## Tag Archives: integral domain

### A fact about the rank of a module over an integral domain

Let $R$ be an integral domain with quotient field $F$ and let $M$ be any (left, unital) $R$-module. Prove that the rank of $M$ equals the dimension of $F \otimes_R M$ over $F$.

Recall that the rank of a module over a domain is the maximal number of $R$-linearly independent elements.

Suppose $B \subseteq M$ is an $R$-linearly independent set, and consider $B^\prime \subseteq F \otimes_R M$ consisting of the tensors $1 \otimes b$ for each $b \in B$. Suppose $\sum \alpha_i (1 \otimes b_i) = 0$. For some $\alpha_i \in F$. Clearing denominators, we have $\sum r_i ( 1 \otimes b_i) = 0$ for some $r_i \in R$. Now $\sum 1 \otimes r_ib_i = 0$, and we have $1 \otimes \sum r_ib_i = 0$. By this previous exercise, there exists a nonzero $r \in R$ such that $\sum rr_ib_i = 0$. Since $B$ is linearly independent, we have $rr_i = 0$ for each $i$, and since $r \neq 0$ and $R$ is a domain, $r_i = 0$ for each $i$. Thus $\alpha_i = 0$ (since the denominators of each $\alpha_i$ are nonzero). So $B^\prime$ is $F$-linearly independent in $F \otimes_R M$. In particular, we have $\mathsf{rank}_R(M) \leq \mathsf{dim}_F(F \otimes_R M)$.

Now note that if $b_i^\prime = \sum \alpha_{i,j} \otimes m_{i,j} \in F \otimes_R M$, is a linearly independent set, with $\alpha_{i,j} = r_{i,j}/s_{i,j}$, then ‘clearing denominators’ by multiplying $b_i^\prime$ by $\prod_j s_{i,j}$, we have a second linearly independent set with the same cardinality whose elements are of the form $1 \otimes m_i$ for some $m \in M$. Suppose there exist $r_i \in R$ such that $\sum r_im_i = 0$; then $\sum r_i(1 \otimes b_i) = 0$, and (since the $1 \otimes b_i$ form a basis) we have $r_i = 0$. So the $m_i$ are $R$-linearly independent in $M$. Thus $\mathsf{rank}_R(M) \geq \mathsf{dim}_F(R \otimes_R M)$.

### Exhibit a nonzero torsion module over an integral domain whose annihilator is trivial

Exhibit an integral domain $R$ and a nonzero torsion module $M$ over $R$ such that $\mathsf{Ann}(M) = 0$. Prove that if $N$ is a finitely generated torsion $R$-module ($R$ a domain), then $\mathsf{Ann}(N) \neq 0$.

First consider the direct sum $M = \bigoplus_{\mathbb{N}^+} \mathbb{Z}/(2^k)$ as a $\mathbb{Z}$-module in the usual way. Every element $\alpha$ of $M$ is annihilated by $2^t$, where $t$ is the largest nonzero component of $\alpha$. However, for all $k \in \mathbb{Z}$, there exist elements not annihilated by $k$. For example, if $2^t \leq k < 2^{t+1}$, then the element with $1$ in the $t+1$ component an 0 elsewhere is not annihilated by $k$.

Suppose now that $N$ is a finitely generated torsion $R$-module; say $N = (A)_R$, with $A = \{a_i\}_{i=1}^n$. For each $a_i$, there exists a nonzero element $r_i \in R$ such that $r_ia_i = 0$, since $N$ is torsion. Let $r = \prod r_i$. Since $R$ is an integral domain, $r \neq 0$, and certainly $rN = 0$. Thus $\mathsf{Ann}(N) \neq 0$.

### Nonprincipal ideals of an integral domain have rank 1 but are not free

Let $R$ be an integral domain, and let $M \subseteq R$ be a nonprincipal ideal of $R$, considered as a (left, unital) $R$-module. Prove that $M$ has rank 1 over $R$ but is not free as an $R$-module.

Let $\{x,y\} \subseteq M$ be a subset containing two nonzero elements. Since $x \cdot y + (-y) \cdot x = 0$, $\{x,y\}$ is not linearly independent. Thus the rank of $M$ as an $R$-module is at most 1. Since $R$ is an integral domain, every (nonzero) singleton set is linearly independent; hence $M$ has rank 1.

Suppose $M$ is free. Then $M$ must have free rank 1, since any free generating set is linearly independent. That is, $M = R\alpha$ for some $\alpha$. But then $M$ is principal, a contradiction.

### Over an integral domain, rank N + rank M/N = rank M

Let $R$ be an integral domain, let $M$ be a (left, unital) $R$-module, and let $N \subseteq M$ be a submodule. Prove that $\mathsf{rank}(N) + \mathsf{rank}(M/N) = \mathsf{rank}(M)$.

Suppose $S = \{s_i\} \subseteq N$ is a cardinality-maximal linearly independent set, and choose $T = \{t_i\} \subseteq M$ such that $\overline{T} \subseteq M/N$ is a cardinality-maximal linearly independent set and $T$ and $\overline{T}$ have the same cardinality. Note in particular that $S \cap T$ is empty. We claim that $S \cup T$ is linearly independent in $M$. To see this, suppose $\sum r_i s_i + \sum r_i^\prime t_i = 0$. Mod $N$, we see that $\sum r_i^\prime \psi(t_i) = 0$, so that $r_i^\prime = 0$. But then $\sum r_is_i = 0$, so that $r_i = 0$. Thus $S \cup T$ is linearly independent in $M$. In particular, $\mathsf{rank}(M) \geq \mathsf{rank}(N) + \mathsf{rank}(M/N)$.

Now let $A \subseteq M$ be the submodule generated by $S \cup T$. We claim that $M/A$ is torsion. Suppose to the contrary that $M/A$ is not torsion; that is, that there exists $y \in M/A$ such that if $ry \in A$, then $r = 0$. In other words, if $ry + \sum r_i s_i + \sum r_i^\prime t_i = 0$, then $r = 0$, and hence $r_i = r_i^\prime = 0$. Thus $S \cup T \cup \{y\}$ is linearly independent. If $y \in N$, then we violate the maximality of $S$, and if $y \notin N$, then we violate the maximality of $T$. So $M/A$ must be torsion. By this previous exercise, $\mathsf{rank}(M) = \mathsf{rank}(N) + \mathsf{rank}(M/N)$.

### Over an integral domain, the rank of a direct sum is the sum of the ranks

Let $R$ be an integral domain and let $A$ and $B$ be (left, unital) $R$-modules. Prove that $\mathsf{rank}(A \oplus B) = \mathsf{rank}(A) + \mathsf{rank}(B)$, where the rank of a module is the largest possible cardinality of a linearly independent subset.

Suppose $A$ has rank $n$ and $B$ has rank $m$. By the previous exercise, there exist free submodules $A_1 \subseteq A$ and $B_1 \subseteq B$ having free ranks $n$ and $m$, respectively, such that the quotients $A/A_1$ and $B/B_1$ are torsion. Note that $A_1 \oplus B_1 \subseteq A \oplus B$ is free. By this previous exercise, we have $(A \oplus B)/(A_1 \oplus B_1) \cong_R (A/A_1) \oplus (B/B_1)$. Note that since $R$ is an integral domain, finite direct sums of torsion modules are torsion. Thus $(A \oplus B)/(A_1 \oplus B_1)$ is torsion. Since $A_1 \oplus B_1$ is free and has free rank $n+m$, by this previous exercise, $A \oplus B$ has rank $n+m$.

### A characterization of rank in modules over an integral domain

Let $R$ be an integral domain and let $M$ be a (left, unital) $R$-module.

1. Suppose $M$ has rank $n$, and that $S = \{m_i\}$ is a linearly independent subset of $M$ having maximal cardinality. Let $N$ be the submodule of $M$ generated by $S$. Prove that $N$ is a free $R$-module having free rank $n$. Prove also that $M/N$ is torsion.
2. Conversely, prove that if there exists a submodule $N \subseteq M$ which is free and has free rank $n$ and such that $M/N$ is torsion, then $M$ has rank $n$.

1. Note that $N$ is free since its generating set $S$ is $R$-linearly independent, and that $S$ is a free generating set. Thus $N$ has free rank $n$. Now let $x+N \in M/N$. Note in particular that $x$ and $S = \{m_i\}$ are linearly dependent in $M$, since $S$ was a linearly independent set of maximal cardinality. Thus there exist $s,r_i \in R$ such that at least one of the $s,r_i$ is nonzero and $sx + \sum r_im_i = 0$. Suppose $s = 0$; then in fact $\sum r_i m_i = 0$, so that $r_i = 0$ for all $i$– a contradiction. Thus $s \neq 0$. Now $sx \in N$, so that $s(x+N) = 0$ in $M/N$. Thus $x+N$ is a torsion element. Since $x+N$ was arbitrary, $M/N$ is torsion.
2. Let $S \subseteq N$ be a free generating set (of cardinality $n$). It is certainly the case that $S$ is linearly independent in $M$, so that the rank of $M$ is at least $n$. Now let $T = \{t_i\}_{i=1}^{n+1}$ be a set of $n+1$ elements in $M$. Since $M/N$ is torsion, for each $t_i$, there exists a nonzero element $r_i \in R$ such that $r_it_i \in N$. If we have $r_it_i = r_jt_j$ for some $i$ and $j$, then $T$ is linearly dependent. Suppose this is not the case, so that $\{r_it_i\} \subseteq N$ is a subset having cardinality $n+1$. Since $N$ is free of free rank $n$, there exist nonzero $s_i$ such that $\sum s_ir_it_i = 0$. In particular, $T$ is linearly dependent in $M$. Hence the rank of $M$ is at most $n$. Thus the rank of $M$ is precisely $n$.

### Over an integral domain, the modules R and R/Tor R have the same rank

Let $R$ be an integral domain and let $M$ be a (left, unital) $R$-module. Recall that the rank of a module is the largest possible cardinality of a linearly independent subset. $\mathsf{Tor}\ M$ is the set of all $m \in M$ such that $rm = 0$ for some nonzero $r \in R$.

1. Prove that $\mathsf{Tor}\ M$ has rank 0.
2. Prove that $M$ and $M/\mathsf{Tor}\ M$ have the same rank.

Let $S \subseteq \mathsf{Tor}\ M$ be a nonempty set. Note that for each $m \in S$, there exists a nonzero element $r \in R$ such that $rs = 0$. In particular, $S$ cannot be linearly independent. Thus $\mathsf{Tor}\ M$ has rank 0.

Let $S = \{m_i\}_I \subseteq M$ be a linearly independent set of maximal order. Note in particular that no element of $S$ can be in $\mathsf{Tor}\ M$ (each subset of $S$ is linearly independent, in particular the singleton subsets are linearly independent.) We claim that the subset $S + \mathsf{Tor}\ M \subseteq M/\mathsf{Tor}\ M$ is linearly independent. To see this, suppose $\sum r_i m_i \in \mathsf{Tor}\ M$ for some $r_i \in R$. Then there exists a nonzero element $a \in R$ such that $\sum ar_im_i = 0$ in $M$. Since $S$ is linearly independent, we have $ar_i = 0$ in $R$. Since $R$ is an integral domain, and $a \neq 0$, we have $r_i = 0$ for all $i$. Thus $S + \mathsf{Tor}\ M$ is linearly independent. We claim also that $S$ and $S + \mathsf{Tor}\ M$ have the same cardinality. It suffices to show that for all $i$ and $j$, $m_i - m_j \notin \mathsf{Tor}\ M$. To that end, suppose $a \in R$ such that $am_i - am_j = 0$. Since $m_i$ and $m_i$ are linearly independent, $a = 0$. Thus $S$ and $S + \mathsf{Tor}\ M$ have the same cardinality. In particular, the rank of $M/\mathsf{Tor}\ M$ is at least as great as the rank of $M$.

Conversely, let $S = \{m_i\}$ and suppose $S + \mathsf{Tor}\ M$ is a linearly independent subset of maximal order. We claim that $S$ is linearly independent in $M$. To see this, suppose $\sum r_i m_i = 0$. Now $\sum r_i \overline{m_i} = 0$ in $M/\mathsf{Tor}\ M$. Thus $r_i = 0$, and so $S \subseteq M$ is linearly independent. Certainly the cardinality of $S$ is at least as great as the cardinality of $S + \mathsf{Tor}\ M$. Thus the rank of $M$ is at least as great as the rank of $M/\mathsf{Tor}\ M$.

Hence $M$ and $M/\mathsf{Tor}\ M$ have the same rank.

### Exterior powers of subdomains of a field of fractions

Let $R$ be an integral domain, and let $F$ be its field of fractions.

1. Considering $F$ as an $R$-module in the usual way, prove that $\bigwedge^2(F) = 0$.
2. Let $I \subseteq F$ be a submodule. Prove that $\bigwedge^k(I)$ is torsion for all $k \geq 2$.
3. Exhibit an integral domain $R$ and an $R$-submodule $I \subseteq F$ such that $\bigwedge^k(I) \neq 0$ for all $k \geq 0$.

Let $\frac{a}{b} \otimes \frac{c}{d} \in \mathcal{T}^2(F)$ be a simple tensor. Note that $\frac{a}{b} \otimes \frac{c}{d} = \frac{ad}{bd} \otimes \frac{cb}{bd}$ $= abcd(\frac{1}{bd} \otimes \frac{1}{bd}) \in \mathcal{A}^2(F)$. In particular, we have $\frac{a}{b} \wedge \frac{c}{d} = 0$ in $\bigwedge^2(I)$.

Suppose $\frac{a_1}{b_1} \wedge \frac{a_2}{b_2} \wedge \cdots \wedge \frac{a_k}{b_k} \in \bigwedge^k(I)$ be nonzero; then the $a_i$ are nonzero, and certainly the $b_i$ are nonzero. Now $a_1a_2b_1b_2 \neq 0$ in $R$, and evidently $a_1a_2b_1b_2(\frac{a_1}{b_1} \wedge \frac{a_2}{b_2} \wedge \cdots \wedge \frac{a_k}{b_k}) = \frac{a_1a_2}{1} \wedge \frac{a_1a_2}{1} \wedge \cdots \wedge \frac{a_k}{b_k} = 0$. So every element of $\bigwedge^k(I)$ is torsion, and thus $\bigwedge^k(I)$ is torsion as an $R$-module.

Now consider $R = \mathbb{Z}[x_1,\ldots,x_n]$, and let $I = (x_1,\ldots,x_n)$. We claim that if $\sum \alpha_ix_i = \sum \beta_ix_i \in I$, then there exist $h_i \in I$ such that $\alpha_i - \beta_i = h_i$. To see this, choose some $j$ and consider $\alpha_jx_j - \beta_jx_j = \sum_{i \neq j} (\beta_i - \alpha_i)x_i$. Since $R$ is a domain, $x_j$ divides the right hand side of this equation; say $\sum_{i \neq j} (\beta_i - \alpha_i)x_i = x_jh_j$. Note that every term on the left hand side is divisible by some $x_i$ other than $x_j$. Hence every term in $h_j$ is divisible by some $x_i$ other than $x_j$. In particular, $h_j \in I$, and $\alpha_i - \beta_i = h_i$.

Now consider the elements of $\prod^k I$ as “column vectors”- that is, write $\sum \alpha_i x_i$ as $[\alpha_1\ \alpha_2\ \ldots\ \alpha_n]^\mathsf{T}$. Note that this does not give a unique representation of the elements of $I$. However, if two column vectors $A$ and $B$ represent the same element in $I$, then there exists a third column vector $H$ such that $A = B+H$, and moreover the entries of $H$ are in $I$.

Now write the elements of $R^k$ as (square) matrices. Let us consider the determinant of such a matrix $A$, as an element of $R$, reduced mod $I$. This is an alternating bilinear map on the set of matrices over $R$; we claim that this is also well-defined up to our nonunique identification of matrices over $R$ with elements of $I^k$. To that end, suppose matrices $A$ and $B$ represent the same element of $I^k$; then we have a matrix $H$ such that $A = B+H$. Consider the determinant of both sides mod $I$. Using the combinatorial formula for computing determinants, we have $\mathsf{det}(B+H) = \sum_{\sigma \in S_n} \epsilon(\sigma) \prod (\beta_{\sigma(i),i} + h_{\sigma(i),i})$. Note that each $h_{i,j}$ is divisible by some $x_i$, and so goes to zero in the quotient $R/I$. So in fact $\mathsf{det}(A) = \mathsf{det}(B+H) \equiv \mathsf{det}(B) \mod\ I$; thus the map $\mathsf{det} : I^k \rightarrow R/I$ is a well-defined alternating bilinear map. Since $\mathsf{det}(x_1 \otimes \cdots \otimes x_n) = 1$ is nonzero, this map is nontrivial. Thus $\bigwedge^k(I) \neq 0$ for all $k$.

### Over an integral domain, every maximal ideal is irreducible

Let $R$ be an integral domain. Prove that every maximal ideal in $R$ is irreducible.

We begin with a lemma. This lemma is actually true (with a minor modification) over a much larger class of rings. However, I am not entirely comfortable with the proof of the general version. I am, however, comfortable with this more specific version. I referred to a discussion at MathForum.org for a proof of the lemma.

Lemma: Let $R$ be an integral domain and let $I \subseteq R$ be an ideal which is finitely generated as a $\mathbb{Z}$-module. If $I^2 = I$, then either $I = 0$ or $I = R$. Proof: Suppose $I \neq 0$. Let $A = \{\alpha_i\}_{i=1}^n$ be a generating set for $A$ over $\mathbb{Z}$; in particular, some $\alpha_i$ is nonzero. Since $I^2 = I$, there exist $u_{i,j} \in A$ such that $\alpha_i = \sum_j u_{i,j} \alpha_j$ for each $i$. In particular, $0 = \sum_j (u_{i,j} - \delta_{i,j})\alpha_j$ for each $i$, where $\delta_{i,j}$ is the Kronecker delta. That is, $[\alpha_1\ \cdots\ \alpha_n]^\mathsf{T}$ is a nontrivial solution to the matrix equation $Mx = 0$, where $M = [u_{i,j} - \delta_{i,j}]$. Thinking of $R$ as embedded in its field $F$ of fractions, we have $\mathsf{det}(M) = 0$. On the other hand, by the Leibniz expansion of $\mathsf{det}(M)$, we have $\mathsf{det}(M) = \beta - 1$, where $\beta \in I$ (using the fact that $I$ is an ideal). In particular, $-1 \in I$, so that $I = R$. $\square$

Now let $M \subseteq R$ be nonzero and maximal. If $M = AB$, then $M \subseteq A$ and $M \subseteq B$, so that $A,B \in \{M, R\}$. If $A = B = M$, then $M^2 = M$. By the lemma, either $M = 0$ (a contradiction) or $M = R$ (also a contradiction). So either $A$ or $B$ is $R$. Since $R$ is a unit in the semigroup of ideals of $R$ under ideal products, $M$ is irreducible.

### The tensor square of a princopal ideal in a domain is torsion free

Let $R$ be an integral domain and let $I \subseteq R$ be a principal ideal. Prove that the (left) $R$-module $I \otimes_R I$ has no nonzero torsion elements; that is, if $m \in I \otimes_R I$ and $r \in R$ is nonzero such that $r \cdot m = 0$ then $m = 0$.

Say $I = (a)$. Note that every simple tensor (hence every element) of $I \otimes_R I$ can be written in the form $ax \otimes ay = xy \cdot (a \otimes a)$. We claim that $I \otimes_R I$ is free on $a \otimes a$. To see this, note that by the universal propery of free modules we have a module homomorphism $R \rightarrow I \otimes_R I$ given by $r \mapsto r \cdot (a \otimes a)$.

Now let $\psi : I \times I \rightarrow R$ be given by $(xa,ya) = xy$. Clearly $\psi$ is bilinear, and so induces a module homomorphism $\Psi : I \otimes_R I \rightarrow R$ with $\Psi(xa \otimes ya) = xy$.

Note that $(\Psi \circ \Phi)(r) = \Psi(ra \otimes a) = r$ and $(\Phi \circ \Psi)(xa \otimes ya) = \Phi(xy) = xa \otimes ya$. Thus $\Phi$ and $\Psi$ are mutual inverses, so that $I \otimes_R I \cong_R R$. Since $R$ is a domain, $R$ is torsion free as an $R$-module, and thus $I \otimes_R I$ has no nonzero torsion elements.