## Maximal and prime ideals of a polynomial ring over a PID (1)

Posted: September 21, 2012 in Elementary Algebra; Problems & Solutions, Rings and Modules
Tags: , , , ,

We know that if $R$ is a field and if $x$ is a variable over $R,$ then $R[x]$ is a PID and a non-zero ideal $I$ of $R[x]$ is maximal if and only if $I$ is prime if and only if $I$ is generated by an irreducible element of $R[x].$ If $R$ is a PID which is not a field, then $R[x]$ could have prime ideals which are not maximal. For example, in $\mathbb{Z}[x]$ the ideal $\langle 2 \rangle$ is prime but not maximal. In this two-part post, we will find prime and maximal ideals of $R[x]$ when $R$ is a PID.

Notation. Throughout this post, $R$ is a PID and $R[x]$ is the polynomial ring in the variable $x$ over $R.$ Given a prime element $p \in R,$ we will denote by $\phi_p$ the natural ring homomorphism $R[x] \to R[x]/pR[x].$

Definition Let $p$ be a prime element of $R.$ An element $f(x) \in R[x]$ is called irreducible modulo $p$ if $\phi_p(f(x))$ is irreducible in $R[x]/pR[x].$ Let $\gamma : R \to R/pR$ be the natural ring homomorphism. Then, since $R[x]/pR[x] \cong (R/pR)[x],$ an element $f(x)=\sum_{i=0}^n a_ix^i \in R[x]$ is irreducible modulo $p$ if and only if $\sum_{i=0}^n \gamma(a_i)x^i$ is irreducible in $(R/pR)[x].$ Note that $R/pR$ is a field because $R$ is a PID.

Problem 1. Prove that if $p \in R$ is prime and if $f(x) \in R[x]$ is irreducible modulo $p,$ then $I:=\langle p, f(x) \rangle$ is a maximal ideal of $R[x].$ If $f =0,$ then $I$ is a prime but not a maximal ideal of $R[x].$

Solution. Clearly $I/pR[x]=\phi_p(I)=\langle \phi_p(f(x)) \rangle.$ So $\phi_p(I)$ is a maximal ideal of $R[x]/pR[x]$ because $\phi_p(f(x))$ is irreducible in $R[x]/pR[x] \cong (R/pR)[x]$ and $R/pR$ is a field. So $I$ is a maximal ideal of $R[x].$ If $f =0,$ then $I=\langle p \rangle=pR[x]$ and so $R[x]/I \cong (R/pR)[x]$ is a domain which implies that $I$ is prime. Finally, $I= \langle p \rangle$ is not maximal because, for example, $I \subset \langle p,x \rangle \ \Box$

Problem 2. Prove that a non-zero ideal $I$ of $R[x]$ is prime if and only if either $I= \langle f(x) \rangle$ for some irreducible element $f(x) \in R[x]$ or $I=\langle p, f(x) \rangle$ for some prime $p \in R$ and some $f(x) \in R[x]$ which is either zero or irreducible modulo $p.$

Solution. If $f(x) \in R[x]$ is irreducible, then $\langle f(x) \rangle$ is a prime ideal of $R[x]$ because $R[x]$ is a UFD. If $f(x)=0$ or $f(x)$ is irreducible modulo a prime $p \in R,$ then $I=\langle p, f(x) \rangle$ is a prime ideal of $R[x]$ by Problem 1.
Conversely, suppose that $I$ is a non-zero prime ideal of $R[x].$ We consider two cases.
Case 1. $I \cap R \neq (0)$ : Let $0 \neq r \in I \cap R.$ Then $r$ is clearly not a unit because then $I$ wouldn’t be a proper ideal of $R[x].$ So, since $r \in I$ and $I$ is a prime ideal of $R[x],$ there exists a prime divisor $p$ of $r$ such that $p \in I.$  So $pR[x] \subseteq I$ and hence $\phi_p(I)=I/pR[x]$ is a prime ideal of $R[x]/pR[x] \cong (R/pR)[x].$ Thus we have two possibilities. The first possibility is that $\phi_p(I)=(0),$ which gives us $I \subseteq \ker \phi_p = pR[x]$ and therefore $I=pR[x]=\langle p \rangle.$ The second possibility is that $\phi_p(I)=\langle \phi_p(f(x)) \rangle= \phi_p(\langle f(x) \rangle)$ for some irreducible element $\phi_p(f(x)) \in R[x]/pR[x],$ which gives us $I=\langle p, f(x) \rangle$ because $\ker \phi_p =pR[x].$
Case 2. $I \cap R = (0)$ : Let $Q$ be the field of fractions of $R$ and put $J:=IQ[x].$ Then $J$ is a non-zero prime ideal of $Q[x]$ because $I$ is a prime ideal of $R[x].$ Note that $J=\{g(x)/r : \ g(x) \in I, \ 0 \neq r \in R \}.$ So, since $Q[x]$ is a PID, $J=q(x)Q[x]$ for some irreducible element $q(x) \in Q[x].$ Obviously, we can write $q(x)=\alpha f(x),$ where $\alpha \in Q$ and $f(x) \in R[x]$ is irreducible and the gcd of the coefficients of $f(x)$ is one. Thus $J = f(x)Q[x]$ and, since $f(x) \in J,$ we have $f(x) = g(x)/r$ for some $0 \neq r \in R$ and $g(x) \in I.$ But then $rf(x)=g(x) \in I$ and so $f(x) \in I$ because $I$ is prime and $I \cap R = (0).$ Hence $\langle f(x) \rangle \subseteq I.$ We will be done if we prove that $I \subseteq \langle f(x) \rangle.$ To prove this, let $h(x) \in I \subseteq J=f(x)Q[x].$ So $h(x)=f(x)q_0(x)$ for some $q_0(x) \in Q[x].$ Therefore, since the gcd of the coefficients of $f(x)$ is one, we must have $q_0(x) \in R[x]$ by Gauss’s lemma. Hence $h(x) \in \langle f(x) \rangle$ and the solution is complete. $\Box$

See the next part here!

## Maximum order of abelian subgroups in a symmetric group

Posted: April 21, 2012 in Elementary Algebra; Problems & Solutions, Groups and Fields
Tags: , ,

For any set $X,$ we denote by ${\rm{Sym}}(X)$ the group of bijective maps from $X$ to $X.$

Problem. Let $\alpha : \mathbb{N} \longrightarrow [1, \infty)$ be any map which satisfies the following two conditions: $\alpha(p) \geq p$ and $\alpha(p+q) \geq \alpha(p) \alpha(q)$ for all $p,q \in \mathbb{N}.$ Let $X$ be a set with $|X|=n.$ Prove that if $G$ is an abelian subgroup of ${\rm{Sym}}(X),$ then $|G| \leq \alpha(n).$

Solution.  The proof is by induction on $n.$ If $n=1,$ then $|G|=1 \leq \alpha(1).$ Let $X$ be a set with $|X|= n \geq 2$ and suppose that the claim is true for any set of size $< n.$ Let $G$ be an abelian subgroup of ${\rm{Sym}}(X).$ Clearly $gx=g(x), \ g \in G, x \in X,$ defines an action of $G$ on $X.$ Let $X_1, \ldots , X_k$ be the orbits corresponding to this action and consider two cases.

Case 1. $k=1$: Fix an element $x_1 \in X.$ Then $X=X_1=Gx_1.$ Suppose that $g_1x_1=x_1$ for some $g_1 \in G$ and let $x \in X.$ Then $x=gx_1$ for some $g \in G.$ Thus, since $G$ is abelian, we have

$x=gx_1=gg_1x_1=g_1gx_1=g_1x.$

Hence $g_1x=x$ for all $x \in X$ and thus $g_1=1.$ So the stabilizer of $x_1$ is trivial and therefore, by the orbit-stabilizer theorem, $|G|=|X|=n \leq \alpha(n).$

Case 2$k \geq 2$: Let $|X_i|=n_i, \ i=1,2, \ldots, k.$ Clearly $\sum_{i=1}^k n_i=n$ and, since $k \geq 2,$ we have $n_i < n$ for all $i.$ For every $g \in G$ and $1 \leq i \leq k$ let $g_i=g|_{X_i},$ the restriction of $g$ to $X_i,$ and put

$G_i=\{g_i: \ g \in G\}.$

Then $g_i \in {\rm{Sym}}(X_i)$ and $G_i$ is an abelian subgroup of ${\rm{Sym}}(X_i).$ Thus, by the induction hypothesis

$|G_i| \leq \alpha(n_i),$

for all $i.$ Now, define $\varphi : G \longrightarrow \bigoplus_{i=1}^k G_i$ by $\varphi(g)=(g_1, g_2, \ldots , g_k)$ for all $g \in G.$ It is obvious that $\varphi$ is one-to-one and so

$|G| \leq |\bigoplus_{i=1}^k G_i|=\prod_{i=1}^k |G_i| \leq \prod_{i=1}^k \alpha(n_i) \leq \alpha(\sum_{i=1}^k n_i)=\alpha(n). \ \Box$

Remark. The map $\alpha: \mathbb{N} \longrightarrow [1, \infty)$ defined by $\alpha(p)=3^{p/3},$ for all $p \in \mathbb{N},$ satisfies both conditions in the above Problem. So if $|X|=n$ and if $G$ is an abelian subgroup of ${\rm{Sym}}(X),$ then $|G| \leq 3^{n/3}.$

## Rings satisfying x^4 = x are commutative

Posted: April 19, 2012 in Elementary Algebra; Problems & Solutions, Rings and Modules
Tags: , ,

Let $R$ be a ring, which may or may not have $1.$ We proved in here that if $x^3=x$ for all $x \in R,$ then $R$ is commutative.  A similar approach shows that if $x^4=x$ for all $x \in R,$ then $R$ is commutative.

Problem. Prove that if $x^4=x$ for all $x \in R,$ then $R$ is commutative.

Solution. Clearly $R$ is reduced, i.e. $R$ has no nonzero nilpotent element. Note that $2x=0$ for all $x \in R$ because $x=x^4=(-x)^4=-x.$ Hence $x^2+x$ is an idempotent for every $x \in R$ because

$(x^2+x)^2=x^4+2x^3+x^2=x^2+x.$

Thus $x^2+x$ is central for all $x \in R,$ by Remark 3 in this post.  Therefore $(x^2+y)^2+x^2+y$ is central for all $x,y \in R.$ But

$(x^2+y)^2+x^2+y=x^2+x+y^2+y+ x^2y+yx^2$

and hence $x^2y+yx^2$ is central. Therefore $(x^2y+yx^2)x^2=x^2(x^2y+yx^2)$ which gives us $xy=yx. \ \Box$

## GK dimension of Weyl algebras

Posted: April 10, 2012 in Gelfand-Kirillov Dimension, Noncommutative Ring Theory Notes
Tags: ,

We defined the $n$-th Weyl algebra $A_n(R)$ over a ring $R$ in here.  In this post we will find the GK dimension of $A_n(R)$ in terms of the GK dimension of $R.$ The result is similar to what we have already seen in commutative polynomial rings (see corollary 1 in here). We will assume that $k$ is a field and $R$ is a $k$-algebra.

Theorem. ${\rm{GKdim}}(A_1(R))=2 + {\rm{GKdim}}(R).$

Proof. Suppose first that $R$ is finitely generated and let $V$ be a frame of $R.$ Let $U=k+kx+ky.$ Since $yx = xy +1,$ we have

$\dim_k U^n = \frac{(n+1)(n+2)}{2}. \ \ \ \ \ \ \ \ \ (*)$

Let $W=U+V.$ Clearly $W$ is a frame of $A_1(R)$ and

$W^n = \sum_{i+j=n} U^i V^j,$

for all $n,$ because every element of $V$ commutes with every element of $U.$ Therefore, since $V^j \subseteq V^n$ and $U^i \subseteq U^n$ for all $i,j \leq n,$ we have $W^n \subseteq U^nV^n$ and $W^{2n} \supseteq U^nV^n.$ Thus $W^n \subseteq U^nV^n \subseteq W^{2n}$ and hence

$\log_n \dim_k W^n \leq \log_n \dim_k U^n + \log_n \dim_k V^n \leq \log_n \dim_k W^{2n}.$

Therefore ${\rm{GKdim}}(A_1(R)) \leq 2 + {\rm{GKdim}}(R) \leq {\rm{GKdim}}(A_1(R)),$ by $(*),$ and we are done.

For the general case, let $R_0$ be any finitely generated $k$– subalgebra of $R.$ Then, by what we just proved,

$2 + {\rm{GKdim}}(R_0)={\rm{GKdim}}(A_1(R_0)) \leq {\rm{GKdim}}(A_1(R))$

and hence $2+{\rm{GKdim}}(R) \leq {\rm{GKdim}}(A_1(R)).$ Now, let $A_0$ be a $k$-subalgebra of $A_1(R)$ generated by a finite set $\{f_1, \ldots , f_m\}.$ Let $R_0$ be the $k$-subalgebra of $R$ generated by all the coefficients of $f_1, \ldots , f_m.$ Then $A_0 \subseteq A_1(R_0)$ and so

${\rm{GKdim}}(A_0) \leq {\rm{GKdim}}(A_1(R_0))=2 + {\rm{GKdim}}(R_0) \leq 2 + {\rm{GKdim}}(R).$

Thus

${\rm{GKdim}}(A_1(R)) \leq 2 + {\rm{GKdim}}(R)$

and the proof is complete. $\Box$

Corollary. ${\rm{GKdim}}(A_n(R))=2n + {\rm{GKdim}}(R)$ for all $n.$ In particular, ${\rm{GKdim}}(A_n(k))=2n.$

Proof. It follows from the theorem and the fact that $A_n(R)=A_1(A_{n-1}(R)). \Box$

## A theorem of Borho and Kraft

Posted: April 2, 2012 in Gelfand-Kirillov Dimension, Noncommutative Ring Theory Notes
Tags: , , ,

As usual, I’ll assume that $k$ is a field. Recall that if a $k$-algebra $A$ is an Ore domain, then we can localize $A$ at $S:=A \setminus \{0\}$ and get the division algebra $Q(A):=S^{-1}A.$ The algebra $Q(A)$ is called the quotient division algebra of $A.$

Theorem (Borho and Kraft, 1976) Let $A$ be a finitely generated $k$-algebra which is a domain of finite GK dimension. Let $B$ be a $k$-subalgebra of $A$ and suppose that ${\rm{GKdim}}(A) < {\rm{GKdim}}(B) + 1.$ Let $S:=B \setminus \{0\}.$ Then $S$ is an Ore subset of $A$ and $S^{-1}A=Q(A).$ Also, $Q(A)$ is finite dimensional as a (left or right) vector space over $Q(B).$

Proof. First note that, by the corollary in this post, $A$ is an Ore domain and hence both $Q(A)$ and $Q(B)$ exist and they are division algebras. Now, suppose, to the contrary, that $S$ is not (left) Ore. Then there exist $x \in S$ and $y \in A$ such that $Sy \cap Ax = \emptyset.$ This implies that the sum $By + Byx + \ldots + Byx^m$ is direct for any integer $m.$ Let $W$ be a frame of a finitely generated subalgebra $B'$ of $B.$ Let $V=W+kx+ky$ and suppose that $A'$ is the subalgebra of $A$ which is generated by $V.$ For any positive integer $n$ we have

$V^{2n} \supseteq W^n(kx+ky)^n \supseteq W^ny + W^nyx + \ldots + W^nyx^{n-1}$

and thus $\dim_k V^{2n} \geq n \dim_k W^n$ because the sum is direct. So $\log_n \dim_k V^{2n} \geq 1 + \log_n \dim_k W^n$ and hence ${\rm{GKdim}}(A) \geq {\rm{GKdim}}(A') \geq 1 + {\rm{GKdim}}(B').$ Taking supremum of both sides over all finitely generated subalgebras $B'$ of $B$ will give us the contradiction ${\rm{GKdim}}(A) \geq 1 + {\rm{GKdim}}(B).$ A similar argument shows that $S$ is right Ore. So we have proved that $S$ is an Ore subset of $A.$ Before we show that $S^{-1}A=Q(A),$ we will prove that $Q(B)A=S^{-1}A$ is finite dimensional as a (left) vector space over $Q(B).$ So let $V$ be a frame of $A.$ For any positive ineteger $n,$ let $r(n) = \dim_{Q(B)} Q(B)V^n.$ Clearly $Q(B)V^n \subseteq Q(B)V^{n+1}$ for all $n$ and

$\bigcup_{n=0}^{\infty}Q(B)V^n =Q(B)A$

because $\bigcup_{n=0}^{\infty}V^n=A.$ So we have two possibilities: either $Q(B)V^n=Q(B)A$ for some $n$ or the sequence $\{r(n)\}$ is strictly increasing. If $Q(B)V^n = Q(B)A,$ then we are done because $V^n$ is finite dimensional over $k$ and hence $Q(B)V^n$ is finite dimensional over $Q(B).$ Now suppose that the sequence $\{r(n)\}$ is strictly increasing. Then $r(n) > n$ because $r(0)=\dim_{Q(B)}Q(B)=1.$ Fix an integer $n$ and let $e_1, \ldots , e_{r(n)}$ be a $Q(B)$-basis for $Q(B)V^n.$ Clearly we may assume that $e_i \in V^n$ for all $i.$ Let $W$ be a frame of a finitely generated subalgebra of $B.$ Then

$(V+W)^{2n} \supseteq W^nV^n \supseteq W^ne_1 + \ldots + W^ne_{r(n)},$

which gives us

$\dim_k(V+W)^{2n} \geq r(n) \dim_k W^n > n \dim_k W^n,$

because the sum $W^ne_1 + \ldots + W^ne_{r(n)}$ is direct. Therefore ${\rm{GKdim}}(A) \geq 1 + {\rm{GKdim}}(B),$ which is a contradiction. So we have proved that the second possibility is in fact impossible and hence $Q(B)A$ is finite dimensional over $Q(B).$ Finally, since, as we just proved, $\dim_{Q(B)}Q(B)A < \infty,$ the domain $Q(B)A$ is algebraic over $Q(B)$ and thus it is a division algebra. Hence $Q(B)A=Q(A)$ because $A \subseteq Q(B)A \subseteq Q(A)$ and $Q(A)$ is the smallest division algebra containing $A. \Box$

## The singular submodule

Posted: December 7, 2011 in Elementary Algebra; Problems & Solutions, Rings and Modules
Tags: , , ,

Throughout $R$ is a ring with 1 and all modules are left $R$-modules. In Definition 2 in this post, we defined $Z(M),$ the singular submodule of a module $M.$

Problem 1. Let $M$ be an $R$-module and suppose that $N_1, \cdots, N_k$ are submodules of $M.$ Prove that $\bigcap_{i=1}^k N_i \subseteq_e M$ if and only if $N_i \subseteq_e M$ for all $i.$

Solution. We only need to solve the problem for $k = 2.$ If $N_1 \cap N_2 \subseteq_e M,$ then $N_1 \subseteq_e M$ and $N_2 \subseteq_e M$ because both $N_1$ and $N_2$ contain $N_1 \cap N_2.$ Conversely, let $P$ be a nonzero submodule of $M.$ Then $N_1 \cap P \neq \{0\}$ because $N_1 \subseteq_e M$ and therefore $(N_1 \cap N_2) \cap P = N_2 \cap (N_1 \cap P) \neq \{0\}$ because $N_2 \subseteq_e M. \ \Box$

Problem 2. Prove that if $M$ is an $R$-module, then $Z(M)$ is a submodule of $M$ and $Z(R)$ is a proper two-sided ideal of $R.$ In particular, if $R$ is a simple ring, then $Z(R)=\{0\}.$

Solution. First note that $0 \in Z(M)$ because $\text{ann}(0)=R \subseteq_e R.$ Now suppose that $x_1,x_2 \in Z(M).$ Then $\text{ann}(x_1+x_2) \supseteq \text{ann}(x_1) \cap \text{ann}(x_2) \subseteq_e M,$ by Problem 1. Therefore $\text{ann}(x_1+x_2) \subseteq_e M$ and hence $x_1+x_2 \in Z(M).$ Now let $r \in R$ and $x \in Z(M).$ We need to show that $rx \in Z(M).$ Let $J$ be a nonzero left ideal of $R.$ Then $Jr$ is also a left ideal of $R.$ If $Jr = \{0\},$ then $J \subseteq \text{ann}(rx)$ and thus $\text{ann}(rx) \cap J = J \neq \{0 \}.$ If $Jr \neq \{0\},$ then $\text{ann}(x) \cap Jr \neq \{0\}$ because $x \in Z(M).$ So there exists $s \in J$ such that $sr \neq 0$ and $srx = 0.$ Hence $0 \neq s \in \text{ann}(rx) \cap J.$ So $rx \in Z(M)$ and thus $Z(M)$ is a submodule of $M.$ Now, considering $R$ as a left $R$-module, $Z(R)$ is a left ideal of $R,$ by what we have just proved. To see why $Z(M)$ is a right ideal, let $r \in R$ and $x \in Z(R).$ Then $\text{ann}(xr) \supseteq \text{ann}(x) \subseteq_e R$ and so $\text{ann}(xr) \subseteq_e R,$ i.e. $xr \in Z(R).$ Finally, $Z(R)$ is proper because $\text{ann}(1)=\{0\}$ and so $1 \notin Z(R). \ \Box$

Problem 3. Prove that if $M_i, \ i \in I,$ are $R$-modules, then $Z(\bigoplus_{i \in I} M_i) = \bigoplus_{i \in I} Z(M_i).$ Conclude that if $R$ is a semisimple ring, then $Z(R)=\{0\}.$

Solution. The first part is a trivial result of Problem 1 and this fact that if $x = x_1 + \cdots + x_n,$ where the sum is direct, then $\text{ann}(x) = \bigcap_{i=1}^n \text{ann}(x_i).$ The second now follows trivially from the first part, Problem 2 and the Wedderburn-Artin theorem. $\Box$

Problem 4. Suppose that $R$ is commutative and let $N(R)$ be the nilradical of $R.$ Prove that

1) $N(R) \subseteq Z(R);$

2) it is possible to have $N(R) \neq Z(R);$

3) if $Z(R) \neq \{0\},$ then $N(R) \subseteq_e Z(R),$ as $R$-modules or $Z(R)$-modules.

Solution. 1) Let $a \in N(R).$ Then $a^n = 0$ for some integer $n \geq 1.$ Now suppose that $0 \neq r \in R.$ Then $ra^n=0.$ Let $m \geq 1$ be the smallest integer such that $ra^m = 0.$ Then $0 \neq ra^{m-1} \in \text{ann}(a) \cap Rr$ and hence $a \in Z(R).$

2) Let $R_i = \mathbb{Z}/2^i \mathbb{Z}, \ i \geq 1$ and put $R=\prod_{i=1}^{\infty}R_i.$ For every $i,$ let $a_i = 2 + 2^i \mathbb{Z}$ and consider $a = (a_1,a_2, \cdots ) \in R.$ It is easy to see that $a \in Z(R) \setminus N(R).$

3) Let $a \in Z(R) \setminus N(R).$ Then $\text{ann}(a) \cap Ra \neq \{0\}$ and thus there exists $r \in R$ such that $ra \neq 0$ and $ra^2=0.$ Hence $(ra)^2 = 0$ and so $ra \in N(R).$ Thus $0 \neq ra \in N(R) \cap Ra$ implying that $N(R)$ is an essential $R$-submodule of $Z(R).$ Now, we view $Z(R)$ as a ring and we want to prove that $N(R)$ as an essential ideal of $Z(R).$ Again,  let $a \in Z(R) \setminus N(R).$ Then $\text{ann}(a) \cap Ra^2 \neq \{0\}$ and thus there exists $r \in R$ such that $ra^2 \neq 0$ and $ra^3 = 0.$ Let $s = ra \in Z(R).$ Then $(sa)^2=0$ and thus $0 \neq sa \in N(R) \cap Z(R)a$ implying that $N(R)$ is an essential ideal of $Z(R). \ \Box$

## A singular module is a quotient of a module by an essential submodule

Posted: December 4, 2011 in Elementary Algebra; Problems & Solutions, Rings and Modules
Tags: , , ,

We will assume that $R$ is a ring (not necessarily commutative) with 1 and all modules are left $R$-modules.

Definition 1. Let $M$ be an $R$-module and $N$ a nonzero submodule of $M.$ We say that $N$ is an essential submodule of $M,$ and we will write $N \subseteq_e M,$ if $N \cap X \neq (0)$ for any nonzero submodule $X$ of $M.$ Clearly, that is equivalent to saying $N \cap Rx \neq (0)$ for any nonzero element $x \in M.$ So, in particular, a nonzero left ideal $I$ of $R$ is an essential left ideal of $R$ if $I \cap J \neq (0)$ for any nonzero left ideal $J$ of $R,$ which is equivalent to the condition $I \cap Rr \neq (0)$ for any nonzero element $r \in R.$

Definition 2. Let $M$ be an $R$-module and $x \in M.$ Recall that the (left) annihilator of $x$ in $R$ is defined by $\text{ann}(x)=\{r \in R: \ rx = 0 \},$ which is obviously a left ideal of $R.$ Now, consider the set $Z(M):=\{x \in M: \ \text{ann}(x) \subseteq_e R \}.$ It is easy to see that $Z(M)$ is a submodule of $M$ (see Problem 2 in this post for the proof!) and we will call it the singular submodule of $M.$ If $Z(M)=M,$ then $M$ is called singular. If $Z(M)=(0),$ then $M$ is called nonsingular. We will not discuss nonsingular modules in this post.

Problem 1. Prove that if $F$ is a free $R$-module, then $Z(F) \neq F,$ i.e. a free module is never singular.

Solution. Let $x \in F$ be any element of an $R$-basis of $F.$ Let $r \in \text{ann}(x).$ Then $rx = 0$ and so $r = 0.$ Thus $\text{ann}(x)=(0)$ and so $x \notin Z(F). \ \Box$

Next problem characterizes singular modules.

Problem 2. Prove that an $R$-module $M$ is singular if and only if $M = A/B$ for some $R$-module $A$ and some submodule $B \subseteq_e A.$

Solution. Suppose first that $M=A/B$ where $A$ is an $R$-module and $B \subseteq_e A.$ Let $x = a + B \in M$ and let $J$ be a nonzero left ideal of $R.$ If $Ja = (0),$ then $Ja \subseteq B$ and so $\text{ann}(x) \cap J = J \neq (0).$ If $Ja \neq (0),$ then $B \cap Ja \neq (0)$ because $B \subseteq_e A.$ So there exists $r \in J$ such that $0 \neq ra \in B.$ That means $0 \neq r \in \text{ann}(x) \cap J.$ So we have proved that $x \in Z(M)$ and hence $Z(M)=M,$ i.e. $M$ is singular. Conversely, suppose that $M$ is singular. We know that every $R$-module is the homomorphic image of some free $R$-module. So there exists a free $R$-module $F$ and a submodule $K$ of $F$ such that $M \cong F/K.$ So we only need to show that $K \subseteq_e F.$ Note that $K \neq (0),$ by Problem 1. Let $\{x_i \}$ be an $R$-basis for $F$ and suppose that $0 \neq x \in F.$ We need to show that there exists $s \in R$ such that $0 \neq sx \in K.$ We can write, after renaming the indices if necessarily,

$x = \sum_{i=1}^n r_ix_i, \ \ \ \ \ \ \ \ \ \ (1)$

where $r_1 \neq 0.$ For any $y \in F,$ let $\overline{y}=y+K \in F/K.$ Now, since $F/K$ is singular, $\text{ann}(\overline{x_1}) \subseteq_e R$ and so $\text{ann}(\overline{x_1}) \cap Rr_1 \neq (0).$ So there exists $s_1 \in R$ such that $s_1r_1 \neq 0$ and $s_1r_1x_1 \in K.$ Hence $(1)$ gives us

$s_1x= s_1r_1x_1 +\sum_{i=2}^n s_1r_ix_i. \ \ \ \ \ \ \ \ \ (2)$

Note that $s_1x \neq 0$ because $s_1r_1 \neq 0.$ Now, if $s_1r_i = 0$ for all $2 \leq i \leq n,$ then $s_1x =s_1r_1x_1 \in K$ and we are done. Otherwise, after renaming the indices in the sum on the right hand side of $(2)$ if necessary, we may assume that $s_1r_2 \neq 0.$ Repeating the above process gives us some $s_2 \in R$ such that $s_2s_1r_2 \neq 0$ and $s_2s_1r_2x_2 \in K.$ Then $(2)$ implies

$s_2s_1x = s_2s_1r_1x_1 + s_2s_1r_2x_2 + \sum_{i=3}^n s_2s_1r_ix_i. \ \ \ \ \ \ \ (3)$

The first two terms on the right hand side of $(3)$ are in $K$ and $s_2s_1x \neq0$ because $s_2s_1r_2 \neq 0.$ If we continue this process, we will eventually have a positive integer $1 \leq m \leq n$ and $s = s_ms_{m-1} \cdots s_1 \in R$ such that $0 \neq sx \in K. \ \Box$