Posts Tagged ‘Jacobson radical’

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


We will assume that R is a commutative ring with unity and S=R[x], the polynomial ring over R. We will denote by J(S) the Jacobson radical of S.

Problem 1. The ring S/J(S) is never Artinian.

Solution. Let I=\langle x + J(S) \rangle, the ideal of S/J(S) generated by the coset x+J(S).

Claim . I^k \neq I^{k+1}, for all integers k \geq 1.

Proof of the claim . Suppose, to the contrary, that I^{k+1}=I^k, for some k \geq 1. Then

x^k + J(S) \in I^k = I^{k+1}=\langle x^{k+1}+J(S) \rangle

and so x^k + J(S)=f(x)x^{k+1} + J(S), for some f(x) \in S. Hence x^k - f(x)x^{k+1} \in J(S) and therefore g(x)=1 -(x^k - f(x)x^{k+1})=1-x^k + f(x)x^{k+1} must be a unit in S. But the coefficient of x^k in g(x) is -1, which is obviously not nilpotent, and so g(x) cannot be a unit (see here). Contradiction!

So, by the claim, we have a strictly descending chain of ideals of S: \ I \supset I^2 \supset I^3 \supset \cdots proving that S/J(S) is not Artinian. \Box.

Problem 2. Prove that S has infinitely many maximal ideals.

Solution. Suppose, to the contrary, that the set of maximal ideals of S is finite. Let \mathfrak{m}_1, \cdots , \mathfrak{m}_n be the maximal ideals of S. Then, by the Chinese remainder theorem, S/J(S) \cong \bigoplus_{i=1}^n S/\mathfrak{m}_i. So, since each S/\mathfrak{m}_i is a field and fields have only two ideals, S/J(S) must have finitely many (2^n in fact) ideals. But then S/J(S) would obviously be Artinian, contradicting Problem 1. \Box

Suppose that R is Noetherian. Then by the Hilbert’s basis theorem, S is Noetherian too. Thus S/J(S) is Noetherian. So S/J(S) is always a non-Artinian Noetherian ring.

Theorem (Maschke, 1899) Let k be a field and let G be a finite group of order n. Let R:=k[G]. Then R is semiprimitive if and only if char(k) \nmid n. 

Proof. Let G = \{g_1, \cdots , g_n \} where g_1=1. Suppose first that char(k) \nmid n and consider the algebra homomorphism \rho : R \longrightarrow End_k(R) defined by \rho(r)(x)=rx for all r,x \in R. Define \alpha : R \longrightarrow k by \alpha(r) = Tr(\rho(r)) for all r \in R. Note that here Tr(\rho(r)) is the trace of the matrix corresponding to the linear transformation \rho(r) with respect to the ordered basis \{g_1, \cdots , g_n \}. We remark a few points about \alpha:

1) \alpha(1)=n because \rho(1) is the identity map of R.

2) If 1 \neq g \in G, then \alpha(g)=0. The reason is that \rho(g)(g_i)=gg_i \neq g_i for all i and thus the diagonal entries of the matrix of \rho(g) are all zero and so \alpha(g)=Tr(\rho(g))=0.

3) If r is a nilpotent element of R, then \alpha(r)=0 because then r^m=0 for some m and thus (\rho(r))^m = \rho(r^m)=0. So \rho(r) is nilpotent and we know that the trace of a nilpotent matrix is zero. 

Now let c \in J(R). Since R is finite dimensional over k, it is Artinian and hence J(R) is nilpotent. Thus c is nilpotent and therefore, by 3), \alpha(c)=0. Let c = \sum_{i=1}^n c_i g_i, where c_i \in k. Then

0=\alpha(c)=\sum_{i=1}^n c_i \alpha(g_i)=c_1n,

by 1) and 2). It follows that c_1=0 because char(k) \nmid n. So the coefficient of g_1=1 of every element in J(R) is zero. But for every i, the coefficient of g_1=1 of the element g_i^{-1}c \in J(R) is c_i and so we must have c_i=0. Hence c = 0 and so J(R)=\{0\}.

Conversely, suppose that char(k) \mid n and put x = \sum_{i=1}^n g_i. Clearly xg_j = x for all j and hence

x^2=x\sum_{j=1}^n g_j = \sum_{j=1}^n xg_j=\sum_{j=1}^n x = nx = 0.

Thus kx is a nilpotent ideal of R and so kx \subseteq J(R). Therefore J(R) \neq \{0\} because kx \neq \{0\}. \ \Box 

Definition 1. Let I be a two-sided ideal of a ring R. We say that an idempotent element a \in R/I can be lifted if a=e + I for some idempotent element e \in R.

Definition 2. A subset S of a ring R is called nil if every element of S is nilpotent.

Problem. If I is a nil ideal of R, then every idempotent of R/I can be lifted.

Solution. Suppose that a=r+ I is an idempotent element of R/I. Then r-r^2 \in I and thus (r-r^2)^n=0, for some integer n \geq 1. Therefore

r^n - r^{n+1} \sum_{i=1}^n (-1)^{i-1} \binom{n}{i}r^{i-1} = 0.

So if we let

s =\sum_{i=1}^n (-1)^{i-1} \binom{n}{i}r^{i-1},

then r^n=r^{n+1}s and rs=sr. Now let e=(rs)^n. Using the fact that r^n=r^{n+1}s we will get e^2=e. Also, since r+ I = r^2+I, we have r+I=r^k + I for all k \geq 1. Therefore



a=r+I=r^n + I=(r+I)^n=(rs+I)^n=(rs)^n+I=e+I. \Box

Example. Let R be a (left) Artinian ring and let J be the Jacobson radical of R. Since J is nilpotent, and hence nil, every idempotent of R/J can be lifted.

LetR be a commutative ring with 1 and J(R) its Jacobson radical. Let S be the set of non-unit elements of R and let T:=S \setminus J(R).

Problem. Suppose that T is non-empty and finite. Then R is finite and R \cong R_1 \times R_2 \times \cdots \times R_k, for some finite local rings R_i and k \geq 2.


\boxed{1} x+y \in T for all y \in T and x \in J(R).

Proof. Clearly x+y \notin J(R). Proving x+y \in S is by contradiction: suppose that x+y \notin S, i.e. (x+y)u=1, for some u \in R. But then yu=1-xu \notin S, because x \in J(R). That means yu, and hence y, is a unit. Contradiction!

\boxed{2} S is finite.

Proof. Let y \in T and define f: J(R) \longrightarrow T by f(x)=x+y. Clearly f is one-to-one. Thus J(R), and therefore S=T \cup J(R), is finite.

\boxed{3} R \cong R_1 \times R_2 \cdots \times R_k, where each R_i is a local ring.

Proof. Since every proper ideal of R is contained in S, our ring has only a finite number of ideals and so it is artinian. Hence J(R)^n=\{0\}, for some positive integer n. Let M_i, \ 1 \leq i \leq k, be the maximal ideals of R and R_i=R/M_i^n.  It is clear that each R_i is a local ring and, by the Chinese remainder theorem, R \cong R_1 \times R_2 \times \cdots \cdot \times R_k.

\boxed{4} k \geq 2 and R, and so each R_i, is finite.

Proof. Since T \neq \emptyset, R is not local and thus k \geq 2. Now suppose, to the contrary, that R is infinite. Then at least one of R_i is infinite and so the set

A=\{ (x_1, x_2, \cdots , x_k) \in R: \ x_i = 0, \ \text{for some} \ i \}

is infinite. But A \subseteq S, which is a contradiction. \Box

Remark. Every 0 \neq s \in S is a zero divisor.

Proof. We have \{s^m : \ m \geq 1 \} \subseteq S and so, since S is finite, there exist some p \neq q \in \mathbb{N} such that s^p = s^q and p+q is minimal. So s(s^{p-1}-s^{q-1})=0 and thus we need to prove that s^{p-1} - s^{q-1} \neq 0. Well, if s^{p-1}=s^{q-1} and p=1 or q=1, then s would be a unit, which is absurd. Otherwise, we’re done by the minimality of p+q.

Fact 1. Let R be a left primitive ring and M a faithful simple left R module. By Schur’s lemma D=\text{End}_R(M) is a division ring and M can be considered as a right vector space over D in the usual way. Let S=\text{End}_D(M) and define \varphi : R \longrightarrow S by \varphi(r)(x)=rx, for all r \in R and x \in M. Then \varphi is a well-defined ring homomorphism. Also \varphi is one-to-one because M is faithful. So R can be viewed as a subring of S.

Fact 2. Every left primitive ring R is prime. To see this, suppose M is a faithful simple left R module and I,J be two non-zero ideals of R with IJ=(0).  Now JM is a submodule of M and M is simple. Therefore either JM=(0) or JM=M. If JM=(0), then we get (0) \neq J \subseteq \text{ann}_R M = (0), which is nonsense. Finally, if JM=M, then we will have (0)=(IJ)M=I(JM)=IM. Thus I \subseteq \text{ann}_R M = (0) and so I=(0), a contradiction!

Fact 3. A trivial result of Fact 2 is that the center of a left primitive ring is a commutative domain. A non-trivial fact is that every commutative domain is the center of some left primitive ring. For a proof of this see: T. Y. Lam,  A First Course in Noncommutative Ring Theory, page 195.

Fact 4. Let R be a prime ring and M a faithful left R module of finite length. Then R is left primitive. To see this, let (0)=M_0 \subset M_1 \subset \cdots \subset M_n=M be a composition series of M. Therefore M_k/M_{k-1} is a simple left R module for every 1 \leq k \leq n. We also let I_k=\text{ann}_R (M_k/M_{k-1}). Then each I_k is an ideal of R and it’s easy to see that I_1I_2 \cdots I_nM = (0). Thus I_1I_2 \cdots I_n = (0), because M is faithful. Hence I_{\ell} = (0), for some \ell, because R is prime. Therefore M_{\ell}/M_{\ell - 1} is a faithful simple left R module.

Fact 5. Every left primitive ring R is semiprimitive. This is easy to see: let M be a faithful simple left R module and J=J(R), as usual, be the Jacobson radical of R. The claim is that J=(0). So suppose that J \neq \{0\}  and choose 0 \neq x \in M. Then Rx=M, because M is simple, and so JM=Jx. Also either JM=(0), which is impossible because then J \subseteq \text{ann}_R M=(0), or JM=M. If Jx=JM=M, then ax =x, for some a \in J. Thus (1-a)x=0, which gives us the contradiction x = 0, because 1-a is invertible in R.

Notation. For a ring R let J(R) be the Jacobson radical of R.

Definition. Recall that if k is a field and G is a group, then the group algebra k[G] has two structures. Firstly, as a vector space over k, it has G as a basis, i.e. every element of k[G] is uniquely written as \sum_{g \in G} x_g g, where x_g \in k. In particular, \dim_k k[G]=|G|, as cardinal numbers. Secondly, multiplication is also defined in k[G]. If x = \sum_{g \in G} x_g g and y = \sum_{g \in G} y_g g are two elements of k[G], then we just multiply xy in the ordinary fashion using distribution law. To be more precise, we define xy = \sum_{g \in G} z_g g, where z_g = \sum_{rs=g} x_r y_s.

We are going to prove that J(\mathbb{C}[G])=0, for every group G.

Lemma. J(\mathbb{C}[G]) is nil, i.e. every element of J(\mathbb{C}[G]) is nilpotent.

Proof. If G is countable, we are done by this theorem. For the general case, let \alpha \in J(\mathbb{C}[G]). So \alpha =\sum_{i=1}^n c_ig_i, for some c_i \in \mathbb{C}, \ g_i \in G. Let H=\langle g_1,g_2, \cdots , g_n \rangle. Clearly \alpha \in H and H is countable. So to complete the proof, we only need to show that \alpha \in J(\mathbb{C}[H]). Write G = \bigcup_i x_iH, where x_iH are the distinct cosets of H in G. Then \mathbb{C}[G]=\bigoplus_i x_i \mathbb{C}[H], which means \mathbb{C}[G]=\mathbb{C}[H] \oplus K, for some right \mathbb{C}[H] module K. Now let \beta \in \mathbb{C}[H]. Since  \alpha \in J(\mathbb{C}[G]), there exists \gamma \in \mathbb{C}[G] such that \gamma (1 - \beta \alpha ) = 1. We also have \gamma = \gamma_1 + \gamma_2, for some \gamma_1 \in \mathbb{C}[H], \ \gamma_2 \in K. That now gives us \gamma_1(1 - \beta \alpha)=1. \ \Box

Theorem. J(\mathbb{C}[G])=0, for any group G.

Proof. For any x =\sum_{i=1}^n c_i g_i\in \mathbb{C}[G] define

x^* = \sum_{i=1}^n \overline{c_i} g_i^{-1}.

It’s easy to see that xx^*=0 if and only if x=0 and for all x,y \in \mathbb{C}[G]: \ (xy)^*=y^*x^*. Now suppose that J(\mathbb{C}[G]) \neq 0 and let 0 \neq \alpha \in J(\mathbb{C}[G]). Put \beta = \alpha \alpha^* \in J(\mathbb{C}[G]). By what I just mentioned \beta \neq 0 and (\beta^m)^* = (\beta^*)^m=\beta^m, for all positive integers m. By the lemma, there exists k \geq 2 such that \beta^k = 0 and \beta^{k-1} \neq 0. Thus \beta^{k-1} (\beta^{k-1})^* = \beta^{2k-2} = 0, which implies that \beta^{k-1} = 0. Contradiction! \Box

Corollary. If G is finite, then \mathbb{C}[G] is semisimple.

Proof. We just proved that J(\mathbb{C}[G])=(0). So we just need to show that \mathbb{C}[G] is Artinian. Let

I_1 \supset I_2 \supset \cdots \ \ \ \ \ \ \ \ \ \ \ \ \ (*)

be a descending chain of left ideals of \mathbb{C}[G]. Obviously each I_j is a \mathbb{C}-subspace of \mathbb{C}[G]. Thus each I_j is finite dimensional because \dim_{\mathbb{C}} \mathbb{C}[G]=|G| < \infty. Hence (*) will stablize at some point because \dim_{\mathbb{C}} I_1 < \infty and \dim_{\mathbb{C}}I_1 > \dim_{\mathbb{C}} I_2 > \cdots . Thus \mathbb{C}[G] is (left) Artinian and the proof is complete because we know a ring is semisimple if and only if it is (left) Artinian and its Jacobson radical is zero. \Box