§ Hilbert basis theorem for polynomial rings over fields (TODO)

Theorem: Every ideal II of k[x1,,xn]k[x_1, \dots, x_n] is finitely generated. First we need a lemma:

§ Monomial ideals

  • Monomial ideals are ideals generated by monomials. in K[x,y]K[x, y], these monomials are of the form xaybx^ay^b.
  • Lemma: Let I=(xα:αA)I = (\vec x^\alpha : \alpha \in A) be a monomial ideal generated by exponent vectors AA. The monomial xβ\vec x^\beta lies in II iff xβx^\beta is divisible by some αA\alpha \in A.
  • Suppose xβ\vec x^\beta lies in II. Thus, xβαzαxαx^\beta \equiv \sum_\alpha z_\alpha \vec x^\alpha for polynomials zαk[x,y]z_\alpha \in k[x, y].
  • Suppose each z[α]jc[α][j]xα[j]z[\alpha] \equiv \sum_j c[\alpha][j] x^{\alpha[j]}for monomials x[α][j]k[x,y]x^{[\alpha][j]} \in k[x, y] and coefficients c[α][j]Kc[\alpha][j] \in K.
  • This makes the equation look like:
\begin{aligned} &x^\beta \equiv \sum \alpha (\sum j c [\alpha ] [j ] x^{\alpha [j ]}) \cdot x^{\alpha} \\ &x^\beta \equiv \sum \alpha \sum j c [\alpha ] [j ] x^{\alpha [j ] + \alpha} \end{aligned}
  • But since xβx^{\beta} occurs on the right hand side, there must a term on the left hand side which is xβx^{\beta} with non-zero coefficient. So we must have some α,j\alpha_*, j_* such that xα[j]+αxβx^{\alpha_*[j_*] + \alpha_*} \sim x^\beta, or α+α[j]=β\alpha_* + \alpha_*[j_*] = \beta, or αβ\alpha_* \leq \beta, which means that xβx^\beta lies in the ideal as it can be generated by scaling xαIx^\alpha_* \in I.

§ Polynomial in monomial ideal is linear combination of ideal elements

  • If fIf \in I, then this means that f=ixαcif = \sum_i x^{\alpha} c_i for polynomials ciK[x,y]c_i \in K[x, y].
  • Expanding cic_i into monomials cijc_{ij}, we see that each of the terms on the RHS is some monomial xαicijx^{\alpha_i}c_{ij} which is a multiple of xαix^{\alpha_i}, and thus lives in the ideal II.
  • So, ff is a linear combination of xαicijx^{\alpha_i} c_{ij} which live in the ideal.
  • MORAL : A monomial ideal is determined by its monomials. Any polynomial in the monomial ideal is generated by monomials in the ideal.

§ Dickson's Lemma: monomial ideals are finitely generated

  • Induction on the number of variables. n=1n = 1 is done since k[x]k[x] is a PID, needs only a single generator.
  • Let's have n+1n+1 variables, which we write as K[x1,,xn,y]K[x_1, \dots, x_n, y] with yy being the new variable we add (for induction).
  • Suppose IK[x1,,xn,y]I \subseteq K[x_1, \dots, x_n, y] is a monomial ideal. We must find a generating set for II.
  • Let JJ be the ideal generated by the ideal II where we set yy to 11. One way to think about this is to write JI/(y1)J \simeq I/(y-1).
  • Alternatively, being very explcit, we define J{xα:k,xαykI}J \equiv \{ x^\alpha : \exists k, x^\alpha y^k \in I\}. That is, JJ consists of all xαx^\alpha such that for some kk, xαykIx^\alpha y^k \in I.
  • Philosophically, JJ is the projection of II onto the {xi}\{ x_i \}.
  • Our inductive hypothesis says that II is finite generated by Ixα(1),,xα(n)I \equiv \langle x^{\alpha(1)}, \dots, x^{\alpha(n)} \rangle.
  • For each i[1,n]i \in [1, n], we know that we have xα(i)ym(i)Ix^{\alpha(i)}y^{m(i)} \in I for some m(i)m(i). Let Mmaxim(i)M \equiv \max_i m(i) be the largest of all mim_i.
  • Now consider the slices of JJ at yky^k. That is, we wish to generate ykIy^k \cdot I for all k[0,m]k \in [0, m]. Define Jkykxα(i)J_k \equiv \langle y^k x^{\alpha(i)} \rangle.
  • By our induction hypothesis, each of the JkJ_k is finitely generated.
  • Thus, the full JJ is generated by the collection of all generators for each JkJ_k for 0kM0 \leq k \leq M. To compute MM, we finitely generated II.
  • See that every monomial in II is divisible by the generator of some JkJ_k for some kk. Suppose some xβybIx^\beta y^b \in I. If bMb \geq M, then we find some xα(i)xβx^\alpha(i)|x^\beta in JJ. Then, we consider xα(i)ym(i)x^\alpha(i) y^{m(i)} which will definitely divide xβybx^\beta y^b since m(i)Mbm(i) \leq M \leq b, and then ym(i)yby^{m(i)}|y^b.
  • If we have b<Mb < M, then we consider the ideal JbJ_b. Then the monomial xβybx^\beta y^b will be generated by monomials in JkJ_k.
  • Thus, since (1) every monomial in II lies in some JkJ_k, and vice versa, (2) monomial ideals are determined by their monomials, and (3) The JkJ_k are finitely generated, we have shown that II is finitely generated by the union of generators of the JkJ_k, gen(Jk)\cup \texttt{gen}(J_k)

§ Ideal of leading terms

  • For any ideal II, define the ideal of leading terms LT(I)LT(I) to be the ideal conisting of elements as the leading term of elements of II. So, LT(I){LT(f):fI}LT(I) \equiv \{ LT(f) : f \in I \}. Check that this is an ideal. ( 0=LT(0)0 = LT(0), 1=LT(1)1 = LT(1), LT(f+g)=LT(f)+LT(g)LT(f+g) = LT(f)+LT(g), and LT(fg)=LT(f)LT(g)LT(fg) = LT(f) LT(g)).
  • Suppose we have an ideal If1,f2,,fnI \equiv \langle f_1, f_2, \dots, f_n \rangle. Now we have two ideals that we wish to compare: LT(I)LT(I), the ideal of leading terms, and LT(f1),,LT(fn)\langle LT(f_1), \dots, LT(f_n) \rangle, the ideal generated by the leading terms of the generators of II.
  • We must always have LT(f1),LT(fs)LT(I)\langle LT(f_1), \dots LT(f_s) \rangle \subseteq LT(I)by the definition of LT(I)LT(I) which contains all leading terms.
  • However, LT(I)LT(I) can be larger.
  • A generating set for II given by If1,f2,,fsI \equiv \langle f_1, f_2, \dots, f_s \rangleis a Grober basis iff it is true that LT(I)LT(I) equals LT(f1),LT(f2),,LT(fs)\langle LT(f_1), LT(f_2), \dots, LT(f_s) \rangle.

§ Proof of hilbert basis theorem

  • We wish to show that every ideal II of k[x1,,xn]k[x_1, \dots, x_n] is finitely generated.
  • If I={0}I = \{ 0 \} then take I=(0)I = (0) and we are done.
  • Pick polynomials gig_i such that (LT(I))=(LT(g1),LT(g2),,LT(gt))(LT(I)) = (LT(g_1), LT(g_2), \dots, LT(g_t)). This is always possible since (LT(I))(LT(I)) is a monomial ideal, which is finitely generated by Dickinson's Lemma.
  • We claim that I=(g1,g2,,gt)I = (g_1, g_2, \dots, g_t).
  • Since each giIg_i \in I, it is clear that (g1,,gt)I(g_1, \dots, g_t) \subseteq I.
  • Conversely, let fIf \in I be a polynomial.
  • Divide ff by g1,,gtg_1, \dots, g_t to get f=iaigi+rf = \sum_i a_i g_i + r where no term of rK[x1,,xn]r \in K[x_1, \dots, x_n] is divisible by any of LT(g1),,LT(gt)LT(g_1), \dots, LT(g_t). We claim that r=0r = 0.
  • See that r=fiaigir = f - \sum_i a_i g_i. We have rIr \in I, since fIf \in I and the gig_i live in II.
  • Thus, we must have LT(r)LT(I)LT(r) \in LT(I) (by the definition of LT(I)LT(I)).
  • If LT(r)LT(r) is nonzero, then since (1) LT(I)=LT(g1),,LT(gn)LT(I) = \langle LT(g_1), \dots, LT(g_n) \rangle, and (2) LT(I)LT(I) is a monomial ideal, LT(r)LT(r) must be divisible by one of the generators!
  • This contradicts the assumpion that rr is a reminader --- a remainder is by definition not divisible by any LT(gi)LT(g_i).
  • Thus, we have shown that if fIf \in I, then f(g1,,gn)f \in (g_1, \dots, g_n).

§ References

  • Cox, Little, O'Shea: computational AG.