## 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 $F[x,y,z]$.