## 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 $R$.