Monthly Archives: February 2011

The ideal quotient of a monomial ideal by a finitely generated monomial ideal is a monomial ideal

Let F be a field and let R = f[x_1, \ldots, x_t]. Let M = (m_i \ |\ i \in I) be a monomial ideal in R, and let n \in R be a monomial. For each i \in I, let d_i be a greatest common divisor of m_i and n, and write m_i = d_iq_i.

  1. Prove that (M : (n)) = (q_i \in I).
  2. Deduce that if N = (n_j \ |\ j \in J) is a finitely generated monomial ideal, then (M : N) is a monomial ideal.

We begin with a lemma.

Lemma: (Euclid) Let R be a unique factorization domain, and let a,b,c \in R such that a and b are relatively prime. Then \mathsf{gcd}(a,bc) = \mathsf{gcd}(a,c). Proof: If d = \mathsf{gcd}(a,c), then certainly d|a and d|bc. Suppose now that e|a and e|bc. If e is a unit, then certainly e|d. Suppose e is not a unit, and that q is an irreducible factor of e of multiplicity k. Since q|a, we have q not dividing b. In particular, q^k|c. By induction, e|c. Hence e|d, and thus d is a greatest common divisor of a and bc. Now let d = \mathsf{gcd}(a,bc). If d is a unit, then certainly d is also a greatest common divisor of a and c. Suppose d is not a unit. We have d|a. If q is an irreducible factor of d of multiplicity k, then q does not divide b; thus q^k|c. Hence d|c. Now suppose e|a and e|c; then e|bc, and so e|d. THus $ltex e$ is a greatest common divisor of a and c. \square

Now we show that (M : (n)) = (t_i \ |\ i \in I).

(\subseteq) Suppose p \in (M : (n)). Then p(n) \subseteq M, and so pn \in M. Say xn = \sum a_im_i, with a_i = \sum b_{i,j} and the b_{i,j} are monomials. Then xn = \sum \sum b_{i,j} m_i = \sum \sum b_{i,j} q_i d_i. Now write p = \sum c_k as a sum of monomials; then pn = \sum c_kn. In particular, each term of pn has the form c_kn = b_{i,j}q_id_i for some i,j,k. Recall that d_i is a greatest common divisor of m_i and n; say n = d_is_i. Then we have c_ks_i = b_{i,j}q_i where s_i and q_i are relatively prime. Since R is a unique factorization domain, by Euclid’s lemma we have q_i|c_k. By this previous exercise, we have p \in (q_i \ |\ i \in I).

\supseteq) Suppose p \in (t_i \ | i \in I); say p = \sum a_it_i. For each i, say n = r_id_i. Now xn = \sum a_it_ir_id_i = \sum a_i m_i r_i, and so xn \in M. Thus x(n) \subseteq M, and we have x \in (M : (n)).

Now suppose N = (n_j \ |\ j \in J) is a finitely generated monomial ideal. By this previous exercise, (M : N) = \bigcap_{j \in J} (M : (n_j)) is a finite intersection of monomial ideals. By this previous exercise, (M : N) is a monomial ideal.

The intersection of two monomial ideals is a monomial ideal

Let F be a field, let R = F[x_1, \ldots, x_t], and let M = (m_i \ |\ i \in I) and N = (n_j \ |\ j \in J) be monomial ideals in R (not necessarily finitely generated). Prove that M \cap N is also a monomial ideal.


Let e_{i,j} be a least common multiple of m_i and n_j for each i \in I and j \in J. We wish to show that M \cap N = (e_{i,j} \ |\ i \in I, j \in J).

(\subseteq) Suppose p \in M \cap N. By this previous exercise, every term in p is divisible by some m_i and also by some n_j; then every term is divisible by some e_{i,j}. Thus p \in (e_{i,j}\ |\ i \in I, j \in J).

(\supseteq) Suppose p \in (e_{i,j} \ |\ i \in I, j \in J). Then each term in p is divisible by some e_{i,j}; hence each term is divisible by some m_i (so that p \in M) and also by some n_j (so that p \in N). So p \in M \cap N.

Note that an easy induction argument shows that in fact any finite intersection of monomial ideals is a monomial ideal.

The union of a chain of monomial ideals is a monomial ideal

Let F be a field, let R = F[x_1, \ldots, x_m], and let K be a linearly ordered set. Let \mathcal{M} = \{M_k\}_{k \in K} be a family of monomial ideals in R such that if i \leq j then M_i \subseteq M_j. Prove that \bigcup_K M_k is a monomial ideal in R.


We showed in this previous exercise that \bigcup_K M_k is an ideal of R. Thus it suffices to show that \bigcup_K M_k has a monomial generating set.

To that end, let G_k be a monomial generating set of M_k. Then \bigcup_K M_k = \bigcup_K (G_k) = (\bigcup_K G_k), so that \bigcup_K G_k is a monomial generating set for \bigcup_K M_k.

The sum and product of two monomial ideals is a monomial ideal

Let F be a field and let R = F[x_1, \ldots, x_m]. Let M = (m_i \ |\ i \in I) and N = (n_j \ |\ j \in J) be monomial ideals in R, where m_i and n_j are monomials. Prove that M+N = (m_i, n_j \ |\ i \in I, j \in J) and MN = (m_in_j \ |\ i \in I, j \in J).


First we show that M+N = (m_i, n_j \ |\ i \in I, j \in J).

(\subseteq) Let x+y \in M+N, where x = \sum_{i \in I} a_im_i and y = \sum_{j \in J} b_jn_j, and all but finitely many of the a_i and b_i are zero. Then x+y = \sum_I a_im_i + \sum_J b_jn_j \in (m_i, n_j \ |\ i \in I, j \in J). (\supseteq) Let x \in (m_i, n_j \ |\ i \in I, j \in J). Collecting the terms in m_i and n_j, respectively, we can write x = \sum_I a_im_i + \sum_J b_jn_j; certainly then x \in M+N.

Now we show that MN = (m_in_j \ |\ i \in I, j \in J).

(\subseteq) Let x \in MN; say x = \sum_K z_kw_k where z_k = \sum_I a_{i,k}m_i \in M and w_k = \sum_J b_{j,k}n_j \in N. Then x = \sum_K \sum_I \sum_J a_{i,k}b_{j,k}m_in_j \in (m_i, n_j \ |\ i \in I, j \in J). (\supseteq) Let x \in (m_i, n_j \ |\ i \in I, j \in J); say x = \sum_{I \times J} c_{i,j} m_i n_j, where all but finitely many c_{i,j} are zero. Certainly then x \in MN.

Compute ideal quotients in a quotient ring

Let R = \mathbb{Q}[y,z] and let K = (y^5-z^4) be an ideal in R. Let bars denote passage to R/K. For each of the following pairs of ideals I and J, compute the ideal quotient (\overline{I} : \overline{J}).

  1. I = (y^3, y^5-z^4), J = (z)
  2. I = (y^3, z, y^5-z^4), J = (y)
  3. I = (y, y^3, z, y^5-z^4), J = (1)

We will let Maple to compute our Gröbner bases for us. Recall from this previous exercise that (\overline{I} : \overline{J}) = \overline{(I:J)}. From this previous exercise, we know that if G = \{g_i\} is a Gröbner basis of I \cap (p), then \{g_i/p\} is a Gröbner basis of (I : (p)). Throughout, fix the lex order induced by t > y > z.

Note also that \{y^5-z^4\} is the reduced Gröbner basis for K. Thus if (a_1, \ldots, a_k) \subseteq R is an ideal, then \overline{A} is precisely (\overline{b_1}, \ldots, \overline{b_k}), where each \overline{b_i} is simply a_i mod y^5-z^4.

  1. First we compute (I : J) in R. Note that tI + (1-t)J = (ty^3, ty^5-tz^4, tz-z). Using Maple, we see that G = \{ z^4, y^3z, tz-z, ty^3 \} is a Gröbner basis for this ideal. Then I \cap J = (z^4, y^3z), and hence (I : J) = (z^3, y^3). Thus (\overline{I} : \overline{J}) = (\overline{y^3}, \overline{z^3}).
  2. First we compute (I : J) in R. Note that tI + (1-t)J = (ty^3, tz, ty^5-tz^4, ty-y). Using Maple, we see that G = \{yz, y^3, tz, ty-y \} is a Gröbner basis for this ideal. Then I \cap J = (yz, y^3), and hence (I:J) = (y^2, z). Thus (\overline{I}, \overline{J}) = (\overline{y^2}, \overline{z}).
  3. First we compute (I : J) in R. Note that tI + (1-t)J = (ty, ty^3, tz, ty^5-tz^4, t-1). Using Maple, we see that G = \{ z, y, t-1 \} is a Gröbner basis for this ideal. Then I \cap J = (y,z), and hence (I : J) = (y,z). Thus (\overline{I} : \overline{J}) = (\overline{y}, \overline{z}).

A basic property of ideal quotients in a quotient ring

Let R be a ring and let I, J, and K be ideals in R such that K \subseteq I. If \overline{I} and \overline{J} denote the images of I and J in R/K via the natural projection, prove that \overline{(I:J)} = (\overline{I}:\overline{J}) where \overline{(I:J)} is the image of the ideal quotient (I:J).


(\subseteq) Suppose r + K \in \overline{(I:J)}. Then r \in (I:J), and thus rJ \subseteq I. Then rJ + K \subseteq I + K, and so (r+K)(J + K) \subseteq I+K. So r+K \in (\overline{I}:\overline{J}).

(\supseteq) Suppose conversely that r+K \in (\overline{I}:\overline{J}). Then (r+K)(J+K) \subseteq I+K, and thus rJ+K \subseteq I+K. Since K \subseteq I, rJ \subseteq I. So r \in (I:J), and r+K \in \overline{(I:J)}.

Compute an ideal quotient in a polynomial ring

Let I = (x^2y+z^3, x+y^3-z, 2y^4z-yz^2-z^3) and J = (x^2y^5, x^3z^4, y^3z^7) be ideals in \mathbb{Q}[x,y,z]. Compute (I : J).


Because the point of this exercise is not to compute Gröbner bases but rather to use them in the solution to another problem, and because computing Gröbner bases is excruciating, we will let Maple compute our Gröbner bases for us. Throughout, we fix the lex order induced by t > x > y > z.

As we saw in this previous exercise, because J is a monomial ideal, we have (I : J) = (I : (x^2y^5)) \cap (I : (x^3z^4)) \cap (y^3z^7)). We will now go about finding Gröbner bases for each of these intersectands.

Recall that if \{g_i\} is a Gröbner basis for I \cap (p), then \{g_i/p\} is a Gröbner basis for (I : (p)). Recall also that I \cap J = (tI + (1-t)J) \cap \mathbb{Q}[x,y,z].

  1. (I : (x^2y^5)): Note that tI + (1-t(x^2y^5)) = (tx^2y+tz^3, tx+ty^3-tz, 2ty^4z-tyz^2-tz^3, tx^2y^5-x^2y^5). Using Maple, we find that G = \{ x^2z^2y^5, x^2y^5z+x^2y^6, -x^2y^5z+x^3y^5, x^2y^5+tz^7, tz^4+tyz^3-2tz^6-12x^2y^5z, 2ty^4z-tyz^2-tz^3, ty^7, tx+ty^3-tz \} is a Gröbner basis for this ideal. Then G \cap \mathbb{Q}[x,y,z] = \{ x^2z^2y^5, x^2y^5z+x^2y^6, -x^2y^5z+x^3y^5 \} is a Gröbner basis for I \cap (x^2y^5). Hence, (I : (x^2y^5)) = (z^2, y+z, x-z).
  2. (I : (x^3z^4)): Note that tI + (1-t)(x^3z^4) = (tx^2y+tz^3, tx+ty^3-tz, 2ty^4z-tyz^2-tz^3, tx^3z^4-x^3z^4). Using Maple, we find that G = \{ x^3z^6, x^3z^5+x^3yz^4, -x^3z^5+x^4z^4, tz^7-x^3z^4, tz^4+tyz^3-2tz^6+12x^3z^5, 2ty^4z-tyz^2-tz^3, ty^7, tx+ty^3-tz \} is a Gröbner basis for this ideal. Then G \cap \mathbb{Q}[x,y,z] = \{ x^3z^6, x^3z^5+x^3yz^4, -x^3z^5+x^4z^4 \} is a Gröbner basis for I \cap (x^3z^4). Hence, (I : (x^3z^4)) = (z^2, y+z, x-z).
  3. (I : (y^3z^7)): Note that tI + (1-t)J = (tx^2y+tz^3, tx+ty^3-tz, 2ty^4z-tyz^2-tz^3, ty^3z^7-y^3z^7). Using Maple, we find that G = \{ y^3z^7, tz^9, tz^4+tyz^3-2tz^6+12tz^8, 2ty^4z-tyz^2-tz^3, ty^7, tx+ty^3-tz \} is a Gröbner basis for this ideal. Then G \cap \mathbb{Q}[x,y,z] = \{ y^3z^7 \} is a Gröbner basis for I \cap (y^3z^7). Hence, (I : (y^3z^7)) = \mathbb{Q}[x,y,z].

Thus we see that (I : J) = (z^2,y+z,x-z) \cap (z^2, y+z, x-z) \cap \mathbb{Q}[x,y,z] = (x-z, y+z, z^2).

Two basic properties of ideal quotients

Recall that if R is a ring with ideals I and J, the ideal quotient (I:J) in R is defined to be (I:J) = \{r \in R \ |\ rJ \subseteq I \}. This set is certainly an ideal of R.

  1. Suppose R is an integral domain, d \neq 0 is in R, and I \subseteq R an ideal. Prove that if G = \{g_i\}_{i=1}^k is a generating set for I \cap (d) where g_i = h_id, then H = \{h_i\}_{i=1}^k is a generating set for (I:(d)).
  2. Let R be a commutative ring, I \subseteq R an ideal, and D = \{d_i\}_{i=1}^k \subseteq R with d_i \neq 0. Prove that (I : (D)) = \bigcap_{d_i \in D} (I : (d_i)).

  1. Suppose r \in (I : (d)). Then r(d) \subseteq I, so rd \in I. Certainly rd \in (d), so that rd \in I \cap (d) = (G). Say rd = \sum s_ig_i = \sum s_ih_id. Since R is an integral domain, r = \sum s_ih_i, and thus r \in (H). Conversely, suppose r \in (H), with r = \sum s_ih_i. Then rd = \sum s_ih_id = \sum s_ig_i, and thus rd \in I. So r(d) \subseteq I, and thus r \in (I : (d)). Hence (I : (d)) = (H).
  2. Suppose r \in (I : (D)). Then r(D) \subseteq I, and in particular rd_i \in I for all i. So r(d_i) \subseteq I for all i, and we have r \in \bigcap_{d_i \in D} (I : (d_i)). Conversely, suppose r \in \bigcap_{d_i \in D} (I : (d_i)). Then r(d_i) \subseteq I for all i, and rd_i \in I for all i. Now r(D) = (rD) \subseteq I, so that r \in (I : (D)). Thus (I : (D)) = \bigcap_{d_i \in D} (I : (d_i)).

Compute the intersection of two polynomial ideals

Use Gröbner bases to compute the intersection of the ideals I = (x^3y-xy^2+1, x^2y^2-y^3-1) and J = (x^2-y^2, x^3+y^3) in F[x,y], where F is a field.


Recall that I \cap J = (tI + (1-t)J) \cap F[x,y], where t is a new indeterminate. Now tI + (1-t)J = (tx^3y-txy^2+t, tx^2y^2-ty^3-t, tx^2-ty^2-x^2+y^2, tx^3+ty^3-x^3-y^3). Using Bichberger’s algorithm (and Maple to keep track of the arithmetic) we find that G = \{ g_1 = xy^2+y^3, g_2 = x^2-y^2, g_3 = ty^4-ty^3-t, g_4 = tx+ty \} is a Gröbner basis of tI + (1-t)J with respect to the lex order induced by t > x > y > z. Proof details are in this Maple worksheet (change file type to .mws) or in plain text here. (Change file type to .txt)

Now G \cap F[x,y] is a Gröbner basis for I \cap J in F[x,y]; and thus we have I \cap J = (xy^2+y^3, x^2-y^2).

Compute the intersection of two polynomial ideals

Let F be a field. Use Gröbner bases to show that (x,z) \cap (y^2, x-yz) = (xy, x-yz) in F[x,y,z].


Recall that I \cap J = (tI + (1-t)J) \cap F[x,y,z], where t is a new indeterminate. Now tI + (1-t)J = (tx, tz, ty^2-y^2, tx-tyz-x+yz). Using Buchberger’s algorithm and Maple to keep track of the arithmetic, we find that this ideal has the reduced Gröbner basis G = \{ g_1 = y^2z, g_2 = x-yz, g_3 = tz, g_4 = ty^2-y^2 \} with respect to the lexicographic order induced by t > x > y > z. The details of this computation can be found in this Maple worksheet (change file type to .mws) or in plain text here (change file type to .txt).

Now G_2 = G \cap F[x,y,z] = \{ g_1 = y^2z, g_2 = x-yz \} is a Gröbner basis of I \cap J with respect to the lex order induced by x > y > z, and moreover is reduced. It is also easy to see using Buchberger’s criterion that G_2 is also the reduced Gröbner basis for (xy, x-yz) with respect to the lex order induced by x > y > z.

Thus (x,z) \cap (y^2, x-yz) = (xy, x-yz) in F[x,y,z].