ComputationalTheoryofPolynomialIdeals Eidgen¨ossischeTechnischeHochschuleZ¨urich

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Computational Theory of Polynomial Ideals

a Bachelor Thesis

written by Paul Steinmann

supervised by Prof. Dr. Richard Pink

Abstract

We provide methods to do explicit calculations in a polynomial ring in finitely many variables over a field. We develop algorithms to compute quotients of ideals, the radical of an ideal and a primary decomposition of an ideal. We also present methods to solve explicit tasks such as testing for prime ideals or identical ideals.

February 2015

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Polynomial rings over fields arise in many areas of mathematics, for instance in algebraic geometry. It is natural to ask to which extent explicit computation in such polynomial rings are possible. Today, computer algebra systems like MAPLE or MATHEMATICA are very useful tools to test conjectures in special cases before trying to prove them. There is a lot of computational theory for algorithms used to compute even the simplest objects in a polynomial ring. Computations in one variable are fairly simple, due to the fact that a polynomial ring in one variable over a field is a principal ideal domain and hence the very powerful Euclidean algorithm can be used. However, computations become more complicated when dealing with multivariate polynomial rings. Some of the computational theory will be presented and explained in this thesis. Certain objects, for example the radical of an ideal, can be computed using different characterisations, which result in different algorithms. There will be a short discussion of this fact and a comparison between algorithms at the end.

A key concept in this setting of computational algebra is the one of Gr¨obner bases. Presented in section 3, this concept will be used throughout the whole thesis. We lay our focus on three objects associated to polynomial ideals: quotients of ideals, the radical of an ideal and a primary decomposition of an ideal. We develop algorithms to compute these objects. There are also several other tasks that we will be able to do computationally. The main algorithms are marked as “Algorithm” and are written in pseudo-code, whereas the simpler tasks such as testing for prime ideals are marked as “Task” and have no special syntactical structure.

In the end we will be able to do almost every computation that a standard computer algebra system can do in the area of polynomial ideals.

The prerequisites for this thesis are a standard algebra course and some basic knowledge of commutative algebra, mainly radical ideals and primary decompositions. Everything needed can be found in the first few chapters of Atiyah and Macdonald’s standard introduction [1].

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2 Notation

We will always consider a field K and denote by K its algebraic closure. Let X₁, . . . , X_n be n independent variables and defineX:={X₁, . . . , X_n}. We denote the polynomial ring inn variables overK by K[X]. Later on, we will invert some of the variables to form a rational function field. In doing so, we will write X = Y tZ and mean that Y and Z are disjoint subsets of X and their union is X. We can then, for instance, form the ring K(Y)[Z] of polynomials in Z over the rational function field K(Y). The integer n > 0 is arbitrary but will always denote the total number of variables.

For computations with an ideal, we need a finite generating set of the ideal. Thus, in the algorithms and tasks we tacitly assume that a finite generating set of I is given. For a finite subsetF ⊂K[X] we denote by hFi ⊂K[X] the ideal generated byF. If the ring in which the ideal is generated is not clear from the context, we indicate the ring as a subscript. For instance, we will write hFi_K(Y_)[_Z_]. For elements f1, . . . , fm ∈ K[X] we write hf₁, . . . , fmi instead ofh{f₁, . . . , f_m}i.

Let R and S be two rings and ϕ :R → S be a ring homomorphism. LetI ⊂ R and J ⊂S be ideals. We use Atiyah and Macdonald’s ([1]) notation and write Iê := hϕ(I)i for the extension ideal of I and J^c := ϕ⁻¹(J) for the contraction ideal of J. We will not state the homomorphism explicitly if it is clear from the context. We also use the notationIêc := (Iê)^c and J^ce:= (I^c)ê.

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This section follows the book of Cox, Little and O’Shea [3].

3.1 Basic definitions

We will generalize the concept of degree and the Euclidean algorithm for polynomial rings in one variable to an algorithm for a ring of multivariate polynomials K[X].

Definition 3.1. Amonomial orderingonK[X] is a binary relationon Zⁿ≥0 with the following properties:

(i) The relationis a total ordering on Zⁿ≥0.

(ii) For allα, β, γ∈Zⁿ_≥0 such thatα β it follows that α+γ β+γ.

(iii) Every non-empty subset has a smallest element, i.e. is a well-ordering on Zⁿ≥0. Example 3.2. The standard ordering relation ≥ on Z≥0 is a total ordering and a well- ordering. Furthermore, for all α, β, γ ∈ Z≥0 such that α ≥ β we have α +γ ≥ β +γ.

Therefore≥ is a monomial ordering onK[X₁].

Example 3.3. We define the monomial ordering _lex on K[X] such that for all α :=

(α₁, . . . , α_n), β := (β₁, . . . , β_n) ∈ Zⁿ≥0 we have α _lex β if and only if the leftmost non- zero entry in (α₁ −β₁, . . . , α_n−β_n) is positive. This is called the lexicographical ordering. For instance, in the lexicographical ordering we have (0,1,5,2) _lex (0,1,2,4) since (0,1,5,2)−(0,1,2,4) = (0,0,3,−2) and the leftmost non-zero entry 3 is positive. Using the result of Example 3.2 we deduce that _lex is a monomial ordering.

Definition 3.4. Let f = P

αa_αX^α ∈ K[X] with multi-index notation and let be a monomial ordering on K[X].

(i) Thedegree off with respect tois deg(f) := max{α∈Zⁿ≥0 |a_α6= 0}iff is non-zero and deg(0) :=−∞. Here, the maximum is taken with respect to .

(ii) Theleading coefficient off is LC(f) :=a_deg(f) ∈K iff is non-zero, and LC(0) := 0.

(iii) Theleading monomialof f is LM(f) :=X^deg(f) iff is non-zero, and LM(0) := 1.

(iv) The leading termof f is LT(f) := LC(f)·LM(f).

Remark 3.5. Later on, we will need to distinguish leading terms in different rings. For this purpose, ifX =Y tZ and f ∈K[X] we write LT_Z(f) for the leading term off in the ring K(Y)[Z] and LTX(f) for the leading term in K[X]. We denote LC(f) analogously.

Definition 3.6. For any subset F ⊂ K[X] we define the set of leading terms LT(F) :={LT(f)|f ∈F} and the set of leading monomials LM(F) :={LM(f)|f ∈F}.

Now we can define Gr¨obner bases. There are several equivalent definitions and the one below is not the most intuitive one. However, this technical property is easier to use:

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Section 3 3.1 Basic definitions Definition 3.7. Let be a monomial ordering onK[X] and let I ⊂K[X] be an ideal. A Gr¨obner basisofI is a finite subsetG⊂I such thathLT(G)i=hLT(I)i.

Remark 3.8. From now on, we will always assume that some monomial orderinghas been chosen and that all Gr¨obner bases and leading terms are taken with respect to this ordering unless stated otherwise.

In order to use Gr¨obner bases we need some basic statements.

Lemma 3.9. Let I ⊂K[X]be an ideal and let f, g1, . . . , gm ∈I. If LT(f) lies in the ideal hLT(g₁), . . . ,LT(g_m)i, thenf = 0or there is some1≤k≤msuch that LT(g_k)divides LT(f).

Proof. By assumption, there exist h1, . . . , hm ∈ K[X] such that LT(f) = Pm

i=1hiLT(gi).

For each 1 ≤ i ≤ m write hi = P

αaⁱ_αX^α with aⁱ_α ∈ K for all α. Then LT(f) = Pm

i=1

P

αaⁱ_αX^αLT(g_i). If f is non-zero, then LT(f) is a non-zero monomial. It follows that, on the right hand side, there is only one non-zero monomial. Thus there is a constant β ∈K such that LT(f) =βa^k_αX^αLT(g_k) for some 1 ≤k ≤m and some α. Hence LT(f) is divisible by LT(g_k) or f = 0.

Proposition 3.10. Every idealI ⊂K[X]has a Gr¨obner basis. Furthermore, every Gr¨obner basis of I is a generating set of I.

Proof. (i) Let M := {hLT(F)i | F ⊂ I is a finite subset}. Since K[X] is a Noetherian ring, the ideal hLT(I)i is Noetherian as a module over K[X]. Therefore M has a maximal element hLT(G)i ∈ M, where G ⊂ I is a finite subset. Let f ∈ I. Since hLT(G)i is a maximal element of M, we have that hLT(G)i = hLT(G∪ {f})i. Hence LT(f)∈ hLT(G)i and Gis thus a Gr¨obner basis of I.

(ii) IfI =h0i, the only possible Gr¨obner bases are∅ and {0}. Both of them generate I.

LetGbe a Gr¨obner basis ofI 6=h0i. We prove thatI =hGiby transfinite induction on the degree of elements inI. Iff = 0∈I thenf ∈ hGi. So letf ∈I be non-zero. Assume that for allf⁰ ∈I the condition deg(f)deg(f⁰) implies that f⁰ ∈ hGi. By definition of a Gr¨obner basis, we have LT(f) ∈ hLT(G)i. By Lemma 3.9 there exists an element g ∈ G such that LT(g) divides LT(f). Thus LT(f) = hLT(g) for some h ∈ K[X].

Since LT(f) and LT(g) consist of only one term and are both non-zero, the polynomial h has the same property. Hence h = LT(h) and LT(f) = LT(h)LT(g) = LT(hg). We have deg(f) deg(f−hg) by definition of the leading term. By induction hypothesis f −hg∈ hGi and hencef ∈ hGi. The claim follows with transfinite induction.

We turn to the generalization of the Euclidean algorithm.

Definition 3.11. Let f₁, . . . , f_m ∈K[X] and define F :={f₁, . . . , f_m}. Let f ∈K[X]. A remainder on division of f with respect to F is an element f^F ∈K[X] such that there exist h₁, . . . , h_m ∈K[X] with the following properties:

(i) The polynomialf can be written asf =Pm

i=1hifi+f^F. (ii) No term of f^F is divisible by any of (LT(fi))^m_i=1.

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(iii) For all 1≤i≤mwe have deg(hifi)deg(f).

Algorithm 3.12. Letf1, . . . , fm ∈K[X] and defineF :={f₁, . . . , fm}. Letf ∈K[X]. The following algorithm computes a remainder on division f^F and elements h₁, . . . , h_m ∈K[X] as in Definition 3.11.

Input: F = (f₁, ..., f_m), f ∈K[X]

Output: r=f^F, some remainder on division; h₁, . . . , h_m begin

hi := 0 for all 1≤i≤m r:= 0

p:=f

whilep6= 0 do i:= 1

divisionoccured := false

whilei≤m and divisionoccured = false do if LT(f_i) divides LT(p) then

h_i :=h_i+ LT(p)/LT(f_i) p:=p−(LT(p)/LT(fi))fi

divisionoccured := true else

i:=i+ 1 end

end

if divisionoccured = false then r:=r+ LT(p)

p:=p−LT(p) end

end end

Proof. We first show that the algorithm terminates after finitely many steps. Note that the inner while-loop has at most m steps and thus terminates in every step of the outer while- loop. So the only possibility that the algorithm does not terminate is if p is non-zero at all times. If LT(f_i) divides LT(p) for some 1 ≤ i ≤ m, then in the next step p is reduced to p−(LT(p)/LT(fi))fi. Then deg(p−(LT(p)/LT(fi))fi) ≺ deg(p). If on the other hand no LT(f_i) divides LT(p), then the inner while-loop terminates with divisionoccured = f alse.

This implies that p is reduced to p−LT(p) for the next step, due to the if-statement at the bottom. Then deg(p−LT(p)) ≺ deg(p). So in each step of the outer while-loop, the degree of p strictly decreases. If the loop did not terminate, we would have an infinitely decreasing sequence. This cannot exist because of the well-ordering condition of monomial orderings. Thus the algorithm terminates after finitely many steps. For later use, note that deg(p)deg(f) during the whole algorithm.

We now show that after every step of the outer while-loop we have f = Pm

i=1h_if_i+r +p with deg(hifi) deg(f) for all 1 ≤ i ≤ m. This is true before the first step. Now fix a

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Section 3 3.2 Computing Gr¨obner Bases step of the outer while-loop. Let r, p,(hi)^m_i=1 denote the elements before the step which have the desired property and let r⁰, p⁰,(h⁰_i)^m_i=1 denote the elements after the step. If no LT(f_i) divides LT(p) then h⁰_i =hi for all 1 ≤i ≤ m. Also r⁰ = r+ LT(p) and p⁰ =p−LT(p). It follows that f = Pm

i=1h⁰_ifi+r⁰ +p⁰ and deg(h⁰_ifi) deg(f) for all 1 ≤ i ≤ m. If, on the other hand, some LT(f_i) divides LT(p), then h⁰_j =h_j for all j 6=i and r⁰ =r. Furthermore h⁰_i = hi + LT(p)/LT(fi) and p⁰ = p −(LT(p)/LT(fi))fi, so h⁰_ifi +p⁰ = hifi +p. Thus f =Pm

i=1h⁰_if_i+r⁰+p⁰ and deg(h⁰_if_i) max{deg(h_if_i),deg(p)} deg(f) for all 1≤i≤m.

Because the property holds after each step, it also holds for the output.

It remains to show that the outputr is not divisible by any of (LT(f_i))^m_i=1. At the beginning r = 0, and the only place where a term is added to r is in the if-statement at the bottom.

But the term is only added if it was not divisible by any of (LT(f_i))^m_i=1. So r has the desired property and is thus a remainder on division.

Remark 3.13. Algorithm 3.12 shows that for all finite subsets F ⊂ K[X] and elements f ∈ K[X] there is a remainder on division of f with respect to F. However, in general it need not be unique. Whenever we use a remainder on division f^F, we assume that some choice has been made. This choice will only affect the explicit calculations and not the general statements.

Proposition 3.14. Let G={g₁, . . . , g_m} be a Gr¨obner basis of an ideal I ⊂K[X]. For all f ∈K[X], there is a unique remainder on division of f with respect to G.

Proof. Letf ∈K[X] and letr, r⁰ ∈K[X] be two polynomials both satisfying the properties of a remainder on division off with respect toG. Then we haver−r⁰ = (f−r⁰)−(f−r)∈I.

Assume that r−r⁰ 6= 0. By Lemma 3.9, there exists an element gi ∈ G such that LT(gi) divides LT(r−r⁰). Sincer−r⁰6= 0, this implies that some term ofr or ofr⁰ must be divisible by LT(g_i), which is not true. Sor−r⁰ = 0.

The following proposition was, historically, the motivation to introduce Gr¨obner bases. In fact, it is one of the equivalent definitions of a Gr¨obner basis.

Proposition 3.15. Let I ⊂K[X]be an ideal and G={g₁, . . . , gm} ⊂I a Gr¨obner basis of I. Then a polynomialf ∈K[X] lies in I if and only if f^G= 0.

Proof. Letf ∈I. Thenf^G∈I by definition of a remainder on division, and no term off^Gis divisible by any element of LT(G). ButGis a Gr¨obner basis, so LT(f^G)∈ hLT(I)i=hLT(G)i.

It follows by Lemma 3.9 that f^G = 0.

Conversely, iff^G= 0, then f is a linear combination of elements of G, sof ∈I.

3.2 Computing Gr¨obner Bases

The key to an algorithm which computes a Gr¨obner basis is a technical criterion for Gr¨obner bases developed by Buchberger.

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Definition 3.16. Let f, g ∈ K[X] be non-zero and let α := deg(f) and β := deg(g). Set γ := (γ₁, . . . , γ_n) whereγ_i:= max(α_i, β_i) for each 1 ≤i≤n. The S-polynomialof f andg is defined by

S(f, g) := X^γ

LT(f)f− X^γ LT(g)g, using multi-index notation.

Lemma 3.17. Let f1, . . . , fs ∈ K[X] be all non-zero with δ := deg(f1) = · · · = deg(fs).

For all i6=j we have deg(S(f_i, f_j))≺deg(f_i). Let c₁, . . . , c_s ∈K satisfy deg(Ps

i=1c_if_i)≺δ.

Then there are coefficientsdij ∈Kfor all1≤i, j≤ssuch thatPs

i=1cifi =P

i6=jdijS(fi, fj).

Proof. For all 1 ≤i≤s definepi := _LC(f^fⁱ

i) and di := ciLC(fi). Then for alli6=j we have S(f_i, f_j) = _LT(f^X^δ

i)f_i − _LT(f^X^δ

j)f_j = p_i−p_j. Since every p_i has leading coefficient 1 and they all have the same degree we find that deg(pi −pj) ≺ deg(pi) = deg(fi), proving the first statement of the lemma. Now consider the following sum:

s

X

i=1

c_if_i=

s

X

i=1

d_ip_i=

s

X

i=1

d_i(p_i−p₁) +

s

X

i=1

d_i

! p₁ (1)

Since deg(Ps

i=1c_if_i) ≺ δ we know that Ps

j=1d_j = 0, so the second sum of the right hand side of equation (1) vanishes. Thus Ps

i=1cifi =Ps

i=1diS(fi, f1).

Theorem 3.18 (Buchberger’s Criterion).

LetI ⊂K[X]be an ideal and letG={g₁, . . . , g_m} ⊂I be a generating set ofI not containing 0. Then Gis a Gr¨obner basis of I if and only if ∀i6=j: S(gi, gj)^G= 0.

Proof. Assume thatG is a Gr¨obner basis. Then by definition of theS-polynomial, we have S(gi, gj)∈I for alli6=j. ThusS(gi, gj)^G = 0 by Proposition 3.15. This proves one direction of the equivalence.

Conversely, assume that S(gi, gj)^G = 0 for all i 6= j. Using Definition 3.7 of a Gr¨obner basis, we need to show that LT(f) ∈ hLT(G)i for all f ∈ I. If f is zero, this is true, so let f ∈ I be non-zero. Since G generates I, there exist h1, . . . , hm ∈ K[X] such that f =P_m

i=1h_ig_i. Since a monomial ordering is a well-ordering, we can choose the h_i such that δ := max{deg(h₁g₁), . . . ,deg(h_mg_m)} is minimal. Set d(i) := deg(h_ig_i) for all 1≤i≤m and write

f = X

d(i)=δ

LT(hi)gi+ X

d(i)=δ

(hi−LT(hi))gi+ X

d(i)≺δ

higi

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Note that only the first sum has terms of degree equal to δ. We claim that deg(f) = δ.

Suppose deg(f) ≺ δ. By Lemma 3.17, there exist coefficients c_ij ∈ K for i 6= j such that P

d(i)=δLT(h_i)g_i = P

i6=jc_ijS(LT(h_i)g_i,LT(h_j)g_j), where c_ij = 0 whenever d(i) ≺ δ

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Section 3 3.2 Computing Gr¨obner Bases ord(j)≺δ. For all i6=j with d(i) =d(j) =δ we have

S(LT(h_i)g_i,LT(h_j)g_j) = X^δ

LT(higi)LT(h_i)g_i− X^δ

LT(hjgj)LT(h_j)g_j =X^δ−γîjS(g_i, g_j) for someγij ∈Zⁿ_≥0. By assumptionS(gi, gj)^G= 0 for alli6=j. Hence there existr_`îj ∈K[X] for all 1 ≤`≤m and i6=j such that deg(r_`îjg_`) deg(S(g_i, g_j)) and S(g_i, g_j) =P_m

`=1r_`^ijg_`. We thus find:

X

d(i)=δ

LT(hi)gi=X

i6=j

cijS(LT(hi)gi,LT(hj)gj) =X

i6=j m

X

`=1

cijX^δ−γ^ijr_`^ijg_` (3)

Furthermore, for all 1≤`≤m and i6=j we have

deg(X^δ−γîjrîj_` g_`)deg(X^δ−γîjS(gi, gj)) = deg(S(LT(hi)gi,LT(hj)gj))≺deg(higi) where the last inequality follows from Lemma 3.17. Combining equation (2) and (3) yieldsf as a linear combination of (g_i)^m_i=1 such that every term has degree strictly less than δ. This contradicts the minimality of δ, and thus the degree of f must equal δ. So LT(gk) divides LT(f) for some 1≤k≤m and therefore LT(f)∈ hLT(G)i.

Using Buchberger’s criterion, we can now check whether a given generating set of an ideal is a Gr¨obner basis. More importantly, we have an algorithm which computes a Gr¨obner basis of a given ideal.

Algorithm 3.19 (Buchberger). Let I ⊂K[X] be an ideal and a monomial ordering on K[X]. The following algorithm computes a Gr¨obner basis of I with respect to.

Input: F = (f₁, ..., f_m), a generating set of I Output: G= a Gr¨obner basis ofI w.r.t.

begin G:=F repeat

G⁰ :=G

foreachpair p, q∈G⁰ with p6=q do S:=S(p, q)^G

0

with respect to (Algorithm 3.12) if S6= 0 then

G:=G⁰∪ {S}

end end untilG=G⁰ end

Proof. Assume that the algorithm never terminates. This is only possible if G and G⁰ never coincide in the outer loop. This implies S = S(p, q)^G

0

6= 0 for some p, q ∈ G⁰ with

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p6=q. By definition of a remainder on division and Lemma 3.9, we have LT(S)∈ hLT(G/ ⁰)i.

But LT(S) ∈ hLT(G)i and G⁰ ⊂ G. Thus hLT(G⁰)i ( hLT(G)i. So if the algorithm never terminates, there is a strictly increasing sequence of ideals in K[X]. But this contradicts K[X] being Noetherian. So the algorithm must terminate after finitely many steps.

Now consider the outputGof the algorithm. The algorithm has terminated, soG⁰ =Gin the outer loop. IfS :=S(p, q)^G

0

6= 0 for some p, q∈G⁰, thenS /∈G⁰ by definition of a remainder on division. Thus G⁰ ( G = G⁰ ∪ {S}. This is a contradiction. So when the algorithm terminates, we have S(p, q)^G = 0 for all p, q ∈ G. Note also that F ⊂ G ⊂ I at all times, so G is always a generating set of I. It follows by Theorem 3.18 that G is a Gr¨obner basis of I.

Task 3.20 (Ideal Membership). Let I ⊂ K[X] be an ideal and let f ∈ K[X]. Is f an element ofI?

Solution. Compute a Gr¨obner basis Gof I using Algorithm 3.19. Compute f^G using Algo- rithm 3.12. Thenf ∈I ⇐⇒ f^G= 0 by Proposition 3.15.

By requiring more structure on a Gr¨obner basis we obtain even stronger statements.

Definition 3.21. A Gr¨obner basis is called reducedif (i) Every elementg∈Gis monic, i.e. LC(g) = 1.

(ii) For allg∈G, no term of g lies inhLT(G\ {g})i.

Algorithm 3.22. LetI ⊂K[X] be an ideal andGa Gr¨obner basis ofI. Then the following algorithm computes a reduced Gr¨obner basis G⁰ ofI.

Input: G, a Gr¨obner basis ofI

Output: G⁰, a reduced Gr¨obner basis of I begin

G⁰ :=G

foreachg∈G do

if LT(g)∈ hLT(G\ {g})ithen G⁰ :=G⁰\ {g}

else

g⁰ :=g^G\{g} (Algorithm 3.12) G⁰ := (G⁰\ {g})∪ {_LC(g^g⁰ 0)} end

end end

Proof. For all g ∈ G the condition LT(g) ∈ hLT(G\ {g})i is equivalent to LT(g⁰) dividing LT(g) for some g⁰ ∈G\ {g}, by Lemma 3.9. This can be checked algorithmically.

Let G={g₁, . . . , gm} be the input Gr¨obner basis ofI. Then the algorithm terminates after m steps. We need to show that the output G⁰ is indeed a reduced Gr¨obner basis of I.

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Section 3 3.2 Computing Gr¨obner Bases Note that for any g ∈Gwith LT(g) ∈ hLT(G\ {g})i the setG\ {g} is still a Gr¨obner basis of I.

Let 1 ≤ k ≤ m with LT(g_k) ∈ hLT(G/ \ {g_k})i and define g⁰ := g_k^G\{g^k^}. Then there exist h_i ∈K[X] for alli6=k such that g_k=P

i6=kh_ig_i+g⁰ and deg(h_ig_i)deg(g_k) for all i6=k.

Since LT(g_k)∈ hLT(G/ \ {g_k})iwe find that deg(higi) is even strictly smaller than deg(g_k) for all i 6= k and hence LT(g_k) = LT(g⁰). Then LT

(G\ {g_k})∪ {_LC(g^g⁰ 0)}

= LT(G) and thus (G\ {g_k})∪ {_LC(g^g⁰ 0)}is still a Gr¨obner basis of I.

Hence G⁰ is a Gr¨obner basis of I at all times in the algorithm. Furthermore, the above argument shows that no term of g lies in hLT(G⁰\ {g})i for allg ∈ G⁰. Also, every element of G⁰ has leading coefficient 1. Hence G⁰ is a reduced Gr¨obner basis ofI.

Proposition 3.23. Let I ⊂ K[X] be an ideal. For every monomial ordering, there is a unique reduced Gr¨obner basis of I.

Proof. The existence follows from the existence of a Gröbner basis and Algorithm 3.22. Let G and G⁰ be two reduced Gröbner bases of I and let g ∈ G. In particular g is a non-zero element of I. Since G⁰ is a Gröbner basis of I, we have LT(g) ∈ hLT(G⁰)i. By Lemma 3.9, there exists an elementg⁰∈G⁰ such that LT(g⁰) divides LT(g). Using the same argument for g⁰ and G we obtain an element ˜g ∈G such that LT(˜g) divides LT(g⁰). Hence LT(˜g) divides LT(g) and thus g= ˜g since G is a reduced Gröbner basis. We deduce that LT(g) = LT(g⁰).

We will show thatg=g⁰.

Note thatg−g⁰∈Iand deg(g−g⁰)≺deg(g) = deg(g⁰). Assume thatg−g⁰is non-zero. By the same argument as before, there exists an element f ∈G such that LT(f) divides LT(g−g⁰).

Then deg(f)deg(g−g⁰). This implies that LT(f) divides some term ofgor some term ofg⁰. The first case is not possible, becauseGis reduced and deg(f)≺deg(g), sof 6=g. Therefore, LT(f) divides some term of g⁰. Repeating once again the above argument, we obtain f⁰∈G⁰ such that LT(f⁰) divides LT(f). Then LT(f⁰) divides some term of g⁰. This implies that f⁰ = g⁰ because G⁰ is reduced. But this is impossible since deg(f⁰) deg(f) ≺ deg(g⁰).

Therefore the assumption that g−g⁰ is non-zero was false. Hence G ⊂G⁰ and analogously G⁰ ⊂G.

Task 3.24 (Ideal Equality). LetI, J be two ideals ofK[X]. Do I andJ coincide?

Solution. Compute reduced Gr¨obner bases G and G⁰ of I and J, respectively, by means of Algorithm 3.19 and Algorithm 3.22. Then I =J ⇐⇒ G=G⁰ by Proposition 3.23.

Task 3.25 (Subideal). LetI, J ⊂K[X] be two ideals. Is I a subideal ofJ?

Solution. For every generator f of I check whether f ∈J using Task 3.20.

Lemma 3.26. LetX =YtZ and letI ⊂K[Y ][Z]be an ideal. Letrdenote the cardinality of Y. Letbe the lexicographical ordering onK[Y ][Z]such thatdeg(Zn−r) · · · deg(Z1) deg(Y_r) · · · deg(Y₁). LetG⊂I be a Gr¨obner basis ofI with respect to. ThenG∩K[Y ] is a Gr¨obner basis ofI ∩K[Y ].

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Proof. Set G⁰ :=G∩K[Y ]. Note that G⁰ ⊂I∩K[Y ] and G⁰ is finite. Letf ∈I∩K[Y ] be non-zero. Since G is a Gr¨obner basis LT(f) is an element of hLT(G)i. By Lemma 3.9, there exists an element g∈Gsuch that LT(g) divides LT(f). In particular, this implies that LT(g)∈K[Y ]. But by our choice of lexicographical ordering this implies that no term ofg contains any variable ofZ. Thus g∈G⁰ and LT(f)∈ hLT(G⁰)i. HenceG⁰ is a Gr¨obner basis of I∩K[Y ].

Task 3.27. LetX =Y tZ and letI ⊂K[Y ][Z] be an ideal. Compute a Gr¨obner basis G of I∩K[Y ] with respect to a monomial ordering⁰.

Solution. Letrdenote the cardinality ofY. Letbe the lexicographical ordering onK[Y][Z] as in Lemma 3.26. Compute a Gröbner basis G⁰ ⊂I ofI with respect to using Algorithm 3.19. Compute ˜G:=G⁰∩K[Y ] by ignoring every polynomial of G⁰ with a term containing any variable of Z. Then ˜G is a Gröbner basis of I ∩K[Y ], by Lemma 3.26, and thus a generating set ofI∩K[Y ]. Compute a Gröbner basisGofI∩K[Y ] with respect to⁰ using Algorithm 3.19.

Lemma 3.28. Let I = hf₁, . . . , f_mi and J = hh₁, . . . , h_`i be two ideals of K[X]. Define L:=htf₁, . . . , tfm,(1−t)h1, . . . ,(1−t)h`i ⊂K[X][t]. Then I∩J =L∩K[X].

Proof. “⊂”: Let f ∈I∩J. Then tf ∈L and (1−t)f ∈L, so f ∈L. Since I ⊂K[X] we have f ∈L∩K[X].

“⊃”: For each c ∈ K consider the homomorphism ϕ_c : K[X][t] → K[X], f(t) 7→ f(c).

Let f ∈L∩K[X]. Then there exist polynomialsa1, . . . , am, b1, . . . , b` ∈K[X][t] such that f =Pm

i=1aitfi+P`

j=1bj(1−t)hj. Note thatϕ0(f) =f =ϕ1(f) sincef is independent oft.

Then ϕ₀(f) =P`

j=1b_jh_j ∈J and similarlyϕ₁(f)∈I. Hence f ∈I∩J.

Task 3.29 (Intersection). LetI1, . . . , Im ⊂K[X] be ideals with given generators. Compute a Gr¨obner basis ofTm

i=1I_i.

Solution. By induction we only need to consider the casem= 2. Assume thatI₁ =hf₁, . . . , f_ki and I2 = hh₁, . . . , h_ì. Set L := htf₁, . . . , tf_k,(1−t)h1, . . . ,(1−t)h_ì ⊂ K[X][t]. Compute a Gröbner basis G of L∩K[X] using Task 3.27. Then G is a Gröbner basis of I₁ ∩I₂ by Lemma 3.28.

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4 Quotients of Ideals

The following section was inspired by the book of Cox, Little and O’Shea [3].

Definition 4.1. LetI, J ⊂K[X] be two ideals. The ideal quotientofI with respect toJ is (I :J) :={a∈K[X]|aJ ⊂I}. Forf ∈K[X] we write (I :f) instead of (I :hfi).

Remark 4.2. For two ideals I, J ⊂K[X] their ideal quotient (I :J) is an ideal of K[X].

Lemma 4.3. Let I, J ⊂ K[X] be two ideals with J = hf₁, . . . , fmi. Then (I : J) = Tm

i=1(I :f_i).

Proof. “⊂”: Let g∈(I :J). Then gJ ⊂I, so in particular gfi ∈I for all 1≤i≤m. Hence g∈Tm

i=1(I :f_i).

“⊃”: Let g ∈Tm

i=1(I :fi). Then gfi ∈I for all 1≤i≤m. Since every h∈J is of the form h = Pm

i=1α_if_i for polynomials α₁, . . . , α_m ∈ K[X], we have gh ∈ I for all h ∈ J. Hence gJ ⊂I.

Lemma 4.4. Let f ∈ K[X] be a non-zero polynomial and let I ⊂K[X] be an ideal. Let g1, . . . , gm ∈I∩ hfi be generators of I∩ hfi. Then hg₁/f, . . . , gm/fi= (I :f).

Proof. “⊂”: First, we note that each gi is divisible by f since g1, . . . , gm ∈ hfi. Hence g₁/f, . . . , g_m/f are indeed polynomials in K[X]. Furthermore, for all 1 ≤ i ≤ m we have (gi/f)hfi=hg_ii ⊂I, so (gi/f)∈(I :f).

“⊃”: Let h ∈ (I : f). Then hf ∈ I ∩ hfi and we can write hf = Pm

i=1αigi for some α₁, . . . , α_m ∈ K[X]. Dividing by f yieldsh = P_m

i=1α_ig_i/f and thus h ∈ hg₁/f, . . . , g_m/fi.

Algorithm 4.5. Let I and J be two ideals of K[X]. The following algorithm computes a Gr¨obner basis Gof (I :J).

Input: a generating set (f₁, . . . , f_m) of I and a generating set (g₁, . . . , g_r) ofJ not containing zero

Output: G, a generating set of (I :J) begin

for1≤i≤r do

H_i := generating set ofI∩ hg_ii(Task 3.29) Gi :={g/g_i |g∈Hi} (Division Algorithm 3.12) end

G:= Gr¨obner basis ofTr

i=1hG_ii (Task 3.29) end

Proof. The for-loop has r steps, so the algorithm terminates after finitely many steps. By Lemma 4.4, the set Gi is a generating set of (I : gi) for all 1 ≤ i ≤ r. By lemma 4.3, the

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intersection Tr

i=1hG_ii = Tr

i=1(I : gi) is precisely the ideal quotient (I : J). Hence G is a Gr¨obner basis of (I :J).

The following task will be useful later.

Task 4.6. Let I ⊂ K[X] be an ideal and f ∈ K[X]. Compute an integer s such that (I :f^s) =S

i≥1(I :fⁱ).

Solution. For all i≤ j we have the inclusion (I :fⁱ) ⊂ (I :f^j). Furthermore, if (I : f^k) = (I :f^k+1) for somek≥1, then (I :f^k+1) = (I :f^k+2). To see this, let a∈(I :f^k+2). Then f a∈(I :f^k+1) = (I :f^k), sof^k+1a∈I. Hence a∈(I :f^k+1).

Compute ideals (I :f),(I :f²), . . . using Algorithm 4.5 until (I :f^s) = (I :f^s+1) for some s ≥ 1. This must occur eventually, since K[X] is Noetherian. The argument above then yieldsS

i≥1(I :fⁱ) =Ss

i=1(I :fⁱ) = (I :f^s).

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5 Radicals

The ideas in this section are mainly found in the book of Becker and Weispfenning [2]. Some ideas around Seidenberg’s Lemma were inspired by the book of Kreuzer and Robbiano [5].

In all of this section K is a field of characteristic zero.

5.1 Basics

Proposition 5.1. Let I = hf₁, . . . , f_mi ⊂ K[X] be an ideal and let f ∈ K[X]. Then f ∈Rad(I) if and only if hf₁, . . . , fm,1−Zfi_K[_X_][Z]=K[X][Z].

Proof. “⇒”: Assume that f ∈ Rad(I). Then there exists n > 0 such that fⁿ ∈ I. Then fⁿ⁻¹ =Zfⁿ+ (1−Zf)fⁿ⁻¹ ∈ hf₁, . . . , f_m,1−Zfi_K[_X_][Z]. Inductively this yields f⁰ = 1∈ hf₁, . . . , fm,1−Zfi_K[X_][Z].

“⇐”: Iff = 0, thenf ∈Rad(I). So assume thatf is non-zero. Leth(Z), h1(Z), . . . , hm(Z)∈ K[X][Z] be such that 1 =Pm

i=1fi·hi(Z) + (1−Zf)·h(Z) and setZ := 1/f in the rational function field K(X). Since f₁, . . . , f_m are independent of Z, we have 1 =Pm

i=1f_i·h_i(1/f)∈ K(X). Let k := max{deg_Z(hi) | 1 ≤ i ≤ m}. Then ehi := f^k ·hi(1/f) ∈ K[X] for all 1≤i≤mand thus f^k=P_m

i=1eh_if_i ∈I.

Task 5.2(Radical Membership). LetI =hf₁, . . . , fmi ⊂K[X] be an ideal and letf ∈K[X].

Is f an element of Rad(I)?

Solution. Compute a Gr¨obner basis of L := hf₁, . . . , f_m,1−Zfi_K[_X_][Z] ⊂ K[X][Z] using Algorithm 3.19. Then check if 1 ∈ L with Task 3.20. By Proposition 5.1, we have f ∈Rad(I) ⇐⇒ 1∈L.

We will see that the key to computing the radical of an ideal is the square-free part of a polynomial:

Definition 5.3. Let f ∈ K[Z] be a non-zero univariate polynomial. Let f = aQ` i=1g_i^rⁱ be the unique factorization of f into monic pairwise non-equivalent irreducible polynomials g₁, . . . , g_` ∈K[Z] and a∈K^×. The square-free part off is defined as Q`

i=1g_i.

A polynomialf ∈K[Z] is said to besquare-freeif it is non-zero and equal to its square-free part.

Lemma 5.4. Let f ∈K[Z]be non-constant. Let f⁰ denote the formal derivative of f. Then f is square-free if and only ifgcd(f, f⁰) = 1.

Proof. “⇐”: Assume that gcd(f, f⁰) = 1 and that f is not square-free. Then f =f₁²f2 for some f₁, f₂ ∈ K[Z], where f₁ ∈/ K. Then f⁰ = 2f₁f₁⁰f₂ +f₁²f₂⁰, so f₁|gcd(f, f⁰). This is a contradiction to gcd(f, f⁰) = 1.

“⇒”: Assume that gcd(f, f⁰) 6= 1 and let h ∈ K[Z] be an irreducible divisor of gcd(f, f⁰).

Then there exists a polynomial a∈K[Z] such that ah=f and thusa⁰h+ah⁰ =f⁰. Since h

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dividesf⁰ it follows thathmust divideah⁰. Note thathdoes not divideh⁰ since char(K) = 0 and h is irreducible, so deg(h⁰) = deg(h)−1 ≥ 0. This implies that h divides a. Hence h² dividesf andf is not square-free.

Claim 5.5. Let f ∈K[Z]be non-constant and monic. Then g:= _gcd(f,f^f 0) is the square-free part of f.

Proof. Note that since f is not constant, it is non-zero. Hence gcd(f, f⁰) 6= 0 andg is thus well-defined. First we will show that g is square-free. Due to Lemma 5.4 we only need to show that gcd(g, g⁰) = 1. We know that f = ggcd(f, f⁰) and thus gcd(g, f⁰) = 1. Hence f⁰ =g⁰gcd(f, f⁰) +g(gcd(f, f⁰))⁰ implies that gcd(g, g⁰) = 1.

Letp∈K[Z] be an irreducible divisor off. Thenf =bp^`for someb∈K[Z] with gcd(b, p) = 1 and some ` >0. We thus have f⁰ =b⁰p^`+`bp^`−1p⁰. Since p is irreducible, it follows that p^` does not divide gcd(f, f⁰). We deduce that gcd(f, f⁰) dividesbp^`−1. Hence p dividesg. This shows that the square-free part of f divides g. Furthermore g is square-free by the above.

Since f and g are both monic, the square-free part of f must be equal tog.

Task 5.6. Let f ∈K[Z] be non-zero. Compute the square-free part off.

Solution. If f ∈ K, then the square-free part of f is 1. So assume that f is not constant.

Then gcd(f, f⁰) is non-zero. Defineg := _LC(f_{) gcd(f,f}^f 0). The square-free part of f and _LC(f)^f are equal. Thus, it follows by Claim 5.5 that g is the square-free part off.

5.2 Ideals of finite codimension

In order to compute the radical of an arbitrary ideal we start with a special case.

Proposition 5.7. An ideal I ⊂K[X] has finite codimension (as a vector space over K) if and only if I ∩K[Xi]6={0} for all 1≤i≤n.

Proof. “⇐”: For each 1 ≤ i ≤ n choose non-zero polynomials fi ∈ I ∩ K[Xi]. Let d:= max{deg(f_i)|1≤i≤n}. ThenK[X] =K[X]deg≤(d,...,d)+I by the division algorithm.

The vector space K[X]deg≤(d,...,d) is finite dimensional. HenceI has finite codimension.

“⇒”: Assume that n := dimK(K[X]/I) < ∞. Let 1 ≤ i ≤ n. The elements (X_i^j)ⁿ⁺¹_j=1 are linearly dependent in this quotient space. This yields a non-trivial K-linear combination of (X_i^j)ⁿ⁺¹_j=1, i.e. a non-zero polynomial f ∈ K[X_i], which maps to zero in the quotient space.

Thus f is a non-zero element ofI ∩K[Xi].

Lemma 5.8. Let I ⊂ K[X] be an ideal such that there is a non-constant square-free g ∈I ∩K[X1]. Let g=Qm

i=1hi be the factorization into pairwise non-equivalent irreducible h₁, . . . , h_m∈K[X₁]. Then I =Tm

i=1(I+hh_ii).

Proof. Clearly I ⊂ Tm

i=1(I +hh_ii). For the other inclusion, let f ∈Tm

i=1(I+hh_ii). There exist r1, . . . , rm ∈ I and q1, . . . , qm ∈ K[X] such that f = ri +qihi for all 1 ≤ i ≤ m.

It follows that fQ

j6=ih_i ∈ I for all 1 ≤ i ≤ m, because g = Qm

i=1h_i ∈ I. Note that

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Section 5 5.2 Ideals of finite codimension gcd(Q

j6=1hj, . . . ,Q

j6=mhj) = 1 and K[X1] is a principal ideal domain. Therefore, there exist p₁, . . . , p_m ∈ K[X₁] such that Pm

i=1p_iQ

j6=ih_j = 1. Together this yields f =Pm

i=1pifQ

j6=ihj ∈I. Hence I ⊃Tm

i=1(I +hh_ii).

Lemma 5.9 (Seidenberg). Let I ⊂ K[X] be an ideal of finite codimension. Then I is a radical ideal if and only if for each1≤i≤nthere is a non-constant square-freeg_i ∈I∩K[X_i].

Proof. “⇒” Let 1≤i≤n and let f ∈I∩K[Xi] be a monic non-constant polynomial. Let f =Q`

j=1h^α_j^j be the unique factorization off into pairwise non-equivalent monic irreducible polynomials h1, . . . , h` ∈K[Xi]. Set α := max{α_j |1≤j ≤`}. Then Q`

j=1h^α_j ∈I ∩K[Xi].

WithI being radical it follows that gi:=Q`

j=1hj ∈I∩K[Xi], which is square-free.

“⇐” For the converse we proceed by induction on the number of variables n for arbitrary fields K. For n= 1 the idealI is a principal ideal, so there is a generator h∈I. Then there is some a∈K[X1] such that g1 =ah. Since g1 is square-free h must also be square-free, by uniqueness of factorization. Now for every f ∈ K[X] with f^k ∈ I we know thath divides f^k. Sinceh is square-free this implies thath dividesf. Hence f ∈I and I is a radical ideal.

Now let n >1. Since g₁ is square-free, there are pairwise non-equivalent irreducible polynomials h1, . . . , hm ∈K[X1] such that g1 =Qm

i=1hi. Let 1 ≤k≤m. We claim that the ideal J := I +hh_ki is radical. Since h_k is irreducible L := K[X₁]/hh_ki is a finite field extension of K. Let ϕ : K[X] → L[X₂, . . . , X_n] be the canonical homomorphism. We know that ϕ is surjective and that kerϕ= hh_ki ⊂ J. Hence L[X2, . . . , Xn]/(Jê) ∼= K[X]/J where Jê is the extension of J with respect to ϕ. Thus Jê is again of finite codimension. Furthermore ϕ(gi) ∈ Jê∩L[Xi] for all 2 ≤i ≤n. Each gi is square-free, thus gcd(gi, g⁰_i) = 1 by Lemma 5.4. This is still true for the image in L[X2, . . . , Xn], so again by Lemma 5.4, the gi are all square-free in L[X_i]. By induction hypothesis, it follows that Jê is radical. This implies that L[X2, . . . , Xn]/(Jê)∼=K[X]/J has no nilpotent elements. Therefore, the ideal J =I +hh_ki is radical. Sincek was arbitrary, the statement holds for all 1≤k≤m.

Lemma 5.8 yields a decompositionI =T_m

i=1(I+hh_ii). Taking the radical of an ideal commutes with intersections, so Rad(I) =Tm

i=1Rad(I+hh_ii) =Tm

i=1(I+hh_ii) =I. HenceI is a radical ideal.

Seidenberg’s Lemma provides a nice method to compute the radical of an ideal of finite codimension:

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Algorithm 5.10. Let I ⊂ K[X] be an ideal of finite codimension. Then the following algorithm computes a generating set Gof Rad(I).

Input: (f₁, . . . , f_m), a generating set ofI Output: G, a finite generating set of Rad(I) begin

G:={f₁, . . . , f_m} for1≤i≤ndo

Ge_i := reduced Gr¨obner basis ofI∩K[X_i] (Task 3.27, Algorithm 3.22) h_i:= the only element of Ge_i

gi := square-free part ofhi (Task 5.6) end

G:={f₁, . . . , f_m, g₁, . . . , g_n} end

Proof. Note that the algorithm terminates after nsteps.

Let 1≤i≤n. SinceK[Xi] is a principal ideal domain there is monic polynomial h∈K[Xi] generatingI ∩K[X_i]. Furthermoreh is non-zero by Proposition 5.7. Then {h} is a reduced Gr¨obner basis ofI∩K[X_i]. Proposition 3.23 implies that this is necessarily the one computed by the algorithm. Thus the Gr¨obner basis contains only one element, so hi in the algorithm is well-defined.

We have f₁, . . . , f_m ∈Rad(I). Let 1≤i≤mand leth_i=aQ`

i=1p^r_iⁱ be the unique factorization of h_i into monic pairwise non-equivalent irreducible polynomialsp₁, . . . , p_` ∈K[X]. By definition, the square-free part of hi is gi =Q`

i=1pi. For r := max{r_i | 1≤ i≤ `} we have that h_i dividesg_i^r=Q`

i=1p^r_i, so g^r_i ∈I. Henceg_i ∈Rad(I). This shows that G⊂Rad(I) at all times in the algorithm.

Now assume that the algorithm has terminated. If I = h1i, then Rad(I) = I = hGi. If not, then by construction of the algorithm, the ideal J :=hGi has the property that for all 1 ≤i ≤ n we have g_i ∈ J ∩K[X_i] and g_i is square-free and non-constant. In particular J is of finite codimension and we can apply Lemma 5.9. It follows that J is radical. Since I ⊂J ⊂Rad(I) we conclude that J = Rad(I).

Remark 5.11. Algorithm 5.10 also works for perfect fields of positive characteristic, provided there is an algorithm to compute the square-free part of any polynomial in K[Z]. However, for non-perfect fields the algorithm might fail. To see this, let p > 0 be a prime number. Consider the ideal I := hX^p−t, Y^p−ti ⊂ Fp(t)[X, Y]. Algorithm 5.10 would give us G = {X^p−t, Y^p −t} as Gr¨obner basis for Rad(I), because both elements are square-free.

Note that (X−Y)^p = X^p−Y^p ∈ I and thus X−Y ∈ Rad(I). But LT(X−Y) is not an element ofhLT(G)i=hX^p, Y^pi. Hence Gis not a Gr¨obner basis of Rad(I).

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Section 5 5.3 The general case 5.3 The general case

Having solved the problem for ideals of finite codimension, we are closer to our goal of computing radicals. However, more theory is needed in order to write an algorithm for the general case.

Lemma 5.12. Let H ⊂K[X]be a subset and f ∈K[X]be a non-zero polynomial. Then hH∪ {1−Zf}i_K[_X_][Z]∩K[X] = [

i≥0

(hHi:fⁱ)

Proof. “⊂”: Let g∈ hH∪ {1−Zf}i_K[_X_][Z]∩K[X]. There existα1(Z), . . . , α`(Z), β(Z)∈ K[X][Z] and h₁, . . . , h_` ∈ H such that g(Z) = P_`

i=1h_i ·α_i(Z) + (1−Zf) ·β(Z). Set Z := 1/f in the field of fractions K(X). Since g, h₁, . . . , h_` are polynomials, we have g = g(1/f) = P`

i=1hi ·αi(1/f). Then there exists a k > 0 such that f^k·αi(1/f) is a polynomial for all 1≤i≤`. Hencegf^k ∈ hHi, sog∈(hHi:f^k).

“⊃”: Let g ∈ S

i≥0(hHi : fⁱ). Let k > 0 such that gf^k ∈ hHi. Then gf^k−1 = Zgf^k+ gf^k−1(1−Zf)∈ hH∪ {1−Zf}i_K[_X_][Z]. Inductively, we obtain g∈ hH∪ {1−Zf}i_K[_X_][Z]. Since g is a polynomial, we haveg∈ hH∪ {1−Zf}i_K[_X_][Z]∩K[X].

Algorithm 5.13. LetX=YtZand letF ⊂K(Y)[Z] be a finite subset. Then the following algorithm computes a Gr¨obner basis of the contraction ideal hFi^c⊂K[X].

Input: F ⊂K(Y)[Z], a finite subset Output: G, a Gr¨obner basis of hFi^c begin

H:= Gr¨obner basis of hFi ⊂K(Y)[Z] (Algorithm 3.19) foreachh∈H do

q := multiple of all denominators of coefficients inK(Y) ofh h:=qh∈K[X]

end

f := lcm{LC_Z(h)|h∈H}

G:= Gr¨obner basis of hH∪ {1−U f}i_K[_X_][U]∩K[X] (Task 3.27) end

Proof. The algorithm terminates after finitely many steps. Assume that the algorithm has terminated. During the algorithm, the setHis only changed by multiplying its elements with some units of K(Y). Therefore, at the endH is still a Gr¨obner basis ofhFi ⊂K(Y)[Z]. Let J :=S

i≥0(hHi_K[X_]:fⁱ). We will show thatJ =hFi^c.

Let g ∈ J. Then there exists a k > 0 such that gf^k ∈ hHi_K[X_] ⊂ hFi^c. But f is a unit in K(Y)[Z] since f ∈ K[Y ]\ {0}. Thus g=gf^kf^−k ∈ hFi and hence g ∈ hFi^c. This implies that J ⊂ hFi^c.

We will show that p ∈ J for all p ∈ hFi^c by transfinite induction on deg_Z(p). Since H is

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a Gr¨obner basis of hFi, there is an element h ∈ H such that LTZ(h) divides LTZ(p). This implies that there exists a monomialα∈K[Z] and an elements∈K(Y) such that LT_Z(p) = sαLTZ(h). The left hand side LTZ(p) lies inK[X] and so must the right hand side. Therefore sLC_Z(h)∈K[Y]. Since f is the lcm of all leading coefficients of elements inH, in particular sf ∈K[Y ]. Definep⁰ :=pf−sαhf ∈K[X]. Thenp⁰ ∈ hFi^c. Furthermore, by construction deg_Z(p⁰)≺deg_Z(p), sincef is a unit inK(Y)[Z]. If deg_Z(p) is minimal (as for the base case) then p⁰ = 0 ∈ J. Otherwise, by induction hypothesis, we deduce that p⁰ ∈ J. Either way there exists ak >0 such thatp⁰f^k∈ hHi_K[_X_]. Thuspf^k+1 =p⁰f^k+sαhf^k+1∈ hHi_K[_X_], so p∈J. Hence J =hFi^c.

It follows by Lemma 5.12 that J =hH∪ {1−U f}i_K[X_][U]∩K[X]. Hence G is a Gr¨obner basis of hFi^c.

Definition 5.14. Let X = Y tZ. Let _Y be a monomial ordering on K[Y] and _Z a monomial ordering on K[Z]. Theblock ordering on K[X] with respect to Z is the monomial ordering defined as follows: for monomialsa, c∈K[Y ] andb, d∈K[Z] letabcd if and only if b_Z dor (b=dand a_Y c).

Lemma 5.15. Let X=Y tZ and consider a block ordering onK[X]with respect toZ. Let g, h∈K[X] such that LTX(g) divides LTX(h). Then LTZ(g) divides LTZ(h) in K(Y)[Z].

Proof. Since LT_X(g) and LT_X(h) are monomials, there are monomials p, q ∈ K[Y ] and m, m⁰ ∈ K[Z] such that LT_X(g) = pmand LT_X(h) = qm⁰. Since LT_X(g) divides LT_X(h), it follows thatm dividesm⁰. Looking at the definition of a block ordering we see that there must be polynomialsa, b∈K[Y ] such that LT_Z(g) =amand LT_Z(h) =bm⁰. Since aand b are units in K(Y)[Z] we conclude that LTZ(g) divides LTZ(h).

Algorithm 5.16. Let I ⊂ K[X] be an ideal and let X = Y tZ. Consider extensions of ideals toK(Y)[Z]. Then the following algorithm computes a non-zero polynomialf ∈K[Y ] such that I = (I+hfi)∩I^ec.

Input: f₁, . . . , f_m∈I, generators of I Output: f ∈K[Y] with I = (I +hfi)∩I^ec begin

:= some block ordering on K[X] with respect toZ ⁰:= restriction oftoK[Z]

H:= Gr¨obner basis ofI with respect to (Algorithm 3.19) g:= lcm{LC_Z(h)|h∈H}, where LC_Z is with respect to⁰ s:= an integer such that (I :g^s) =S

i≥1(I :gⁱ) (Task 4.6) f :=g^s

end

Proof. Clearly the algorithm terminates and we therefore will only have to prove correctness.

Let f be the output of the algorithm and g, s as in the algorithm such thatf =g^s. We will prove that H⊂K[X] is not only a Gr¨obner basis ofI butH ⊂K(Y)[Z] is also a Gr¨obner

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Section 5 5.3 The general case basis of I^e. Letp∈I^e. Then there exist a1, . . . , am∈K(Y)[Z] andg1, . . . , gm ∈I such that p = Pm

i=1a_ig_i. Let r ∈K[Y ] be the least common multiple of all denominators of (a_i)^m_i=1. Then rp=Pm

i=1raigi ∈I. Since H is a Gröbner basis of I, there is an elementh ∈H such that LT_X(h) divides LT_X(rp). Hence LT_Z(h) divides LT_Z(rp) by Lemma 5.15. Sincer is a unit in K(Y)[Z] it follows that LT_Z(h) divides LT_Z(p). Thus H ⊂K(Y)[Z] is a Gröbner basis of Iê.

We will next prove that Iêc = (I :f). We show that p∈(I :f) for allp∈Iêc by transfinite induction on the degree deg_Z(p). SinceH is a Gröbner basis ofIê, there is an elementh∈H such that LT_Z(h) divides LT_Z(p) in K(Y)[Z]. Thus there exists α ∈ K(Y)[Z] such that LTZ(p) = αLTZ(h). Since LTZ(p) is a polynomial in K[X] the denominator in K[Y ] of α has to divide LC_Z(h) and thus has to divide g. Hence αg ∈ K[X] and gp−gαh ∈ Iêc. Furthermore deg_Z(gp−gαh) ≺ deg_Z(p). If deg_Z(p) is minimal (as in the base case) then gp−gαh= 0∈(I :g^s). Otherwise, the induction hypothesis implies thatgp−gαh∈(I :g^s).

So either way g^s(gp−gαh) = g^s+1p−g^s+1αh ∈ I. Hence g^s+1p ∈ I. This implies that p∈S

i≥1(I :gⁱ) = (I :g^s) = (I :f). Thus I^ec⊂(I :f).

For the other inclusion, let p ∈ (I : f). Then f p ∈ I. Since f is a non-zero polynomial in K[Y ], it is a unit in K(Y)[Z]. It follows that p = pf f⁻¹ ∈ Iê, so p ∈ Iêc. Hence Iêc= (I :f).

We now only need to prove that I = (I+hfi)∩(I :f). The inclusion⊂ is clearly true. Let p ∈ (I +hfi)∩(I : f). Then there exists h ∈ I and α ∈ K[X] such that p = h +αf.

Furthermore f p = f h + f²α ∈ I and thus f²α ∈ I. Since (I : f²) = (I : g^2s) ⊂ T∞

i=1(I : gⁱ) = (I : g^s) = (I : f) we have f α ∈ I. Hence p ∈ I. This concludes the proof.

Claim 5.17. Let I ⊂ K[X] be an ideal. Let Y ⊂ X be a maximal set of variables whose image in K[X]/I is algebraically independent. Let Z := X\Y. Then the extension ideal I^e⊂K(Y)[Z] is of finite codimension.

Proof. Let ` denote the cardinality of Z. By choice ofY, for each 1≤i≤` there exists a non-zero polynomialf ∈K[Y ][Zi] such thatf ∈I. Thenf ∈I^e∩K(Y)[Zi] for all 1≤i≤`.

It follows the claim by Proposition 5.7.

Finally we are able to formulate an algorithm to compute the radical of an ideal:

ComputationalTheoryofPolynomialIdeals Eidgen¨ossischeTechnischeHochschuleZ¨urich