Tag Archives: intersection

Composites and intersections of finite dimensional splitting fields are splitting fields

Let K_1,K_2 be finite dimensional splitting fields over a field F and contained in a field K. Show that K_1K_2 and K_1 \cap K_2 are also finite dimensional splitting fields over F (and contained in K).


Say K_1 and K_2 are splitting fields over F for the finite sets S_1,S_2 \subseteq F[x].

We claim that K_1K_2 is a splitting field over F for S_1 \cup S_2. To see this, note that each polynomial in S_1 \cup S_2 splits over K_1K_2. Now if E is a field over which S_1 \cup S_2 split, then the polynomials in S_1 split over E, so that K_1 \subseteq E. Likewise K_2 \subseteq E, and so K_1K_2 \subseteq E. So K_1K_2 is a splitting field for S_1 \cup S_2 over F.

Now suppose q(x) is irreducible over F with a root in K_1 \cap K_2. This root is in the splitting field K_1, so q splits over K_1 by this previous exercise. Likewise, q splits over K_2. Thus q splits over K_1 \cap K_2. Again using this exercise, K_1 \cap K_2 is a splitting field over F.

The classes of left, right, and two-sided ideals of a semigroup are closed under finite intersection and arbitrary union

Let S be a semigroup, and let L(S), R(S), and I(S) denote the sets of left- right- and two-sided ideals of S. Let X \in \{L(S), R(S), I(S)\} and let Y \subseteq X be nonempty. Show that \bigcup Y \in X and that if \bigcap Y \neq \emptyset, then \bigcap Y \in X. Further, show that if Y \subseteq I(S) is finite, then \bigcap Y is not empty (and so is an ideal).


Suppose Y \subseteq L(S) is a nonempty collection of left ideals of S. Certainly \bigcup Y \neq \emptyset. If a \in \bigcup Y, then we have a \in T for some T \in Y. Now if s \in S, then sa \in T \in \bigcup Y. So S(\bigcup Y) \subseteq \bigcup Y, and thus \bigcup Y is a left ideal of S. Now suppose \bigcup Y \neq \emptyset, and say a \in \bigcap Y. So a \in T for all T \in Y. If s \in S, then sa \in T for all T \in Y, and so sa \in \bigcap Y. Thus \bigcap Y is a left ideal.

Likewise, the results hold for R(S). Since every two-sided ideal is also a left and a right ideal, the results also follow for I(S).

Now suppose Y = \{T_i\}_{i=1}^n is a finite collection of ideals of S. Now \prod T_i \subseteq T_k for each k, so that \prod T_i \subseteq \bigcap Y. Since each T_i is nonempty, \prod T_i is nonempty, and so \bigcap Y is nonempty.

In an algebraic integer ring, two ideals whose intersection contains 3 may be comaximal

Let K be an algebraic number field with ring of integers \mathcal{O}, and let A,B \subseteq \mathcal{O} be nontrivial ideals. If 3 \in A \cap B, must it be the case that (A,B) \neq (1)?


Consider K = \mathbb{Q}(\sqrt{-2}), so that \mathcal{O} = \mathbb{Z}[\sqrt{-2}]. Consider A = (1+\sqrt{-2}) and B = (1-\sqrt{-2}). Now AB = (3), so that 3 \in A \cap B. However, note that 1 = (1+\sqrt{-2})(1-\sqrt{-2}) - (1+\sqrt{-2}) - (1-\sqrt{-2}) = (A,B), so that (A,B) = (1).

The nonempty intersection of submodules is a submodule

Let R be a ring with 1 and let M be a left R-module. Let \{M_t\}_T be a family of left R-submodules of M with T \neq \emptyset. Prove that \bigcap_T M_t is a submodule of M.


We use the submodule criterion.

Note that for all t, 0 \in M_t. Because T \neq \emptyset, we have 0 \in \bigcap_T M_t. Now suppose a,b \in \bigcap M_t and r \in R. Since each M_t is a submodule, we have a+rb \in M_t for all t, and thus a+rb \in \bigcap M_t.

So \bigcap M_t \subseteq M is a submodule.

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.

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].

The complete homomorphic preimage of a prime ideal is a prime ideal

Let R and S be commutative rings and let \varphi : R \rightarrow S be a ring homomorphism.

  1. Prove that if P \subseteq S is a prime ideal then \varphi^\ast[P] \subseteq R is a prime ideal. Apply this to the special case when R \subseteq S to deduce that if P is prime in S then P \cap R is prime in R.
  2. Prove that if M \subseteq S is a maximal ideal and if \varphi is surjective, then \varphi^\ast[M] \subseteq R is maximal. Give an example to show that this need not be the case if \varphi is not maximal.

  1. By §7.3 #24, \varphi^\ast[P] is an ideal of R. Now suppose ab \in \varphi^\ast[P]. Then \varphi(ab) = \varphi(a)\varphi(b) \in P, so that since P is prime, either \varphi(a) \in P or \varphi(b) \in P. Thus either a \in \varphi^\ast[P] or b \in \varphi^\ast[P]. Hence \varphi^\ast[P] is a prime ideal of R.

    Note that if \iota : R \rightarrow S is the inclusion map, then \iota^\ast[P] = R \cap P.

  2. Let M \subseteq S be maximal, and note that \varphi^\ast[M] \subseteq R is an ideal. Note that \varphi^\ast[M] \neq R since \varphi is surjective. Let \pi : S \rightarrow S/M denote the natural projection. Since \varphi is surjective, \pi \circ \varphi : R \rightarrow S/M is a surjective ring homomorphism and S/M is a field. Moreover, \varphi^\ast[M] \subseteq \mathsf{ker}\ \pi \circ \varphi. Now R/\mathsf{ker}\ (\pi \circ \varphi) \cong S/M is a field, and thus has only the trivial ideals. Using the Lattice Isomorphism Theorem and since \varphi^\ast[M] \neq R, we have \varphi^\ast[M] = \mathsf{ker}\ \pi \circ \varphi. Since R/\varphi^\ast[M] is a field, \varphi^\ast[M] is maximal in R.

    Now let M \subseteq R be a maximal ideal and consider the inclusion map \iota : M \rightarrow R. Then \iota^\ast[M] = M is not maximal in M.

Some more properties of ideal arithmetic

Let R be a ring and let I,J,K \subseteq R be ideals.

  1. Prove that I(J+K) = IJ+IK and (I+J)K = IK+JK.
  2. Prove that if J \subseteq I, then (J+K) \cap I = J+(K \cap I).

  1. We show that I(J+K) = IJ + IK; the proof of the other equality is similar. (\subseteq) Let \alpha \in I(J+K). Then \alpha = \sum a_i(b_i+c_i) for some a_i \in I, b_i \in J, and c_i \in K. Then \alpha = \sum (a_ib_i + a_ic_i) = (\sum a_ib_i) + (\sum a_ic_i) \in IJ + IK. (\supseteq) Note that since J \subseteq J+K, IJ \subseteq I(J+K). Similarly, since K \subseteq J+K, IK \subseteq I(J+K). By this previous exercise, IJ+IK \subseteq I(J+K).
  2. (\subseteq) Let x \in (J+K) \cap I. Then x \in I and x = y+z for some y \in J and z \in K. Since J \subseteq I, x-y = z \in I. Thus z \in K \cap I, and x = y+z \in J+(K \cap I). (\supseteq) Let x \in J+(K \cap I). Then x = y+z where y \in J and z \in K \cap I. Again because J \subseteq I, we have x = y+z \in I. Moreover, x = y+z \in J+K. Thus x \in (J+K) \cap I.

The set of ideals of a ring is closed under arbitrary intersections

Let R be a ring.

  1. Prove that if I and J are ideals of R, then I \cap J is also an ideal of R.
  2. Prove that if I_k is an ideal of R for each k \in A, A an arbitrary set, then \bigcap_A I_k is also an ideal of R.

  1. In this previous exercise, we showed that I \cap J is a subring of R. Thus it suffices to show that I \cap J absorbs R on the right and the left. To that end, let r \in R and x \in I \cap J. Now x \in I, so that rx, xr \in I. Likewise, rx, xr \in J. Thus rx, xr \in I \cap J, and I \cap J \subseteq R is an ideal.
  2. Again, we showed in this previous exercise that \bigcap_A I_k is a subring of R, os it suffices to show absorption. Let r \in R and x \in \bigcap_A I_k. Now x \in I_k for each k \in A, so that rx, xr \in I_k for all k \in A. Thus xr, rx \in \bigcap_A I_k. Hence \bigcap_A I_k is an ideal of R.