Archive for the ‘Simple Rings’ Category

Quaternion algebras

Posted: March 23, 2022 in Noncommutative Ring Theory Notes, Simple Rings
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We have seen the division algebra of real quaternions \mathbb{H} several times in this blog; here for example. The algebra \mathbb{H} is just an example of a large class of algebras called quaternion algebras, which are the subject of this post.

Definition. Let F be a field, and let a,b \in F \setminus \{0\}. The quaternion algebra, with respect to F,a,b, denoted by (a,b)_F, is the F-algebra on two generators \bold{i}, \bold{j} subject only to the relations

\bold{i}^2=a, \ \ \ \bold{j}^2=b, \ \ \ \bold{i}\bold{j}=-\bold{j}\bold{i}.

For a concrete description of (a,b)_F, see Remark ii).

Example. The quaternion algebra (-1,-1)_{\mathbb{R}} is just the division ring of real quaternions \mathbb{H}. Similarly, the quaternion algebra (-1,-1)_{\mathbb{Q}} is the division ring of rational quaternions.

Remarks. Let F be a filed, a,b \in F \setminus \{0\}, and A:=(a,b)_F.

i) We usually write \bold{k}=\bold{i}\bold{j}. Then from the relations defined on A, we get that

\bold{k}^2=(\bold{i}\bold{j})^2=\bold{i}(\bold{j}\bold{i})\bold{j}=-\bold{i}^2\bold{j}^2=-ab, \ \ \ \ \bold{i}\bold{k}=-\bold{k}\bold{i}=a\bold{j}, \ \ \ \ \bold{k}\bold{j}=-\bold{j}\bold{k}=b\bold{i}.

ii) So, what is A, really? Well, by definition, A=F \langle x,y \rangle/I, where F \langle x,y\rangle is the ring of polynomials in non-commuting variables x,y, where elements of F commute with x,y, and I is the two-sided ideal of F\langle x,y \rangle generated by the polynomials x^2-a, \ y^2-b, \ xy+yx. Then \bold{i},\bold{j} are just x+I, \ y+I, respectively. It should be clear now that for every z \in A, there exist unique c_i \in F, \ 1 \le i \le 4, such that

z=c_1 +c_2 x +c_3 y+ c_4 xy+I =c_1 + c_2 \bold{i}+c_3 \bold{j}+c_4 \bold{ij}=c_1 + c_2 \bold{i}+c_3 \bold{j}+ c_4 \bold{k}.

iii) By ii), \{1,\bold{i},\bold{j}, \bold{k}\} is a basis for A, as an F-vector space, and so \dim_F A=4.

iv) Let z_1,z_2 \in A. By iii), there exist (unique) c_i,d_i \in F, \ 1 \le i \le 4, such that

z_1= c_1 + c_2 \bold{i}+c_3 \bold{j}+ c_4 \bold{k}, \ \ \ \ z_2=d_1 + d_2 \bold{i}+d_3 \bold{j}+ d_4 \bold{k}.

Then, using the relations on A and i), we get the following formula for z_1z_2

z_1z_2=c_1d_1+ac_2d_2+bc_3d_3-abc_4d_4 + (c_1d_2+c_2d_1-bc_3d_4+bc_4d_3)\bold{i} \ +

(c_1d_3+ac_2d_4+c_3d_1-ac_4d_2)\bold{j}+(c_1d_4+c_2d_3-c_3d_2+c_4d_1)\bold{k}.

v) Let z \in A. By iii), z=c_1 + c_2 \bold{i}+c_3 \bold{j}+c_4 \bold{k}, for some (unique) c_i \in F. The conjugate \overline{z} of z is defined by \overline{z}=c_1 -c_2 \bold{i}-c_3 \bold{j}-c_4 \bold{k}. Let N(z):=z\overline{z}. Using the product formula given in iv), we see that

N(z)=\overline{z}z=c_1^2-ac_2^2-bc_3^2+abc_4^2 \in F.

Also, if z_1,z_2 \in A, then iv) gives \overline{z_1}\overline{z}_2=\overline{z}_2\overline{z}_1 and so

N(z_1z_2)=z_1z_2\overline{z_1}\overline{z}_2=z_1z_2\overline{z}_2 \overline{z}_1=z_1N(z_2)\overline{z}_2=N(z_2)z_1\overline{z}_2=N(z_1)N(z_2).

So the map N: A \to F is multiplicative.

vi) Let N: A \to F be the map defined in iv). We show that A is a division ring if and only if N(z) \ne 0 for all 0 \ne z \in A. Suppose first that the condition is satisfied and 0 \ne z \in A. Then, 0 \ne N(z)=z\overline{z} \in F and z\overline{z} is invertible in F. Thus z\overline{z}(z\overline{z})^{-1}=1 and hence z^{-1}=(z\overline{z})^{-1}\overline{z}. Conversely, suppose that A is a division ring, and 0 \ne z \in A. Then z is invertible and so zz'=1, for some z' \in A, and hence, by iv), 1=N(1)=N(z)N(z'), giving N(z) \ne 0.

vii) By vi), (-1,-1)_{\mathbb{R}}=\mathbb{H} is a division ring because for any 0 \ne z=c_1 + c_2 \bold{i}+c_3 \bold{j}+c_4 \bold{k} \in \mathbb{H}, we have, N(z)=c_1^2+c_2^2+c_3^2+c_4^2 \ne 0, because c_i \in \mathbb{R}. For the same reason, (-1,-1)_F is a division ring for every field F \subseteq \mathbb{R}. But (1,1)_F is not a division ring for any field F \subseteq \mathbb{R} because N(1+\bold{i})=0 even though 1+\bold{i} \ne 0.

viii) If \text{char}(F)=2, then \bold{i}\bold{j}=-\bold{j}\bold{i}=\bold{j}\bold{i} and so A is commutative in this case. If \text{char}(F) \ne 2, the center of A is F. To see this, first note that, by definition, F is in the center of A. Now let z \in A be in the center of A. By iii), z=c_1 + c_2 \bold{i}+c_3 \bold{j}+c_4 \bold{k}, for some c_i \in F. Then z\bold{i}=\bold{i}z and i) give 2c_3=2ac_4\bold=0 and hence c_3=c_4=0 because \text{char}F) \ne 2, \ a \ne 0. Now z\bold{j}=\bold{j}z gives 2c_2=0 and so c_2=0. Thus z=c_1 \in F.

Proposition 1. Let F be a field of characteristic \ne 2, and let a,b \in F \setminus \{0\}. Let A:=(a,b)_F. Then A is a division ring if and only if ax^2+by^2 \ne 1 for all x,y \in F.

Proof. Suppose first that there exist x,y \in F such that ax^2+by^2=1. Let z=1+x\bold{i}+y\bold{j}. Then z \ne 0 but by Remark v), N(z)=1-ax^2-by^2=0 and so, by Remark vi), A is not a division ring. Conversely, suppose, to the contrary, that ax^2+by^2 \ne 1, for all x, y \in F, but A is not a division ring. So, by Remarks v), vi),

\exists \ z=c_1+c_2\bold{i}+c_3\bold{j}+c_4\bold{k} \in A \setminus \{0\}: \ \ N(z)=c_1^2-ac_2^2-bc_3^2+abc_4^2=0. \ \ \ \ \ \ \ \ \ \ \ \ (1)

Claim, (at^2+b)(t^2-a) \ne 0, \ \ \ \forall x \in F.

Proof of the Claim. If t^2=a for some t \in F, then ax^2+by^2=1 for x=t^{-1}, y=0, contradicting our assumption that ax^2+by^2 \ne 1, for all x,y \in F. If at^2+b=0, for some t \in F, then t \ne 0, because b \ne 0, and ax^2+by^2=1 for \displaystyle x=\frac{1+a}{2a}, \ y=\frac{1-a}{2at}, which is again a contradiction. The proof of the Claim is now complete.

Now, c_1^2-ac_2^2=b(c_3^2-ac_4^2), by (1), and so (c_1^2-ac_2^2)(c_3^2-ac_4^2)=b(c_3^2-ac_4^2)^2, which gives

a(c_1c_4+c_2c_3)^2+b(c_3^2-ac_4^2)^2=(c_1c_3+ac_2c_4)^2. \ \ \ \ \ \ \ \ \ \ \ \ \ (2)

If c_1c_3+ac_2c_4 \ne 0, then dividing both sides of (2) by (c_1c_3+ac_2c_4)^2 will give ax^2+by^2=1 for some x,y \in F, which is a contradiction. So c_1c_3+ac_2c_4=0 and now (2) gives

a(c_1c_4+c_2c_3)^2+b(c_3^2-ac_4^2)^2=0. \ \ \ \ \ \ \ \ \ \ \ \ (3)

If c_3^2-ac_4^2 \ne 0, then dividing (3) by (c_3^2-ac_4^2)^2 will give at^2+b =0, for some t \in F, contradicting the Claim. So

c_3^2-ac_4^2 =0. \ \ \ \ \ \ \ \ \ \ \ \ (4)

If c_4 \ne 0, then dividing (4) by c_4^2 will give t^2-a=0, for some t \in F, contradicting the Claim. So c_4=0 and hence, by (4), \ c_3=0. Thus, by (1),

ac_2^3+bc_3^2=0. \ \ \ \ \ \ \ \ \ \ \ (5)

If c_3 \ne 0, then dividing (5) by c_3^2 will give at^2+b=0, contradicting the Claim. So c_3=0 and hence, by (5), \ c_2=0. So we have shown that c_1=c_2=c_3=c_4=0 and thus z=0, which contradicts (1). \ \Box

By Remark vii), quaternion algebras may or may not be division rings but, as the following theorem shows, they are simple if the base field has characteristic \ne 2.

Proposition 2. Every quaternion algebra over a field F of characteristic \ne 2 is simple.

Proof. Let A=(a,b)_F, where a,b \in F \setminus \{0\}, and let (0) \ne I be a two-sided ideal of A. We must show that I=A, i.e. 1 \in I. Let 0 \ne z \in I. Then, by Remark iii),

z=c_1 + c_2 \bold{i}+c_3 \bold{j}+c_4 \bold{k},

for some (unique) c_i \in F, where not all c_i are zero. Replacing z with z\bold{i}, z\bold{j} or z\bold{k}, if necessary, we may assume that c_1 \ne 0. Now, since 2a(c_1+c_2\bold{i})=\bold{i}z\bold{i}+az \in I and 2a \ne 0, because \text{char}(F) \ne 2, a \ne 0, we get that z_1:=c_1+c_2\bold{i} \in I. So 2bc_1=\bold{j}z_1\bold{j}+bz_1 \in I and hence c_1 \in I, because 2b \ne 0. Therefore 1=c_1^{-1}c_1 \in I, because c_1 \ne 0. \ \Box

Corollary. Let F be a field of characteristic \ne 2, and consider the quaternion algebra A=(a,b)_F. Then either A is a division ring or A \cong M_2(F), the ring of 2 \times 2 matrices with entries from F.

Proof. By Remark iii), \dim_F A=4 and by Remark viii), the center of A is F. Also, by the above theorem, A is simple. So A is a central simple F-algebra of dimension 4 and hence, by the Artin-Wedderburn theorem, A \cong M_n(D) for some integer n \ge 1 and some division F-algebra D. Thus

4=\dim_F A=\dim_F M_n(D)=n^2\dim_F D

and so either n=1, \dim_F D=4, or n=2, \dim_FD=1. Therefore either A \cong D or A \cong M_2(F). \ \Box

Rings in which the equation ax + xa = 1 has a solution for every non-zero element a

Posted: June 6, 2020 in Noncommutative Ring Theory Notes, Simple Rings
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In this post, we take a look at the rings R with 1 that satisfy the following property

\forall a \in R \setminus \{0\}, \ \exists x \in R: \ \ \ ax+xa=1. \ \ \ \ \ \ \ \ \ \ \ \ (*)

In other words, rings in which the equation ax+xa=1 has a solution for every 0 \ne a \in R.

Remark 1. It is clear that a commutative ring satisfies (*) if and only if it is a field of characteristic \ne 2 because in this case (*) becomes 2ax=1. It is also clear that any division ring R of characteristic \ne 2 satisfies (*) because if 0 \ne a \in R and x=2^{-1}a^{-1}, then ax+xa=1.

Remark 2. The equation ax+xa=1 could have more than one solution in a ring. For example, in the division ring of real quaterninos \mathbb{H}=\{\alpha+\beta i + \gamma j + \delta k, \ \ \alpha, \beta, \gamma, \delta \in \mathbb{R}\}, choose a=i+j and x=\beta i-\left(\frac{1}{2}+\beta\right)j, where \beta is any real number, and see that ax+xa=1.
However, if a=\alpha+\beta i + \gamma j + \delta k \in \mathbb{H} and \alpha \ne 0, then the equation ax+xa=1 has the unique solution x=2^{-1}a^{-1} (why ?).

We now show that any ring that satisfies (*) is a simple domain of characteristic \ne 2.

Proposition. Let R be a ring and suppose that R satisfies (*). Then R is a simple domain and its center is a field of chracateristic \ne 2.

Proof. Let I \ne (0) be an ideal of R and choose 0 \ne a \in I. There exists x \in R such that ax+xa=1 and so 1=ax+xa \in I implying that I=R. So R is a simple ring and thus its center is a field (see Remark 1 in this post!). For a=1, there exists x \in R such that 2x=ax+xa=1 and so 2 \ne 0 in R, i.e. the characteristic of the center of R is \ne 2.

It remains to show that R is a domain. We’ll do that by first proving two claims.

Claim 1. If a,x \in R and ax+xa =1, then a^2x=xa^2.

Proof. a^2x=a(ax)=a(1-xa)=a-axa=(1-ax)a=(xa)a=xa^2. \ \Box

Claim 2. If a \in R and a^2=0, then a=0.

Proof.  Suppose, to the contrary, that there exists 0 \ne a \in R such that a^2=0 and let x \in R be such that ax+xa=1. Then

a=a(ax+xa)=axa, \ \ \ (ax)^2=ax \ \ \ \ \ \ \ \ \ \ \ \ (1)

and, as a result, ax \ne 0. So there exists y \in R such that

axy+yax=1. \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)

But then, by (1) and Claim 1, axy=(ax)^2y=y(ax)^2=yax and so by (2), \ \ 2axy=1 and hence a=2a^2xy=0, contradiction! \Box

We are now ready to show that R is a domain. Suppose that ab=0 for some a,b \in R and a \ne 0. To show that R is a domain, we just need to prove that b=0. We have (ba)^2=b(ab)a=0 and so, by Claim 2,

ba=0. \ \ \ \ \ \ \ \ \ \ \ \ (3)

Also, since a \ne 0, there exists x \in R such that ax+xa=1 and so

b^2=b(ax+xa)b=(ba)xb+bx(ab)=0,

by (3). Hence, by Claim 2, b=0 and that completes the proof. \Box

Remark 3. Since an artinian domain is a division ring, it follows from the above problem and Remark 1 that an artinian ring satisfies (*) if and only if it is a division ring of characteristic \ne 2.

Question. Is it true that a ring R satisfies (*) if and only if R is a division ring of characteristic \ne 2 ? In other words, is there an example of a  ring that satisfies (*) but the ring is not a division ring?

Miklós Schweitzer 2019, Problem 2

Posted: December 29, 2019 in Noncommutative Ring Theory Notes, Simple Rings
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Problem (Miklós Schweitzer Competition, 2019). Let R be a noncommutative finite ring with multiplicative identity element 1. Show that if the subring generated by I \cup \{1\} is R for each nonzero two-sided ideal I, then R is simple.

Solution. I show that, more generally, the result holds true for any noncommutative (left) artinian ring R. Let Z be the center of R. For any two-sided ideal I \ne (0) of R, the subring Z+I contains I \cup \{1\} and so

Z+I=R. \ \ \ \ \ \ \ \ \ (*)

As a result, non-zero ideals of R are noncommutative because R is noncommutative.
We now show that Z is a domain. Suppose, to the contrary, that ab=0 for some a,b \in Z \setminus \{0\}. Then, by (*), we have Z+Ra=R and so Zb=Rb implying that Rb \ne (0) is a commutative ideal of R, contradiction. So Z is a domain.
Now let J be the Jacobson radical of R. We show that J=(0). So suppose, to the contrary, that J \ne (0). Since R is artinian, J is nilpotent. Thus there exists the smallest integer n \ge 2 such that J^n=(0). Let x \in J^{n-1} \setminus \{0\}. We have Z+J=R, by (*), and so Rx=xR=Zx, i.e. Rx \ne (0) is a commutative ideal of R and that’s a contradiction. So J=(0) and thus, by the Artin-Wedderburn’s theorem, R is a finite direct product of some matrix rings over division rings

R \cong \prod_{i=1}^kM_{n_i}(D_i).

Let F_i be the center of D_i. Then Z \cong \prod_{i=1}^k F_i which is possible only if k=1 because, as we showed, Z is a domain. So R \cong M_n(D) for some integer n and some division ring D and hence R is simple. \Box

Remark. The above result is not true for commutative artinian rings. For example, let R:=\mathbb{Z}/n\mathbb{Z}, where n > 1 is not prime. Clearly R is not simple and the only subring of R that contains 1 is R itself. So the subring generated by I \cup \{1\} is R for any non-zero ideal I of R.

Reduced norm and reduced trace (2)

Posted: October 16, 2011 in Noncommutative Ring Theory Notes, Simple Rings
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Here you can see part (1).

Introduction. Let’s take a look at a couple of properties of the trace map in matrix rings. Let k be a field and B=M_m(k). Let \{e_{ij}: \ 1 \leq i,j \leq m\} be the standard basis for B. Let b = \sum_{i,j}\beta_{ij}e_{ij} \in B, where \beta_{ij} \in k. Now, \sum_{r,s}e_{rs}be_{sr}=\sum_{i,j,r,s} \beta_{ij}e_{rs}e_{ij}e_{sr}=\sum_{i,j} \beta_{ii}e_{jj} = \text{Tr}(b)I, where I is the identity element of B. Now let’s define \nu_B : = \sum_{ij} e_{ij} \otimes_k e_{ji} \in B \otimes_k B. See that \nu_B^2=I \otimes_k I = 1_{B \otimes_k B} and (b_1 \otimes_k b_2)\nu_B=\nu_B(b_2 \otimes_k b_1) for all b_1,b_2 \in B. We are going to extend these facts to any finite dimensional central simple algebra.

Notation. I will assume that A is a finite dimensional central simple k-algebra.

Theorem. There exists a unique element \nu_A = \sum b_i \otimes_k c_i \in A \otimes_k A such that \text{Trd}_A(a)=\sum b_iac_i for all a \in A. Moreover, \nu_A^2=1 and (a_1 \otimes_k a_2) \nu_A = \nu_A(a_2 \otimes_k a_1) for all a_1, a_2 \in A.

Proof. As we saw in this theorem, the map g : A \otimes_k A^{op} \longrightarrow \text{End}_k(A) \cong M_n(k) defined by g(a,a')(b)=aba', \ a,a',b \in A, is a k-algebra isomorphism. Let’s forget about the ring structure of A^{op} for now and look at it just as a k-vector space.  Then A \cong A^{op} and so we have a k-vector space isomorphism g: A \otimes_k A \longrightarrow \text{End}_k(A) defined by g(a \otimes_k a')(b)=aba', \ a,a',b \in A. Since \text{Trd}_A \in \text{End}_k(A), there exists a unique element

\nu_A = \sum b_i \otimes_k c_i \in A \otimes_k A

such that g(\nu_A) = \text{Trd}_A. Then \text{Trd}_A(a)=g(\nu_A)(a) = \sum g(b_i \otimes_k c_i)(a) = \sum b_iac_i.

To prove \nu_A^2=1, we choose a splitting field of K of A. Then B=A \otimes_k K \cong M_n(K) for some integer n, which is the degree of A. Let’s identify A and K with A \otimes_k 1 and 1 \otimes_k K respectively. Then A and K become subalgebras of B and B=AK. Let a \in A and \gamma \in K. Recall that, by the last part of the theorem in part (1), \text{Trd}_B(a) = \text{Trd}_A(a), for all a \in A. Let a \in A and \gamma \in K. Then, since K is the center of B, we have

\sum b_i(\gamma a) c_i = \gamma \sum b_i a c_i= \gamma \text{Trd}_A(a)= \gamma \text{Trd}_B(a)= \text{Trd}_B(\gamma a).

Thus \nu_B = \sum b_i \otimes_k c_i = \nu_A, by the uniqueness of \nu_B. Hence \nu_B^2 = \nu_A^2. But B \cong M_n(K) is a matrix ring and so, as we mentioned in the introduction, \nu_B^2=1. So \nu_A^2=1.

To prove the last part of the theorem, let s,t,a \in A. Then

g((a_1 \otimes_k a_2) \nu_A)(a)=g(\sum a_1b_i \otimes_k a_2c_i)(a) = \sum g(a_1b_i \otimes_k a_2c_i)(a)

= \sum a_1b_iaa_2c_i = a_1 \text{Trd}_A(aa_2)= \text{Trd}_A(a_2a)a_1.

The last equality holds by the second part of the theorem in part (1). Also, the image of \text{Trd}_A is in k, the center of A, and so \text{Trd}_A(a_2a) commutes with a_1. Now, we also have

g(\nu_A(a_2 \otimes_k a_1))(a) = g(\sum b_i a_2 \otimes_k c_ia_1)(a) = \sum g(b_ia_2 \otimes_k c_i a_1)(a)

= \sum b_ia_2ac_ia_1 = \text{Trd}_A(a_2a)a_1.

Thus g( (a_1 \otimes_k a_2) \nu_A) = g(\nu_A( a_2 \otimes_k a_1)) and so (a_1 \otimes_k a_2) \nu_A= \nu_A(a_2 \otimes_k a_1). \ \Box

Definition. The element \nu_A \in A \otimes_k A in the theorem is called the Goldman element for A.

Remark. David Saltman in a short paper used the properties of \nu_A to give a proof of Remark 2 in this post.

The reduced Whitehead group

Posted: October 14, 2011 in Noncommutative Ring Theory Notes, Simple Rings
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Let A be a finite dimensional central simple k-algebra. We proved in this theorem that \text{Nrd}_A: A^{\times} \longrightarrow k^{\times} is a group homomorphism. The image of \text{Nrd}_A is an abelian group because it lies in k^{\times}. So if a,b \in A^{\times}, then \text{Nrd}_A(aba^{-1}b^{-1})=\text{Nrd}_A(aa^{-1}) \text{Nrd}_A(bb^{-1})=1. Therefore (A^{\times})', the commutator subgroup of A^{\times}, is contained in \ker \text{Nrd}_A=\{a \in A: \ \text{Nrd}_A(a) = 1\} and so the following definition makes sense.

Definition. Let A be a finite dimensional central simple algebra. The reduced Whitehead group of A is the factor group \text{SK}_1(A) = (\ker \text{Nrd}_A)/(A^{\times})'.

Remark. Let D be a finite dimensional central division k-algebra of degree n. By the theorem in this post, for every a \in D there exists v \in (D^{\times})' such that \text{Nrd}_D(a)=a^nv. So if \text{Nrd}_D(a)=1, then a^n \in (D^{\times})'. Therefore g^n=1 for all g \in \text{SK}_1(D).

Example 1. \text{SK}_1(M_n(k))=\{1\} if k is a field and either n > 2 or n=2 and |k| > 3.

Proof. Let A=M_n(k). Then A^{\times} = \text{GL}(n,k) and

\ker \text{Nrd}_A = \{a \in M_n(k): \ \det(a) = 1 \}=\text{SL}(n,k).

We proved in here that (\text{GL}(n,k))' = \text{SL}(n,k). Thus \text{SK}_1(A)=\{1\}. \Box

Example 2. \text{SK}_1(\mathbb{H})=\{1\}, where \mathbb{H} is the division algebra of quaternions over \mathbb{R}.

Proof. Let z = \alpha + \beta i + \gamma j + \delta k \in \mathbb{H} and suppose that \text{Nrd}_{\mathbb{H}}(z)=\alpha^2+ \beta^2 + \gamma^2 + \delta^2=1. We need to prove that z is in the commutator subgroup of \mathbb{H}^{\times}. We are going to prove a stronger result, i.e. z = aba^{-1}b^{-1} for some a,b \in \mathbb{H}^{\times}. If z \in \mathbb{R}, then \beta = \gamma = \delta = 0 and z = \alpha = \pm 1. If z = 1, then we can choose a=b=1 and if z=-1, we can choose a=i, \ b=j. So we may assume that z \notin \mathbb{R}. We also have z^2 - 2 \alpha z + 1 = 0 (you may either directly check this or use this fact that every element of a finite dimensional central simple algebra satisfies its reduced characteristic polynomial).  Thus (z-\alpha)^2 = \alpha^2 - 1. Note that since \text{Nrd}_{\mathbb{H}}(z)=1 and z \notin \mathbb{R} we have \alpha^2 < 1. So 1 - \alpha^2 = \alpha_0^2 for some 0 \neq \alpha_0 \in \mathbb{R}. Let

w = \alpha_0^{-1}(z - \alpha). \ \ \ \ \ \ \ \ \ (1)

So, since the real part of z is \alpha, the real part of w is zero and hence

w^2 = -1, \ \ \overline{w} = -w. \ \ \ \ \ \ \ \ \ \ (2).

Since the center of \mathbb{H} is \mathbb{R} and w \notin \mathbb{R}, there exists c \in \mathbb{H} such that u = cw-wc \neq 0. Therefore

uw = -wu = \overline{w}u. \ \ \ \ \ \ \ \ \ \ (3)

Now, by (1), we have z = \alpha + \alpha_0 w and \alpha^2+\alpha_0^2=1. So we can write \alpha = \cos \theta and \alpha_0 = \sin \theta. Let v = \cos(\theta/2) + \sin(\theta/2)w. Then (2) gives us

v^2=z, \ \ \ \overline{v}=\cos(\theta/2) + \sin(\theta/2)\overline{w} = \cos(\theta/2) - \sin(\theta/2)w=v^{-1}. \ \ \ \ \ \ \ \ (4)

We also have u v^{-1} = \overline{v^{-1}}u=vu, by (3) and (4). Thus, by (4) again, uv^{-1}u^{-1}v = v^2=z. \ \Box

Representations of central simple algebras

Posted: October 12, 2011 in Noncommutative Ring Theory Notes, Simple Rings
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We will assume that k is a field.

Definition. Let A be a k-algebra. A representation of A is a k-algebra homomorphism A \longrightarrow M_m(k), for some integer m \geq 1. If \ker \phi =(0), then \phi is called faithful.

If \phi is a representation of a k-algebra A and a \in A, then \phi(a) is a matrix and so it has a characteristic polynomial. Things become interesting when A is a finite dimensional central simple k-algebra. In this case, the characteristic polynomial of \phi(a) is always a power of the reduced characteristic polynomial of a. This fact justifies the name “reduced characteristic polynomial”!

Theorem. Let A be a finite dimensional central simple k-algebra of degree n and let a \in A. Let \phi : A \longrightarrow M_m(k) be a representation of A and suppose that p(x) is the characteristic polynomial of \phi(a). Then m =rn for some integer r \geq 1 and p(x) = (\text{Prd}_A(a,x))^r.

Proof. Let K be a splitting field of A. We now define \psi : A \otimes_k K \longrightarrow M_m(K) by \psi(b \otimes_k \alpha) = \alpha \phi(b) for all b \in A and \alpha \in K and then extend it linearly to all A \otimes_k K. We now show that \psi is a K-algebra homomorphism. Clearly \psi is additive because \phi is so. Now let \alpha, \beta \in K and b,c \in A. Then

\psi(\beta (b \otimes_k \alpha)) = \psi(b \otimes_k \alpha \beta)= \beta \alpha \phi(b)=\beta \psi(b \otimes_k \alpha)

and

\psi((b \otimes_k \alpha)(c \otimes_k \beta))=\psi(bc \otimes_k \alpha \beta)=\alpha \beta \phi(bc) = \alpha \phi(b) \beta \phi(c)

=\psi(b \otimes_k \alpha) \psi(c \otimes_k \beta).

So, by Lemma 2, m=rn for some integer r \geq 1 and

p(x)=\det(xI - \phi(a))=\det(xI - \psi(a \otimes_k 1))=(\text{Prd}_A(a,x))^r. \ \Box

Example. Let A be a finite dimensional central simple k-algebra of degree n. So \dim_k A = n^2 and  we have a faithful representation \phi : A \longrightarrow \text{End}_k(A) \cong M_{n^2}(k) defined by \phi(a)(b)=ab, for all a,b \in A. Let a \in A and suppose that p(x) is the characteristic polynomial of \phi(a). Then p(x) = (\text{Prd}_A(a,x))^n, by the above theorem.

Reduced norm and reduced trace (1)

Posted: October 11, 2011 in Noncommutative Ring Theory Notes, Simple Rings
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Throughout this post, k is a field and A is a finite dimensional central simple k-algebra of degree n.

Let a \in M_m(k) and suppose that p(x) = x^m + \alpha_{m-1}x^{m-1} + \ldots + \alpha_1x + \alpha_0 \in k[x] is the characteristic polynomial of a. We know from linear algebra that \text{Tr}(a)= - \alpha_{m-1}, \ \det(a)=(-1)^m \alpha_0. We also know that \text{Tr} is k-linear, \det is multiplicative and \text{Tr}(bc)=\text{Tr}(cb) for all b,c \in M_m(k). We’d like to extend the concepts of trace and determinant to any finite dimensional central simple algebra.

Definition. Let \text{Prd}_A(a,x)=x^n + \alpha_{n-1}x^{n-1} + \ldots + \alpha_1 x + \alpha_0 \in k[x] be the reduced characteristic polynomial of a \in A (see the definition of reduced characteristic polynomials in here). The reduced trace and the reduced norm of a are defined, respectively, by \text{Trd}_A(a) = - \alpha_{n-1} and \text{Nrd}_A(a) = (-1)^n \alpha_0.

Remark. Let K be  a splitting field of A with a K-algebra isomorphism f : A \otimes_k K \longrightarrow M_n(K). So \text{Trd}_A(a)=\text{Tr}(f(a \otimes_k 1)) and \text{Nrd}_A(a)=\det(f(a \otimes_k 1)) because \text{Prd}_A(a,x) is the characteristic polynomial of f(a \otimes_k 1).

Theorem. 1) The map \text{Trd}_A: A \longrightarrow k is k-linear and the map \text{Nrd}_A: A \longrightarrow k is multiplicative.

2) \text{Trd}_A(ab)=\text{Trd}_A(ba) for all a,b \in A.

3) If a \in k, then \text{Tr}_A(a)=na and \text{Nrd}_A(a)=a^n.

4) \text{Nrd}_A(a) \neq 0 if and only if a is a unit of A. So \text{Nrd}_A : A^{\times} \longrightarrow k^{\times} is a group homomorphism.

5) \text{Trd}_A and \text{Nrd}_A are invariant under isomorphism of algebras and extension of scalars.

Proof. Fix a splitting field K of A and a K-algebra isomorphism f: A \otimes_k K \longrightarrow M_n(K).

1) We have already proved that the values of \text{Trd}_A and \text{Nrd}_A are in k. Now, let a_1,a_2 \in A and \alpha \in k. Then

\text{Trd}_A(\alpha a_1 + a_2) = \text{Tr}(f((\alpha a_1 + a_2) \otimes_k 1)) = \text{Tr}(\alpha f(a_1 \otimes_k 1) + f(a_2 \otimes_k 1))

=\alpha \text{Tr}(f(a_1 \otimes_k 1)) + \text{Tr}(f(a_2 \otimes_k 1)) = \alpha \text{Trd}_A(a_1) + \text{Trd}_A(a_2)

and

\text{Nrd}_A(a_1a_2) = \det(f(a_1a_2 \otimes_k 1)) = \det(f(a_1 \otimes_k 1)f(a_2 \otimes_k 1)

=\det(f(a_1 \otimes_k 1)) \det (f(a_2 \otimes_k 1)) = \text{Nrd}_A(a_1) \text{Nrd}_A(a_2).

2) This part is easy too:

\text{Trd}_A(ab)=\text{Tr}(f(ab \otimes_k 1))=\text{Tr}(f((a \otimes_k 1)(b \otimes_k 1)))=\text{Tr}(f(a \otimes_k 1)f(b \otimes_k 1))

=\text{Tr}(f(b \otimes_k 1)f(a \otimes_k 1))= \text{Tr}(f(ba \otimes_k 1))=\text{Trd}_A(ba).

3) Let I be the identity element of M_n(K). Then

\text{Trd}_A(a)=a \text{Trd}_A(1) = a \text{Trd}_A(f(1 \otimes_k 1)) = a \text{Tr}(I)=na

and \text{Nrd}_A(a)=\det(f(a \otimes_k 1))=\det(a f(1 \otimes_k 1)) = \det(aI) \det(I) = a^n.

4) If ab=1 for some b \in A, then 1 = \text{Nrd}_A(ab)=\text{Nrd}_A(a) \text{Nrd}_A(b) and thus \text{Nrd}_A(a) \neq 0. Conversely, if \text{Nrd}_A(a) \neq 0, then \det(f(a \otimes_k 1)) \neq 0 and so f(a \otimes_k 1) is invertible in M_n(K). Let U be the inverse of f(a \otimes_k 1). Then U = f(u), for some u \in A \otimes_k K because f is surjective. Since f is injective, it follows that (a \otimes_k 1)u=1. Now if a is not a unit of A, then it is a zero divisor because A is artinian. So ba = 0 for some 0 \neq b \in A. But then b \otimes_k 1 = (b \otimes_k 1)(a \otimes_k 1)u = (ba \otimes_k 1)u=0, contradiction!

5) By Prp 3 and Prp 4 in this post, reduced characteristic polynomials are invariant under those things. \Box

Basic properties of reduced characteristic polynomials

Posted: October 11, 2011 in Noncommutative Ring Theory Notes, Simple Rings
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Prp1. Let K be a field and S=M_n(K). The characteristic polynomial and the reduced characteristic polynomial of an element of S are equal.

Proof. Well, f: S \otimes_K K \longrightarrow S = M_n(K) defined by f(a \otimes_K \alpha) = \alpha a, \ a \in S, \alpha \in K, is a K-algebra isomorphism. Thus \text{Prd}_S(a,x) = \det(xI - f(a \otimes_K 1)) = \det(xI - a). \ \Box

Prp2. (Cayley-Hamilton) Let A be a finite dimensional central simple k-algebra, and let a \in A. Then \text{Prd}_A(a,a)=0.

Proof. Let K be a splitting field of A with a K-algebra isomorphism f: A \otimes_k K \longrightarrow M_n(K). We have \text{Prd}_A(a,x)=\det(xI-f(a \otimes_k 1))=x^n + \alpha_{n-1}x^{n-1} + \ldots + \alpha_1 x + \alpha_0 \in k[x]. Since \text{Prd}_A(a,x) is just the characteristic polynomial of f(a \otimes_k 1) \in M_n(K), we may apply the Cayley-Hamilton theorem from linear algebra to get (f(a \otimes_k 1))^n + \alpha_{n-1}(f(a \otimes_k 1))^{n-1} + \ldots + \alpha_1 f(a \otimes_k 1) + \alpha_0=0. Thus, since f is a K-algebra homomorphism, we get

f((a^n + \alpha_{n-1}a^{n-1} + \ldots + \alpha_1 a + \alpha_0) \otimes_k 1)=0.

Hence, since f is injective, we must have (a^n + \alpha_{n-1}a^{n-1} + \ldots + \alpha_1 a + \alpha_0) \otimes_k 1 = 0 which implies

\text{Prd}_A(a,a)=a^n + \alpha_{n-1}a^{n-1} + \ldots + \alpha_1 a + \alpha_0 = 0. \ \Box

Prp3. Reduced characteristic polynomials are invariant under extension of scalars.

Proof. First let’s understand the question! We have a finite dimensional central simple k-algebra A with a \in A. We are asked to prove that if K/k is a field extension and B= A \otimes_k K, then \text{Prd}_B(a \otimes_k 1, x) = \text{Prd}_A(a,x). Note that B is a central simple K-algebra and \deg A = \deg B = n. Now, let L be a splitting field of B with an L-algebra isomorphism f : B \otimes_K L \longrightarrow M_n(L). But

B \otimes_K L =(A \otimes_k K) \otimes_K L \cong A \otimes_k (K \otimes_K L) \cong A \otimes_k L.

So we also have an isomorphism g : A \otimes_k L \longrightarrow M_n(L). Clearly f((a \otimes_k 1) \otimes_K 1)=g(a \otimes_k 1) and hence \text{Prd}_B(a \otimes_k 1, x) = \text{Prd}_A(a,x). \ \Box

Prp4. Reduced characteristic polynomials are invariant under isomorphism of algebras.

Proof. So A_1,A_2 are finite dimensional central simple k-algebras and f : A_1 \longrightarrow A_2 is a k-algebra isomorphism. We want to prove that \text{Prd}_{A_1}(a_1,x) = \text{Prd}_{A_2}(f(a_1),x) for all a_1 \in A_1. Let K be a splitting field of A_2 with a K-algebra isomorphism g : A_2 \otimes_k K \longrightarrow M_n(K). The map f \otimes \text{id}_K: A_1 \otimes_k K \longrightarrow A_2 \otimes_k K is also a K-algebra isomorphism. Thus we have a K-algebra isomorphism h=g \circ (f \otimes \text{id}_K) : A_1 \otimes_k K \longrightarrow M_n(K). So if a_1 \in A_1, then

\text{Prd}_{A_2}(f(a_1),x)=\det (xI - g(f(a_1) \otimes_k 1))=\det (xI - h(a_1 \otimes_k 1)) = \text{Prd}_{A_1}(a_1,x). \ \Box

Reduced characteristic polynomials; an example

Posted: October 11, 2011 in Noncommutative Ring Theory Notes, Simple Rings
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Recall that \mathbb{H}, the real quaternion algebra, as a vector space over \mathbb{R} has a basis \{1,i,j,k\} and multiplication of two elements in \mathbb{H} is done using distributive law and the relations i^2=j^2=k^2=ijk=-1. It is easy to see that \mathbb{H} is a division algebra and its center is \mathbb{R}. So, by the lemma in this post, \mathbb{H} \otimes_{\mathbb{R}} \mathbb{C} \cong M_2(\mathbb{C}). This isomorphism is also a result of this fact that \mathbb{C} is a maximal subfield of \mathbb{H} (see corollary 3 in this post!). In the following example, we are going to find the reduced characteristic polynomial of an element of \mathbb{H} by giving a \mathbb{C}-algebra isomorphism \mathbb{H} \otimes_\mathbb{R} \mathbb{C} \longrightarrow M_2(\mathbb{C}) explicitely. We will learn later how to find the reduced characteristic polynomial of an element of any finite dimensional central division algebra.

Example. If a = \alpha + \beta i + \gamma j + \delta k \in \mathbb{H}, then \text{Prd}_{\mathbb{H}}(a,x)=x^2 - 2 \alpha x + \alpha^2+\beta^2+\gamma^2 + \delta^2.

Solution. We define f : \mathbb{H} \otimes_{\mathbb{R}} \mathbb{C} \longrightarrow M_2(\mathbb{C}) as follows: for every a_1 = \alpha_1 + \beta_1 i + \gamma_1 j + \delta_1 k \in \mathbb{H} and z \in \mathbb{C} we define

f(a_1 \otimes_{\mathbb{R}} z) = \begin{pmatrix} (\alpha_1 + \beta_1 i)z & (\gamma_1 + \delta_1 i)z \\ (-\gamma_1 + \delta_1 i)z & (\alpha_1 - \beta_1 i)z \end{pmatrix}.

Then of course we extend f linearly to all elements of \mathbb{H} \otimes_{\mathbb{R}} \mathbb{C}. See that f is a \mathbb{C}-algebra homomorphism. Since \mathbb{H} \otimes_{\mathbb{R}} \mathbb{C} is simple, f is injective. Therefore f is an isomorphism because

4=\dim_{\mathbb{C}} M_2(\mathbb{C}) = \dim_{\mathbb{C}} \mathbb{H} \otimes_{\mathbb{R}} \mathbb{C}.

It is easy now to find \text{Prd}_{\mathbb{H}}(a):

\text{Prd}_{\mathbb{H}}(a,x)=\det(xI - f(a \otimes_{\mathbb{R}} 1)) = \det \left ( \begin{pmatrix} x & 0 \\ 0 & x \end{pmatrix} - \begin{pmatrix} \alpha + \beta i & \gamma + \delta i \\ -\gamma + \delta i & \alpha - \beta i \end{pmatrix} \right).

The rest of the proof is straightforward. \Box

In the above example, the coefficients of \text{Prd}_{\mathbb{H}}(a,x) are all in \mathbb{R}. This is not an accident. We have already proved that the reduced characteristic polynomial of an element of a finite dimensional central simple k-algebra is always in k[x] (see the theorem in this post!).

Reduced characteristic polynomials (2)

Posted: October 2, 2011 in Noncommutative Ring Theory Notes, Simple Rings
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See part (1)  here. We will assume that k is a field and A is a finite dimensional central simple k-algebra of degree n.

Lemma 3. Let K/k be a Galois splitting field of A with a K-algebra isomorphism f: A \otimes_k K \longrightarrow M_n(K). Then \det(xI - f(a \otimes_k 1)) \in k[x] for all a \in A.

Proof. Recall that A always have a Galois splitting field (see the corollary in this post). Let a \in A and put q(x) = \det(xI-f(a \otimes_k 1))=\sum_{i=0}^n c_i x^i. Clearly q(x) \in K[x] but we want to prove that p(x) \in k[x]. Let \phi \in \text{Gal}(K/k). By Lemma 1 in part (1), there exists a K-algebra isomorphism g : A \otimes_k K \longrightarrow M_n(K) such that \det(xI - g(a \otimes_k 1))=\phi_p(q(x)). By Lemma 2 in part (1), \det(xI - g(a \otimes_k 1))=q(x) and so q(x)=\phi_p(q(x)), i.e. \sum_{i=0}^n c_i x^i = \sum_{i=0}^n \phi(c_i)x^i. Thus \phi(c_i)=c_i for all i and all \phi \in \text{Gal}(K/k). So c_i \in k, because K/k is Galois, and thus q(x) \in k[x]. \ \Box

Lemma 4. Let F/k and E/k be splitting fields of A with algebra isomorphisms g : A \otimes_k F \longrightarrow M_n(F) and h : A \otimes_k E \longrightarrow M_n(E). Then \det(xI - g(a \otimes_k 1))= \det(xI - h(a \otimes_k 1)) \in k[x] for all a \in A.

Proof. Fix a Galois splitting field K/k of A and a K-algebra isomorphism f : A \otimes_k K \longrightarrow M_n(K). Let a \in A and put

q(x)=\det(xI - f(a \otimes_k 1)).

By Lemma 3, q(x) \in k[x]. Let

s(x) = \det(xI - h(a \otimes_k 1)).

Then, in order to prove the lemma, we only need to show that q(x) =h(x). Let L = (K \otimes_k E)/\mathfrak{M}, where \mathfrak{M} is a maximal ideal of K \otimes_k E. So L is a field. Define the k-algebra homomorphisms \phi : K \longrightarrow L and \psi : E \longrightarrow L by \phi(u) = u \otimes_k 1 + \mathfrak{M} and \psi(v) = 1 \otimes_k v + \mathfrak{M} for all u \in K and v \in E. By Lemma 1 in part (1), there exist L-algebra isomorphisms

g_1,g_2: A \otimes_k L \longrightarrow M_n(L)

such that \det(xI - g_1(a \otimes_k 1))=\phi_p(q(x)) and \det(xI - g_2(a \otimes_k 1))=\psi_p(s(x)). By Lemma 2, \det(xI - g_1(a \otimes_k 1))=\det(xI - g_2(a \otimes_k 1)) and so \phi_p(q(x))=\psi_p(s(x)). We also have \phi_p(q(x)) = \psi_p(q(x))=q(x) because q(x) \in k[x]. Hence \psi_p(q(x))=\psi_p(s(x)) and therefore q(x)=s(x) because \psi_p is injective. \Box

Definition. Let K/k be a splitting field of A with a K-algebra isomorphism f: A \otimes_k K \longrightarrow M_n(K). For a \in A let \text{Prd}_A(a,x)=\det(xI - f(a \otimes_k 1)). The monic polynomial \text{Prd}_A(a,x) is called the reduced characteristic polynomial of a.

Theorem. Let a \in A. Then \text{Prd}_A(a,x) \in k[x] and \text{Prd}_A(a,x) does not depend on K or f.

Proof. By Lemma 2 in part (1), \text{Prd}_A(a,x) does not depend on f(x). By Lemma 4, \text{Prd}_A(a,x) \in k[x] and \text{Prd}_A(a,x) does not depend on K. \ \Box

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