-sequence.
If a1, . . . , am are points belonging to A, then the convex combination λ1a1+ · · · + λmam also belongs to A.
Proof. It is obvious for m = 1 and m = 2. Assume that it is true for some m > 2. Let a1, . . . , am, am+1 ∈ A and x = λ1a1 + · · · + λmam + λm+1am+1, where (λ, . . . , λm, λm+1) is an (m + 1)-sequence. Put β = λ1 + · · · + λm. If β = 0, then we are done because x = am+1 ∈ A.
Assume that β 6= 0. Put γj = λβj for j = 1, . . . , m, and let b = γ1a1+ · · · + γmam. Since (γ1, . . . , γm) is an m-sequence, the point b (by an induction) belongs to A. Observe that (β, λm+1) is a 2-sequence, so βb + λm+1am+1 ∈ A. But βb + λm+1am+1 = x, and hence x ∈ A.
Proposition 1.2. Let S be a set in Rn and let A be the set of all convex combinations of points belonging to S. Then A is convex.
Proof. Let x, y ∈ A. We will show that hx, yi ⊂ A. Let z ∈ hx, yi. Then z = px + qy, where (p, q) is a 2-sequence. Since x, y ∈ A, we have
x = α1a1+ · · · + αuau, y = β1b1+ · · · + βvbv,
where u > 1, v > 1 are integers, a1, . . . , au, b1, . . . , bv ∈ S, (α1, . . . , αu) is a u-sequence, and (β1, . . . , βv) is a v-sequence. Hence,
z = px + qy = (pα1)a1+ · · · + (pau)au+ (qb1)b1+ · · · + (qβv)bv.
Observe that (pα1, . . . , pαu, qβ1, . . . , qβv) is a (u + v)-sequence. Hence, z is a convex combination of points from S, and this means that z ∈ A. Therefore, hx, yi ⊂ A.
Clearly, every intersection of convex sets is convex, and Rn itself is convex. Thus, for any S ⊂ Rn there exists the smallest convex set containing S, called the convex hull of S. We denote this convex hull by hSi. Usually it is denoted by conv(S) (see for example [2] or [1]). Thus, hSi is the intersection of all convex sets in Rn that contain S. As an immediate consequence of Proposition 1.2 we obtain
Proposition 1.3. If S is a subset of Rn, then the convex hull hSi is the set of all convex combinations of points belonging to S.
If S = {a1, . . . , am} is a finite set, then we write ha1, . . . , ami instead ofD
{a1, . . . , am}E . By definition, a polytope is the convex hull of a finite set in Rn. Thus, in this note every polytope is convex. We do not consider nonconvex polytopes. Note the following special case of Proposition 1.3.
Proposition 1.4. If a1, . . . , am ∈ Rn, then the polytope ha1, . . . , ami is the set of all points in Rn of the form λ1a1 + · · · + λmam, where (λ1, . . . , λm) is an m-sequence.
2 Minkowski sums
The Minkowski sum of two sets A, B ⊂ Rn is defined to be A + B :=n
a + b; a ∈ A, b ∈ Bo .
Proposition 2.1. The Minkowski sum of two convex sets is a convex set.
Proof. Let A, B ⊂ Rn be convex sets. Let x, y ∈ A + B, and let (p, q) be a 2- sequence. We will show that px + qy ∈ A + B. We have: x = a1+ b1, y = a2+ b2, where a1, a2 ∈ A, b1, b2 ∈ B. Put a = pa1+ qa2, b = pb1+ qb2. Since A, B are convex, a ∈ A and b ∈ B, and we have px+qy = p(a1+b1)+q(a2+b2) = (pa1+qa2)+(pb1+qb2) = a+b.
Thus px + qy ∈ A + B.
If A is a set in Rn and λ is an arbitrary real number, then the set λA is defined to be {λa; a ∈ A}.
Proposition 2.2. If A is a convex set in Rn and λ ∈ R, then the set λA is convex.
Proof. Let x, y ∈ λA, and let (p, q) be a 2-sequence. We will show that px + qy ∈ λA. Let x = λa1, y = λa2, where a1, a2 ∈ A, and put a = pa1+ qa2. Since A is convex, the point a belongs to A. Now we have: px +qy = p(λa1) +q(λa2) = λ(pa1+qa2) = λa.
Thus px + qy ∈ λA.
As a consequence of the above propositions we obtain
Proposition 2.3. Let A, B be convex sets in Rn, and let α, β ∈ R. Then the set αA + βB is convex. In particular, the set A − B := {a − b; a ∈ A, b ∈ B} is convex.
Let A be a set Rn. Consider the sets A + A and 2A. Clearly, we have always the inclusion 2A ⊂ A + A. For example, let n = 1 and A = {0, 1}. Then 2A = {0, 2} and A + A = {0, 1, 2}. Thus, in this case A + A 6= 2A.
Proposition 2.4. If A is a convex set in Rn, then A + A = 2A.
Proof. Let x ∈ A + A; x = a + b, where a, b ∈ A. Then 2a, 2b ∈ 2A, and x = 12(2a) + 12(2b).
Since 12,12 is a 2-sequence and the set 2A is convex, the point x belongs to 2A. Thus, A + A ⊂ 2A. The opposite inclusion is obvious.
By the same way we obtain the following generalization of the above proposition.
Proposition 2.5. If A is a convex set in Rn and m > 1 is an integer, then A + A + · · · + A
| {z }
m
= mA.
Proof. Put B = A + A + · · · + A
| {z }
m
. Let x ∈ B; x = a1 + · · · + am, where a1, . . . , am ∈ A. Then ma1, . . . , mam ∈ mA, and
x = m1(ma1) + · · · +m1(mam).
Since m1, . . . , m1 is an m-sequence and the set mA is convex, by Proposition 1.1 the point x belongs to mA. Thus, B ⊂ mA. The opposite inclusion is obvious.
In the next proposition we examine convex hulls.
Proposition 2.6. If S is a set in Rn and λ ∈ R, then λhSi = hλSi.
Proof. Let x ∈ λhSi. Then x = λu, for some uhSi. It follows from Proposition 1.3 that u = αa1+ · · · + αmam, for some m > 1, a1, . . . , am ∈ S, and an m-sequence (α1, . . . , αm). Thus,
x = λu = α(λa1) + · · · + αm(λam),
λa1, . . . , λam ∈ λS, and (α1, . . . , αm) is an m-sequence. Using Proposition 1.3 we see that x ∈ hλSi. Hence, λhSi ⊂ hλSi. The opposite inclusion is obvious, because λhSi is a convex set (by Proposition 2.2) containing λS.
Proposition 2.7. If S, T are subsets of Rn, then hSi + hT i = h S + T i.
Proof. Clearly, S + T ⊂ hSi + hT i. So hSi + hT i is a convex set in Rn containing S + T . This implies that h S + T i ⊂ hSi + hT i. Now we will show that the opposite inclusion also holds.
Let z ∈ hSi + hT i. Let z = x + y, where x ∈ hSi and y ∈ hT i. Then (by Proposition 1.3) x = α1a1+ · · · + αsas, y = β1b1+ · · · + βtbt, where a1, . . . , as ∈ S, b1, . . . , bt ∈ T , (α1, . . . , αs) is an s-sequence, and (β1, . . . , βt) is a t-sequence. Then we have
z = x + y = (α1a1+ · · · + αsas) + (β1b1+ · · · + βtbt)
=
s
P
i=1
αiai+
t
P
j=1
βjbj =
s
P
i=1
αi
t
P
j=1
βj
! ai +
t
P
j=1
βj
s P
i=1
αi
bj
=
s
P
i=1 t
P
j=1
αiβjai+
t
P
j=1 s
P
i=1
αiβjbj =
s
P
i=1 t
P
j=1
αiβj(ai+ bj) .
Thus, z =
s
P
i=1 t
P
j=1
αiβj(ai+ bj).. It is clear that every point of the form ai + bj be- longs to S + T . Moreover,
s
P
i=1 t
P
j=1
αiβj =
s P
i=1
αi
t
P
j=1
βj
!
= 1 · 1 = 1 and hence (α1β1, α1β2, . . . , αsβt) is an (s + t)-sequence. Thus, by Proposition 1.3, the point z belongs to the convex hull h S + T i. Therefore, hSi + hT i ⊂ h S + T i.
Note the following consequences of the above propositions.
Proposition 2.8. If S, T are subsets of Rn and α, β ∈ R, then In particular, αhSi + βhT i = h αS + βT i.
In particular, hSi − hT i = h S − T i.
Proposition 2.9. The Minkowski sum of two polytopes is a polytope.
3 Vertices
Let A be a convex set in Rn and let z ∈ A. We say that the point z is a vertex of A if
a,b∈A∀
z ∈ ha, bi =⇒ z = a or z = b.
Proposition 3.1. Let A be a convex set in Rn and let z ∈ A. Then z is a vertex of A if and only if the set A r {z} is convex.
Proof. Assume that z is a vertex of A. Let x, y ∈ A r {z}, and suppose that the segment hx, yi is not contained in A r{z}. But x, y ∈ A and A is convex, so hx, yi ⊂ A.
This means that z ∈ hx, yi. Then x = z or y = z, and we have a contradiction. Thus, the set A r {z} is convex.
Now assume that A r {z} is convex. Let x, y ∈ A and z ∈ hx, yi. Suppose that z 6= x and z 6= y. Then x, y belong to the convex set A r {z}, so z ∈ hx, yi ⊂ A r {z};
a contradiction. Therefore, z is a vertex of A.
Proposition 3.2. Let A = hb, a1, . . . , ami be a polytope in Rn. If the set A r {b} is nonconvex, then A = ha1, . . . , ami.
Proof. Since A r {b} is nonconvex, there exist points x, y ∈ A r {b} such that hx, yi 6⊂ A r {b}. But x, y ∈ A and A is convex, so hx, yi ⊂ A. This means that b ∈ hx, yi. Thus, b = px + qy for some 2-sequence (p, q). Moreover, x, y ∈ A = hb, a1, . . . , ami, so
x = ub + α1a1+ · · · + αmam, y = vb + β1a1+ · · · + βmam,
where (u, α1, . . . , αm) and (v, β1, . . . , βm) are (m + 1)-sequences. Note that u < 1 and v < 1, because x 6= b and y 6= b. Now we have b = px+qy = p (ub + α1a1+ · · · + αmam)+
q (vb + β1a1+ · · · + βmam) , and so,
(1 − pu − pv) b = (pα1+ qβ1) a1+ · · · + (pαm+ qβm) am.
Put γ = 1 − pu − qv. Since u < 1 and v < 1, we have 0 6 pu + qv < p + q = 1, that is, γ > 0. Let γ1 = pα1+qβγ 1, . . . , γm = pαm+qβγ m. Then b = γ1a1+ · · · + γmam. Observe that (γ1, . . . , γm) is an m-sequence. In fact,
γ1+ · · · + γm = 1γ
p(α1+ · · · + αm) + q(β1+ · · · + βm)
= 1γ
p(1 − u) + q(1 − v)
= γ1
(p + q) − pu − qv)
= 1γ(1 − pu − qv) = 1γγ = 1.
Hence, b ∈ ha1, . . . , ami (by Proposition 1.4) and hence, A = ha1, . . . , ami.
Proposition 3.3. Let A = hb, a1, . . . , ami be a polytope in Rn, and b 6∈ {a1, . . . , am}.
If A = ha1, . . . , ami, then b is not a vertex of A.
Proof. Assume that A = ha1, . . . , ami, and suppose that b is a vertex of A. Then, by Proposition 3.1, the set A r {b} is convex. This implies that ha1, . . . , ami ⊂ A r {b}, because {a1, . . . , am} ⊂ A r {b}. Thus, we have a contradiction: b ∈ A ⊂ A r {b}.
Let S = {a1, . . . , am} be a finite set in Rn, and consider the polytope A = hSi = ha1, . . . , ami. In this case we say that S is a set of generators of A. We say that S is a minimal set of generators of A if
D
S r {aj}E 6= A
for any j = 1, . . . , m. It is clear that if S = {a1, . . . , am} is an arbitrary set of generators of a polytope A, then there exists a subset T of S such that T is a minimal set of generators of A.
Proposition 3.4. Let A = ha1, . . . , ami be a polytope in Rn. If {a1, . . . , am} is a minimal set of generators of A, then any point aj, for j = 1, . . . , m, is a vertex of A.
Proof. Put S = {a1, . . . , am}. We have A = hSi. Suppose that some aj is not a vertex of A. Then, by Proposition 3.1, the set A r {aj} is nonconvex. Hence, by Proposition 3.2, the polytope generated by S r {aj} is equal to A. Thus, we have a contradiction: A = hS r {aj}i 6= A.
Proposition 3.5. Every polytope A in Rn has a unique minimal set of generators.
This set form all vertices of A.
Proof. Let T = {a1, . . . , at} and S = {b1, . . . , bs} be minimal sets of generators of A. We will show that S = T . For this aim we will prove that bs ∈ T .
Suppose bs6∈ T . Since bs ∈ A = hT i = ha1, . . . , ati, we have bs= α1a1+ · · · + αtat,
where (α1, . . . , αt) is a t-sequence, and α1 < 1, . . . , αt < 1 (because bs 6∈ T ). Each point ai, for i = 1, . . . , t, belongs to A = hb1, . . . , bsi. Hence
a1 = β11b1+ · · · + β1sbs, a2 = β21b1 + · · · + β2sbs, . . . , at = βt1b1 + · · · + βtsbs, where each (βi1, . . . , βis) is an s-sequence with βis < 1. Hence, bs = α1a1+ · · · + αtat, = α1(β11b1+ · · · + β1sbs) + · · · + αt(βt1b1+ · · · + βtsbs) and hence,
cbs= γ1b1+ · · · + γs−1bs−1,
where c = 1 − α1β1s − α2β2s − · · · − αtβts, and γi = α1βi1 + · · · + αtβti for any i = 1, . . . , s − 1. Since all the numbers α1, . . . , αt are smaller than 1, we have
1 − c = α1β1s+ · · · + αtβts < β1s+ · · · + βts = 1
and hence, c > 0. Now we have bs = γc1b1 + · · · + γs−1c bs−1 and it is clear that the sequence γc1, . . . , γs−1c is a (s − 1)-sequence. This means, by Proposition 1.4, that hb1, . . . , bs−1, bsi = hb1, . . . , bs−1i. But it is a contradiction, because {b1, . . . , bs} is a minimal set of generators of A.
Therefore, bs ∈ T and, by the same way we see that each bj belongs to T . Hence, S ⊂ T and similarly T ⊂ S so, S = T . Thus, we proved that every polytope A has a unique minimal set of generators. It follows from Proposition 3.4 that this unique set of generators is equal to the set of all vertices of A.
Proposition 3.6. Let A, B be polytopes in Rn, and α, β ∈ R. If c is a vertex of the polytope αA + βB, then c = αa + βb, where a is a vertex of A and b i a vertex of B.
Proof. Put C = αA + βB. Let {a1, . . . , ap} be the set of all vertices of A, and let {b1, . . . , bq} be the set of all vertices of B. Then A = ha1, . . . , api, B = hb1, . . . , bqi and, by Proposition 2.8, we know that U := {ai+ bj; i = 1, . . . , p, j = 1, . . . , q} is a set of generators of the polytope C. Let U0 be a subset of U which is a minimal set of generators of C. It follows from Proposition 3.5 that U0 is the set of all vertices of C. Thus, every vertex of C is of the form αai+ βbj for some i ∈ {1, . . . , p} and some j ∈ {1, . . . , q}.
Note the following special case of the above proposition.
Proposition 3.7. If A, B are polytopes in Rn, then every vertex of the Minkowski sum A + B is of the form a + b, where a is a vertex of A and b is a vertex of B.
Note also the following useful proposition.
Proposition 3.8. Let A, B be polytopes in Rn. Let a, c ∈ A, b, d ∈ B, a 6= c and b 6= d. If a + b = c + d, then the point a + b is not a vertex of the polytope A + B. More exactly, if a + b = c + d, then a + b is the centre of the line segment ha + d, c + bi.
Proof. Let a + b = c + d. Then a + b = 12(2a + 2b) = 12(a + d) + 12(c + b) so, a + b is the centre of the line segment ha + d, c + bi. Suppose thata + b is a vertex of A + B. Then, by definition, a + b = a + d or a + b = c + b and so, b = d or a = c; a contradiction.
4 Newton polytopes
Let k be a commutative ring with unity, and let n > 1 be a fixed natural number. We denote by k [X, X−1] the ring kx1, . . . , xn, x−11 , . . . , x−1n , that is, k [X, X−1] is the ring of Laurent polynomials in variables x1, . . . , xn over k. If α = (α1, . . . , αn) ∈ Zn, then we denote by Xα the rational monomial xα11· · · xnαn. In particular, X0 = x01· · · x0n = 1.
Every Laurent polynomial f ∈ k [X, X−1] has a unique presentation in a finite sum
f = X
α∈Zn
fαXα,
where each fα is an element of k. The support of f , denoted S(f ), is the set of all α ∈ Zn such that fα 6= 0. For every f ∈ k [X, X−1] the support S(f ) is a finite subset of Zn so, it is a finite subset of Rn. In particular, if f = 0 then the set S(f ) is empty.
The Newton polytope of f , denoted N (f ), is the convex hull of the support S(f ). Thus, in our notations
N (f ) = D S(f )E
.
We will prove that if the commutative ring k is without zero divisors and f, g ∈ k [X, X−1], then the Newton polytope of the product f g is the Minkowski sum of the Newton polytopes of f and g. It is quite obvious that we have always the following proposition.
Proposition 4.1. If k is an arbitrary commutative ring and f, g ∈ k [X, X−1], then N (f g) ⊂ N (f ) + N (g).
Proof. If f = 0 or g = 0, then N (f g) = ∅ and nothing to prove. Assume that f 6= 0 and g 6= 0. Let f =P
α
aαXα and g =P
β
bβXβ. Then f g =P
γ
cγXγ, where cγ= X
α+β=γ
aαbβ
for any γ ∈ Zn. Assume that γ ∈ S(f g). Then cγ 6= 0, and then there exist α, β ∈ Zn such that α + β = γ and aαbβ 6= 0. In this case aα 6= 0 and bβ 6= 0, that is, α ∈ S(f ) and β ∈ S(g). Hence, γ = α + β ∈ S(f ) + S(g). Thus, we proved that
S(f g) ⊂ S(f ) + S(g).
Now we use Proposition 2.7 and we have N (f g) =D
S(f g)E
⊂D
S(f ) + S(g)E
=D S(f )E
+D S(g)E
= N (f ) + N (g).
This completes the proof.
Theorem 4.2. If k is a commutative ring without zero divisors and f, g ∈ k [X, X−1], then
N (f g) = N (f ) + N (g).
Proof. If f = 0 or g = 0, then N (f g) = ∅ and N (f ) + N (g) = ∅, so in this case nothing to prove. Assume that f 6= 0 and g 6= 0. Let f = P
µ
aµXµ and g =P
ν
bνXν. Then f g =P
γ
cγXγ, where cγ = P
µ+ν=γ
aµbν.
First observe that, by Proposition 2.7, we have N (f ) + N (g) =D
S(f ) + S(g)E
Let γ ∈ Zn be a vertex of the polytope N (f ) + N (g). Then γ ∈ S(f ) + S(g) and, by Proposition 3.7, there exist a vertex α of N (f ) and a vertex β of N (g) such that γ = α + β. Let us recall that for each set of generators of a polytope we may choose a minimal set of generators. Thus, it follows from Propositions 3.5 and 3.4, that α ∈ S(f ) and β ∈ S(g). Hence, aα 6= 0 and bβ 6= 0. Consequently, aαbβ 6= 0, because k is without zero divisors. Now we will prove that cγ 6= 0.
Suppose that cγ = 0. Then we have
0 6= aαbβ = − X
(µ,ν)∈W
aµbν,
where W is the set of all pairs (µ, ν) such that µ, ν ∈ Zn, (µ, ν) 6= (α, β) and µ + ν = γ.
This implies that there exists a pair (µ, ν) belonging to W such that aµbν 6= 0. Then aµ 6= 0 and bν 6= 0 and hence, µ ∈ S(f ), ν ∈ S(g). Thus, we have
α, µ ∈ N (f ), β, ν ∈ N (g), α + β = γ = µ + ν, (µ, ν) 6= (α, β).
Observe that λ 6= µ and β 6= ν. Hence, by Proposition 3.8, the point α + β is not a vertex of the polytope N (f ) + N (g). But α + β = γ and, by our assumption, γ is a vertex of N (f ) + N (g). Hence, if cγ = 0, then we have a contradiction. Therefore cγ 6= 0, that is, γ ∈ S(f g).
We proved that each vertex of N (f ) + N (g) belongs to S(f g). Let γ1, . . . , γm be all the vertices of N (f ) + N (g). Then we have
N (f ) + N (g) =D
γ1, . . . , γmE
⊂D
S(f g)E
= N (f g).
Thus, N (f ) + N (g) ⊂ N (f g). The opposite inclusion follows from Proposition 4.1. Note an immediate consequence of this theorem.
Proposition 4.3. If k is a commutative ring without zero divisors and f ∈ k [X, X−1], then
N (fm) = mN (f ) for any integer m > 1.
5 Rational points and rational polytopes
A point x = (x1, . . . , xn) ∈ Rn is called rational if all the coordinates x1, . . . , xn are rational. Since Q is a dense subset of R, every line segment ha, bi in R1, where a, b ∈ R and a < b, contains a rational point. If n > 2, then there exist line segments in Rn without rational points.
Proposition 5.1. Let {a1, . . . , an, b1, . . . , bn}, where n > 2, be a subset of R which is algebraically independent over Q. Such a subset exists, because the transcendence degree of R over Q is infinite. Let a = (a1, . . . , an), b = (b1, . . . , bn). Then the line segment ha, bi in Rn is without rational points.
Proof. Suppose that q = (q1, . . . , qn) is a rational point belonging to ha, bi . Then q = pa + (1 − p)b for some p ∈ R with 0 6 p 6 1. Then q1 = p(a1− b1) + b1, . . . , qn= p(an− bn) + bn and, in particular, we have the equalities
p = q1− b1
a1− b1 = q2 − b2 a2− b2.
Clearly a1 − b1 6= 0 and a2 − b2 6= 0. Thus, (q1 − b1)(a2− b2) + (q2− b2)(b1− a1) = 0, and q1, q2 ∈ Q. Hence F (a1, a2, b1, b2) = 0, where
F (x1, x2, y1, y2) = (q1− y1)(x2− y2) + (q2− y2)(y1− x1)
is a nonzero polynomial belonging to Q[x1, x2, y1, y2]. But it is a contradiction, because the set {a1, a2, b1, b2} is algebraically independent over Q.
A polytope ha1, . . . , ami in Rm is called rational if all the points a1, . . . , am are rational. A polytope ha1, . . . , ami in Rm is called lattice if all the points a1, . . . , amhave integer coordinates. All vertices of rational polytopes are rational points. Similarly, all vertices of lattice polytopes have integer coordinates. Since hSi + hT i = h S + T i (see Proposition 2.7), we have
Proposition 5.2. The Minkowski sum of two rational polytopes is a rational polytope.
The Minkowski sum of two lattice polytopes is a lattice polytope.
It is well known (see for example [2] or [1]) that a nonempty subset A of Rn is a polytope in Rn if and only if A is bounded and A is the intersection of a finite set of closed halfspaces. A polytope is rational if and only if its determining closed halfspaces are defined by hyperplanes with rational coefficients. As an immediate consequence of these facts we obtain
Proposition 5.3. Let A, B be polytopes in Rn such that A ∩ B 6= ∅. Then A ∩ B is a polytope in Rn. If the polytopes A, B are rational, then the polytope A ∩ B is also rational.
References
[1] D. Cox, J. Little, D. O’Shea, Using Algebraic Geometry, Graduate Texts in Math- ematics 185, Springer, 1998.
[2] G. M. Ziegler, Lectures on Polytopes, Graduate Texts in Mathematics 152, Springer, 2006.
Nicolaus Copernicus University, Faculty of Mathematics and Computer Science, 87-100 Toru´n, Poland, (e-mail: anow@mat.uni.torun.pl).
