## Jacobson radical of polynomial rings

Posted: February 21, 2015 in Jacobson Radical, Noncommutative Ring Theory Notes
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Let $R$ be a ring (not necessarily commutative) with $1$. Let $S=R[x]$ be the ring of polynomials in the central indeterminate $x,$ i.e. $x$ is in the center of $S.$ Let $J(S)$ be the Jacobson radical of $S.$ A few decades ago, Amitsur proved that $J(S)=(R \cap J(S))[x].$ In fact, Amitsur’s result holds for polynomial rings in any number of indeterminates, finite or infinite. The proof of this result is not hard but we will not give it here. What we are going to prove is the following simple fact.

Problem. Prove that if $R$ is commutative, then the leading coefficient of every element of $J(S)$ is nilpotent.

Solution. Suppose that there exists $f=\sum_{i=0}^n a_i x^i \in J(S)$ such that $a_n$ is not nilpotent. Then, by Zorn’s lemma, there exists a prime ideal $P$ of $R$ such that $\{a_n, a_n^2, \cdots \} \cap P = \emptyset.$ Let $\overline{R}=R/P.$ For every element $h = \sum_i c_ix^i \in S$ let $\overline{h}=\sum_i \overline{c_i}x^i \in \overline{R}[x].$ Note that $\deg \overline{f} = \deg f$ because $\overline{a_n} \neq 0.$ Now, since $f \in J(S),$ there exists $g \in S$ such that $1=(1-xf)g$ and thus $\overline{1}=(\overline{1}-x\overline{f})\overline{g}.$ But then, since $\overline{R}$ is an integral domain, we’ll get the contradiction $0= \deg \overline{1} = \deg (\overline{1}-x\overline{f}) + \deg \overline{g} \geq 1. \ \Box$

## Rank of Hermitian matrices

Posted: October 2, 2012 in Elementary Algebra; Problems & Solutions, Linear Algebra
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For $a \in \mathbb{C}$ let $\overline{a}$ denote the complex conjugate of $a.$ Recall that a matrix $[a_{ij}] \in M_n(\mathbb{C})$  is called Hermitian if $a_{ij}=\overline{a_{ji}},$ for all $1 \leq i,j \leq n.$ It is known that if $A$ is Hermitian, then $A$ is diagonalizable  and every eigenvalue of $A$ is a real number. In this post, we will give a lower bound for the rank of a Hermitian matrix. In fact, the lower bound holds for any diagonalizable complex matrix whose eigenvalues are real numbers. To find the lower bound, we first need an easy inequality.

Problem 1. Prove that if $a_1, \ldots , a_m \in \mathbb{R},$ then $(a_1 + \ldots + a_m)^2 \leq m(a_1^2 + \ldots + a_m^2).$

Solution.  We have $a^2+b^2 \geq 2ab$ for all $a,b \in \mathbb{R}$ and so

$(m-1)\sum_{i=1}^m a_i^2=\sum_{1 \leq i < j \leq m}(a_i^2+a_j^2) \geq \sum_{1 \leq i < j \leq m}2a_ia_j.$

Adding the term $\sum_{i=1}^m a_i^2$ to both sides of the above inequality will finish the job. $\Box$

Problem 2. Prove that if $0 \neq A \in M_n(\mathbb{C})$ is Hermitian, then ${\rm{rank}}(A) \geq ({\rm{tr}}(A))^2/{\rm{tr}}(A^2).$

Solution. Let $\lambda_1, \ldots , \lambda_m$ be the nonzero eigenvalues of $A.$ Since $A$ is diagonalizable, we have ${\rm{rank}}(A)=m.$ We also have ${\rm{tr}}(A)=\lambda_1 + \ldots + \lambda_m$ and ${\rm{tr}}(A^2)=\lambda_1^2 + \ldots + \lambda_m^2.$ Thus, by Problem 1,

$({\rm{tr}}(A))^2 \leq {\rm{rank}}(A) {\rm{tr}}(A^2)$

and the result follows. $\Box$

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
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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 $I:=\langle 2 \rangle$ 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
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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 $xy=yx. \ \Box$