## Archive for the ‘Rings and Modules’ Category

Throughout this post, $U(R)$ and $J(R)$ are the group of units and the Jacobson radical of a ring $R.$ Assuming that $U(R)$ is finite and $|U(R)|$ is odd, we will show that $|U(R)|=\prod_{i=1}^k (2^{n_i}-1)$ for some positive integers $k, n_1, \ldots , n_k.$ Let’s start with a nice little problem.

Problem 1. Prove that if $U(R)$ is finite, then $J(R)$ is finite too and $|U(R)|=|J(R)||U(R/J(R)|.$

Solution. Let $J:=J(R)$ and define the map $f: U(R) \to U(R/J))$ by $f(x) = x + J, \ x \in U(R).$ This map is clearly a well-defined group homomorphism. To prove that $f$ is surjective, suppose that $x + J \in U(R/J).$ Then $1-xy \in J,$ for some $y \in R,$ and hence $xy = 1-(1-xy) \in U(R)$ implying that $x \in U(R).$ So $f$ is surjective and thus $U(R)/\ker f \cong U(R/J).$
Now, $\ker f = \{1-x : \ \ x \in J \}$ is a subgroup of $U(R)$ and $|\ker f|=|J|.$ Thus $J$ is finite and $|U(R)|=|\ker f||U(R/J)|=|J||U(R/J)|. \Box$

Problem 2. Let $p$ be a prime number and suppose that $U(R)$ is finite and $pR=(0).$ Prove that if $p \nmid |U(R)|,$ then $J(R)=(0).$

Solution. Suppose that $J(R) \neq (0)$ and $0 \neq x \in J(R).$ Then, considering $J(R)$ as an additive group, $H:=\{ix: \ 0 \leq i \leq p-1 \}$ is a subgroup of $J(R)$ and so $p=|H| \mid |J(R)|.$ But then $p \mid |U(R)|,$ by Problem 1, and that’s a contradiction! $\Box$

There is also a direct, and maybe easier, way to solve Problem 2: suppose that there exists $0 \neq x \in J(R).$ On $U(R),$ define the relation $\sim$ as follows: $y \sim z$ if and only if $y-z = nx$ for some integer $n.$ Then $\sim$ is an equivalence relation and the equivalence class of $y \in U(R)$ is $[y]=\{y+ix: \ 0 \leq i \leq p-1 \}.$ Note that $[y] \subseteq U(R)$ because $x \in J(R)$ and $y \in U(R).$ So if $k$ is the number of equivalence classes, then $|U(R)|=k|[y]|=kp,$ contradiction!

Problem 3. Prove that if $F$ is a finite field, then $|U(M_n(F))|=\prod_{i=1}^n(|F|^n - |F|^{i-1}).$ In particular, if $|U(M_n(F))|$ is odd,  then $n=1$ and $|F|$ is a power of $2.$

Solution. The group $U(M_n(F))= \text{GL}(n,F)$ is isomorphic to the group of invertible linear maps $F^n \to F^n.$ Also, there is a one-to-one correspondence between the set of invertible linear maps $F^n \to F^n$ and the set of (ordered) bases of $F^n.$ So $|U(M_n(F))|$ is equal to the number of bases of $F^n.$ Now, to construct a basis for $F^n,$ we choose any non-zero element $v_1 \in F^n.$ There are $|F|^n-1$ different ways to choose $v_1.$ Now, to choose $v_2,$ we need to make sure that $v_1,v_2$ are not linearly dependent, i.e. $v_2 \notin Fv_1 \cong F.$ So there are $|F|^n-|F|$ possible ways to choose $v_2.$ Again, we need to choose $v_3$ somehow that $v_1,v_2,v_3$ are not linearly dependent, i.e. $v_3 \notin Fv_1+Fv_2 \cong F^2.$ So there are $|F|^n-|F|^2$ possible ways to choose $v_3.$ If we continue this process, we will get the formula given in the problem. $\Box$

Problem 4. Suppose that $U(R)$ is finite and $|U(R)|$ is odd. Prove that $|U(R)|=\prod_{i=1}^k (2^{n_i}-1)$ for some positive integers $k, n_1, \ldots , n_k.$

Solution. If $1 \neq -1$ in $R,$ then $\{1,-1\}$ would be a subgroup of order 2 in $U(R)$ and this is not possible because $|U(R)|$ is odd. So $1=-1.$ Hence $2R=(0)$ and $\mathbb{Z}/2\mathbb{Z} \cong \{0,1\} \subseteq R.$ Let $S$ be the ring generated by $\{0,1\}$ and $U(R).$ Obviously $S$ is finite, $2S=(0)$ and $U(S)=U(R).$ We also have $J(S)=(0),$ by Problem 2. So $S$ is a finite semisimple ring and hence $S \cong \prod_{i=1}^k M_{m_i}(F_i)$ for some positive integers $k, m_1, \ldots , m_k$ and some finite fields $F_1, \ldots , F_k,$ by the Artin-Wedderburn theorem and Wedderburn’s little theorem. Therefore $|U(R)|=|U(S)|=\prod_{i=1}^k |U(M_{m_i}(F_i))|.$ The result now follows from the second part of Problem 3. $\Box$

## Maximal and prime ideals of a polynomial ring over a PID (2)

Posted: September 21, 2012 in Elementary Algebra; Problems & Solutions, Rings and Modules
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See part (1) here! Again, we will assume that $R$ is a PID and $x$ is a varibale over $x.$ In this post, we will take a look at the maximal ideals of $R[x].$ Let $I$ be a maximal ideal of $R[x].$ By Problem 2, if $I \cap R \neq (0),$ then $I=\langle p, f(x) \rangle$ for some prime $p \in R$ and some $f(x) \in R[x]$ which is irreducible modulo $p.$ If $I \cap R =(0),$ then $I=\langle f(x) \rangle$ for some irreducible element $f(x) \in R[x].$ Before investigating maximal ideals of $R[x]$ in more details, let’s give an example of a PID $R$ which is not a field but $R[x]$ has a maximal ideal $I$ which is principal. We will see in Problem 3 that this situation may happen only when the number of prime elements of $R$ is finite.

Example 1. Let $F$ be a filed and put $R=F[[t]],$ the formal power series in the variable $t$ over $F.$ Let $x$ be a variable over $R.$ Then $I:=\langle xt - 1 \rangle$ is a maximal ideal of $R[x].$

Proof. See that $R[x]/I \cong F[[t,t^{-1}]]$ and that $F[[t,t^{-1}]]$ is the field of fractions of $R.$ Thus $R[x]/I$ is a field and so $I$ is a maximal ideal of $R[x]. \ \Box$

Problem 3. Prove that if $R$ has infinitely many prime elements, then an ideal $I$ of $R[x]$ is maximal if and only if $I=\langle p, f(x) \rangle$ for some prime $p \in R$ and some $f(x) \in R[x]$ which is irreducible modulo $p.$

Solution. We have already proved one direction of the problem in Problem 1. For the other direction, let $I$ be a maximal ideal of $R[x].$ By the first case in the solution of Problem 2 and the second part of Problem 1, we  only need to show that $I \cap R \neq (0).$ So suppose to the contrary that $I \cap R=(0).$ Then, by the second case in the solution of Problem 2, $I=\langle f(x) \rangle$ for some $f(x) \in R[x].$ We also know that $R[x]/I$ is a field because $I$ is a maximal ideal of $R[x].$ Since $R$ has infinitely many prime elements, we can choose a prime $p \in R$ such that $p$ does not divide the leading coefficient of $f(x).$ Now, consider the natural ring homomorphism $\psi : R[x] \to R[x]/I.$ Since $I \cap R=(0),$ $\psi(p) \neq 0$ and so $\psi(p)$ is invertible in $R[x]/I.$ Therefore $pg(x)-1 \in \ker \psi = I$ for some $g(x) \in R[x].$ Hence $pg(x)-1=h(x)f(x)$ for some $h(x) \in R[x].$ If $p \mid h(x),$ then we will have $p \mid 1$ which is non-sense. So $h(x)=pu(x) + v(x)$ for some $u(x),v(x) \in R[x]$ where $p$ does not divide the leading coefficient of $v(x).$ Now $pg(x) - 1 =h(x)f(x)$ gives us $p(g(x)-u(x)f(x)) - 1 =v(x)f(x)$ and so the leading coefficient of $v(x)f(x)$ is divisible by $p.$ Hence the leading coefficient of $f(x)$ must be divisible by $p,$ contradiction! $\Box$

Example 2. The ring of integers $\mathbb{Z}$ is a PID and it has infinitely many prime elements. So, by Problem 3, an ideal $I$ of $\mathbb{Z}[x]$ is maximal if and only if $I=\langle p, f(x) \rangle$ for some prime $p \in \mathbb{Z}$ and some $f(x)$ which is irreducible modulo $p.$ By Problem 2, the prime ideals of $\mathbb{Z}[x]$ are the union of the following sets:
1) all maximal ideals
2) all ideals of the form $\langle p \rangle,$ where $p \in \mathbb{Z}$ is a prime
3) all ideals of the form $\langle f(x) \rangle,$ where $f(x)$ is irreducible in $\mathbb{Z}[x].$

## 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!

## 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. Thus $(x^2y+yx^2)x^2=x^2(x^2y+yx^2),$ which gives $xy=yx. \ \Box$

## The singular submodule

Posted: December 7, 2011 in Elementary Algebra; Problems & Solutions, Rings and Modules
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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 $0 \neq sx \in K. \ \Box$

## The Krull dimension of a commutative ring; another view

Posted: November 3, 2011 in Elementary Algebra; Problems & Solutions, Rings and Modules
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We will assume that $R$ is a commutative ring with $1.$ We will allow the case $R=(0),$ i.e. where $1_R=0_R.$ The goal is to describe  the Krull dimension of $R$ in terms of elements of $R$ and not prime ideals of $R.$ The reference is this paper: A Short Proof for the Krull Dimension of a Polynomial Ring. Recall that the Krull dimension of $R$ is the largest integer $n \geq 0$ for which there exist prime ideals $P_i, \ 0 \leq i \leq n,$ of $R$ such that $P_0 \subset P_1 \subset \ldots \subset P_n.$ Then we write $\text{K.dim}(R)=n.$ If there is no such integer, then we define $\text{K.dim}(R)=\infty$ and if $R=(0),$ we define $\text{K.dim}(R)=-1.$

Notation. Let $x \in R$ and set $S_x = \{x^n(1+ax): \ n \geq 0, \ a \in R \}.$ Clearly $S_x$ is multiplicatively closed. Let $R_x$ be the localization of $R$ at $S_x.$ If $0 \in S_x,$ then we define $R_x =(0).$

Problem 1. Let $x \in R$ and suppose that $\mathfrak{m}$ is a maximal ideal of $R.$ Then $\mathfrak{m} \cap S_x \neq \emptyset.$ Moreover, if $P \subset \mathfrak{m}$ is a prime ideal and $x \in \mathfrak{m} \setminus P,$ then $P \cap S_x = \emptyset.$

Solution. if $x \in \mathfrak{m},$ then $x \in \mathfrak{m} \cap S_x$ and we are done. Otherwise, $\mathfrak{m} + Rx = R$ and thus $1 + ax \in \mathfrak{m}$ for some $a \in R.$ Then $1+ax \in \mathfrak{m} \cap S_x.$ For the second part, suppose to the contrary that $x^n(1+ax) \in P$ for some integer $n \geq 0$ and $a \in R.$ Then, since $x \notin P,$ we have $1+ax \in P \subset \mathfrak{m}$ and thus $1 \in \mathfrak{m}$ because $x \in \mathfrak{m}. \Box$

Problem 2. $\text{K.dim}(R)=0$ if and only if $R_x=\{0\}$ for all $x \in R.$

Solution. As we already mentioned, $R_x = \{0\}$ means $0 \in S_x.$ Suppose that $0 \notin S_x$ for some $x \in R.$ Then there exists a prime ideal $P$ of $R$ such that $P \cap S_x = \emptyset,$ because $S_x$ is multiplicatively closed. By Problem 1, $P$ is not a maximal ideal and thus $\text{K.dim}(R) > 0.$ Conversely, suppose that $0 \in S_x$ for all $x \in R$ and $P$ is a prime ideal of $R.$ Let $\mathfrak{m}$ be  a maximal ideal of $R$ which contains $P.$ Choose $x \in \mathfrak{m} \setminus P.$ By Problem 1, $P \cap S_x =\emptyset,$ contradicting $0 \in P \cap S_x. \ \Box$

Problem 3. Let $n \geq 0$ be an integer. Then $\text{K.dim}(R) \leq n$ if and only if $\text{K.dim}(R_x) \leq n-1$ for all $x \in R.$

Solution. The case $n=0$ was done in Problem 2. So we’ll assume that $n \geq 1.$ Suppose that $\text{K.dim}(R) \leq n$ and let $x \in R.$ A prime ideal of $R_x$ is in the form $PR_x,$ where $P$ is a prime ideal of $R$ and $P \cap S_x = \emptyset.$ Also, if $PR_x$ and $QR_x$ are two prime ideals of $R_x$ with $PR_x \subset QR_x,$ then $P \subset Q.$ Now, suppose to the contrary that $\text{K.dim}(R_x) \geq n$ and consider the chain $P_0R_x \subset P_1R_x \subset \ldots \subset P_n R_x$ of prime ideals of $R_x.$ By Problem 1, $P_n$ is not a maximal ideal and so there exists a prime ideal $P_{n+1}$ such that $P_n \subset P_{n+1}$ and that will give us the chain $P_0 \subset P_1 \subset \ldots \subset P_{n+1}$ of prime ideals of $R,$ contradicting $\text{K.dim}(R) \leq n.$ Conversely, suppose that $\text{K.dim}(R_x) \leq n-1$ for all $x \in R.$ Suppose also, to the contrary, that there exists a chain $P_0 \subset P_1 \subset \ldots \subset P_{n+1}$ of prime ideals of $R.$ Let $x \in P_{n+1} \setminus P_n.$ By the second part of Problem 1, $P_n \cap S_x = \emptyset$ and thus $P_0R_x \subset P_1R_x \subset \ldots \subset P_nR_x$ is a chain of prime ideals of $R_x,$ contradicting $\text{K.dim}(R_x) \leq n-1. \ \Box$

Problem 4. Let $n \geq 0$ be an integer. Then $\text{K.dim}(R) \leq n$ if and only if for every $x_0, x_1, \ldots , x_n \in R$ there exist integers $k_0, k_1, \ldots , k_n \geq 0$ and $a_0, a_1, \ldots ,a_n \in R$ such that

$x_n^{k_n}( \ldots (x_2^{k_2}(x_1^{k_1}(x_0^{k_0}(1+a_0x_0)+a_1x_1) + a_2x_2) + \ldots) + a_nx_n)=0.$

Solution. By Problem 3, $\text{K.dim}(R) \leq n$ if and only if $\text{K.dim}((\ldots ((R_{x_0})_{x_1})_{x_2} \ldots)_{x_n})=-1.$ Therefore $\text{K.dim}(R) \leq n$ if and only if $(\ldots ((R_{x_0})_{x_1})_{x_2} \ldots)_{x_n}=\{0\}. \ \Box$