Other Rings & Algebras | Abstract Algebra

Archive for the ‘Other Rings & Algebras’ Category

Clifford algebras

Throughout this post, $k$ is a field of characteristic $\ne 2$ and $k\langle x_1, \ldots ,x_n\rangle$ is the $k$ -algebra of polynomials in noncommuting indeterminates $x_1, \ldots ,x_n$ and with coefficients from $k.$ So we’re assuming the elements of $k$ are central.

Note. The reference for this post is Exercises 34, 35, in Part III of the Benson Farb & R. Keith Denis’ book Noncommutative Algebra.

Here we defined the exterior algebra $\Lambda_n(k)$ as the $k$ -algebra generated by $x_1, \cdots , x_n$ subject only to the relations $x_ix_j=-x_jx_i, \ 1 \le i,j \le n.$ Note that since $\text{char}(k) \ne 2,$ we may replace the relations $x_ix_j=-x_jx_i, \ 1 \le i,j \le n,$ with $x_ix_j =-x_jx_i, \ 1 \le i \ne j \le n, \ x_i^2=0, \ 1 \le i \le n.$ In other words, $\Lambda_n(k)=k\langle x_1, \ldots ,x_n\rangle/I,$ where $I$ is the ideal of $k\langle x_1, \ldots ,x_n\rangle$ generated by the set

$\{x_ix_j+x_jx_i, \ 1 \le i \ne j \le n, \ \ x_i^2, \ 1 \le i \le n\}.$

In this post, we defined the quaternion algebras $(a_1,a_2)_k, \ a_1,a_2 \in k \setminus \{0\},$ as the $k$ -algebra generated by $x_1,x_2$ subject only to the relations $x_1x_2=-x_2x_1, \ x_1^2=a_1, \ x_2^2=a_2.$ So $(a_1,a_2)_k=k\langle x_1, x_2\rangle/I,$ where $I$ is the ideal of $k\langle x_1, x_2\rangle$ generated by the set $\{x_1x_2+x_2x_1, x_1^2-a_1, x_2^2-a_2\}.$

Exterior and quaternion algebras are just special cases of a large class of algebras known as Clifford algebras.

Definition. Let $a_1, \cdots , a_n \in k.$ The Clifford algebra $\text{Cl}_k(a_1, \cdots , a_n)$ is the $k$ -algebra generated by $x_1, \cdots , x_n$ subject only to the relations $x_ix_j=-x_jx_i, \ 1 \le i \ne j \le n, \ \ x_i^2=a_i, \ 1 \le i \le n.$

In other words, $\text{Cl}_k(a_1, \cdots, a_n)=k\langle x_1, \cdots , x_n\rangle/I,$ where $I$ is the ideal of $k\langle x_1, \cdots , x_n\rangle$ generated by

$\{x_ix_j+x_jx_i, \ 1 \le i \ne j \le n, \ \ x_i^2-a_i, \ 1 \le i \le n\}.$

Example. Clearly $\Lambda_n(k)= \text{Cl}_k(0, \cdots ,0)$ and $(a_1,a_2)_k= \text{Cl}_k(a_1,a_2).$

We now characterize semisimple Clifford algebras.

Theorem. Let $R:=\text{Cl}_k(a_1, \cdots , a_n),$ and let $J(R)$ be the Jacobson radical of $R.$

i) As a vector space, $\dim_k R=2^n.$ In particular, $R$ is Artinian.

ii) If $a_i=0$ for some $i,$ then $x_i \in J(R)$ and so $J(R) \ne (0).$

iii) if $a_i \ne 0$ for all $i,$ then $J(R)=(0).$

iv) $R$ is semisimple if and only if $a_i \ne 0$ for all $i.$

Proof. i) Every element of $R$ is in the form $c_0+c_1z_1+ \cdots +c_mz_m,$ where $c_i \in k$ and $z_i$ is a monomial in $x_1, \cdots , x_n,$ for all $i.$ Since $x_ix_j=-x_jx_i$ for all $i \ne j,$ each $z_i$ has the form $\pm x_1^{r_1}x_2^{r_2} \cdots x_n^{r_n}$ for some non-negative integers $r_i.$ So since $x_i^2=a_i \in k,$ for all $i,$ each $z_i$ is in the form $c x_1^{r_1}x_2^{r_2} \cdots x_n^{r_n}$ for some $c \in k$ and some integers $r_i \in \{0,1\}.$ Therefore, since there are no other relations between $x_1, \cdots , x_n,$ the set

$\mathcal{B}:=\{1\} \cup \{x_{i_1}x_{i_2} \cdots x_{i_s}: \ 1 \le s \le n, \ 1 \le i_1 < i_2 < \cdots < i_s \le n\} \ \ \ \ \ \ \ (*)$

is a $k$ -basis for $R$ and hence $\dim_k R=2^n.$ So since $R$ is a finite dimensional $k$ -algebra, it is Artinian.

ii) If $a_i=0,$ then $x_i^2=a_i=0$ and thus $(rx_i)^2=0$ for all $r \in R.$ Therefore $1-rx_i$ is a unit of $R$ (its inverse is $1+rx_i$ ), and hence $x_i \in J(R).$

iii) The proof of this part is almost identical to the proof of Maschke’s theorem that we gave here. Let $\mathcal{B}$ be the $k$ -basis of $R$ given in $(*).$ We can make $\mathcal{B}$ ordered by writing it as

$\mathcal{B}=\{1, x_1, \cdots , x_n, x_1x_2, x_1x_3, \cdots , x_1x_n, x_1x_2x_3, \cdots , x_1x_2 \cdots x_n\}.$

Note that, since $a_i \ne 0$ for all $i,$ each $x_i,$ and hence each element of $\mathcal{B},$ is a unit of $R$ because $x_i^2=a_i,$ which gives $x_i^{-1}=a_i^{-1}x_i.$ Now, consider the $k$ -algebra homomorphism $\rho : R \longrightarrow \text{End}_k(R)$ defined by $\rho(r)(z)=rz$ for all $r,z \in R.$ Define the map $\alpha : R \longrightarrow k$ by $\alpha(r) = \text{tr}(\rho(r)), \ r \in R,$ where $\text{tr}(\rho(r))$ is the trace of the matrix corresponding to the linear map $\rho(r)$ with respect to the ordered basis $\mathcal{B}$ given in $(*).$ We now make three simple observations.

1) $\alpha(1)=|\mathcal{B}|=2^n$ because $\rho(1)$ is the identity map of $R.$

2) If $1 \neq r \in \mathcal{B},$ then $\alpha(r)=0.$ That’s because $\rho(r)(z)=rz \ne z$ for all $z \in \mathcal{B} \setminus \{r\}$ and so the diagonal entries of the matrix of $\rho(r)$ are all zero hence $\alpha(r)=\text{tr}(\rho(r))=0.$

3) If $r \in R$ is nilpotent, then $\alpha(r)=0.$ That’s because $r^m=0$ for some $m$ and so $(\rho(r))^m = \rho(r^m)=0.$ Thus $\rho(r)$ is nilpotent and we know that the trace of a nilpotent matrix is zero.

Now let $r \in J(R).$ Then $r \ne 1$ and since, by i), $R$ is Artinian, $r$ is nilpotent hence $\alpha(r)=0,$ by 3). Let $r = \sum_{i=1}^{2^n} c_i z_i,$ where $c_i \in k, \ z_i \in \mathcal{B}, \ z_1=1.$ So, by 1), 2),

$0=\alpha(r)=\sum_{i=1}^{2^n} c_i \alpha(z_i)=c_1\alpha(z_1)=c_1\alpha(1)=2^nc_1$

and hence $c_1=0$ because $\text{char}(k) \ne 2.$ So the coefficient of $z_1$ of every element in $J(R)$ is zero. But for every $i,$ the coefficient of $z_1$ of the element $z_i^{-1}r \in J(R)$ is $c_i$ and so $c_i=0$ for all $i.$ Hence $r = 0$ and so $J(R)=(0).$

iv) Recall that a ring is semisimple if and only if it is Artinian and its Jacobson radical is zero. By i), $R$ is Artinian, and by ii), iii), $J(R)=(0)$ if and only if $a_i \ne 0$ for all $i. \ \Box$

The exterior algebra

Posted: April 4, 2022 in Noncommutative Ring Theory Notes, Other Rings & Algebras
Tags: Amitsur-Levitzki theorem, Artinian rings, exterior algebra, finite dimensional algebras, finitely generated algebras, grassmann algebra, group of units, nilpotent, Noetherian ring, Weyl algebra, zero-divisor

Throughout this post, $k$ is a field and $k\langle x_1, \ldots ,x_n\rangle$ is the $k$ -algebra of polynomials in non-commuting indeterminates $x_1, \ldots ,x_n$ and with coefficients from $k.$ So we’re assuming the elements of $k$ are central.

Let $A=k[a_1, \ldots , a_n]$ be a finitely generated $k$ -algebra which may or may not be commutative. The map $f: k\langle x_1, \ldots ,x_n\rangle \to A$ defined by $f(c)=c, \ f(x_i)=a_i,$ for all $c \in k, \ 1 \le i \le n,$ is clearly an onto $k$ -algebra homomorphism and so $A \cong k\langle x_1, \ldots ,x_n\rangle/I,$ where $I=\ker f,$ which is a two-sided ideal of $k\langle x_1, \ldots ,x_n\rangle.$ So, for example, if $I=\langle x_ix_j-x_jx_i, \ 1 \le i, j \le n\rangle,$ then $A$ is simply the commutative polynomial $k$ -algebra in $n$ indeterminates. As another example, if $n=2, \ I=\langle x_2x_1-x_1x_2-1 \rangle,$ then $A$ is just the first Weyl algebra. In this post, we take a look at another well-known example that we have not seen in this blog, i.e. where $I=\langle x_ix_j+x_jx_i, \ 1 \le i,j \le n\rangle.$ This algebra is called the exterior algebra. Let’s make the definition official.

Definition. The algebra $k\langle x_1, \ldots ,x_n\rangle/I, \ n \ge 2,$ where $I=\langle x_ix_j+x_jx_i, \ 1 \le i,j \le n\rangle,$ is called the exterior algebra, a.k.a. Grassmann algebra, and is denoted by $\Lambda_n(k).$

Ok, let’s make the definition a little more friendly. If in $\Lambda_n(k)$ we write $x_i$ instead of the coset $x_i+I,$ for every $i,$ then in $\Lambda_n(k)$ we get $x_ix_j+x_jx_i=0$ and so we’ll have a simpler definition of $\Lambda_n(k).$

A Simpler Definition of Exterior Algebras. The exterior algebra $\Lambda_n(k)$ is the $k$ -algebra generated by $x_1, \ldots ,x_n$ subject only to the relations $x_ix_j=-x_jx_i, \ 1 \le i,j \le n.$

Remark. If $\text{char}(k)=2,$ then the relations $x_ix_j=-x_jx_i$ become $x_ix_j=x_jx_i$ and so $\Lambda_n(k)$ is just the commutative polynomial ring $k[x_1, \ldots , x_n].$

The case $\text{char}(k) \ne 2$ is much more interesting. As an example, let’s see how $\Lambda_2(k)$ looks like in this case.

Example. If $\text{char}(k) \ne 2,$ then $\Lambda_2(k)=k+kx_1+kx_2+kx_1x_2.$

Proof. By definition, $\Lambda_2(k)$ is the $k$ -algebra generated by $x_1,x_2$ subject only to the relations

$x_ix_j=-x_jx_i, \ 1 \le i,j \le 2.$

So $x_1^2=-x_1^2, \ x_2^2=-x_2^2,$ which give $x_1^2=x_2^2=0$ because $\text{char}(k) \ne 2,$ and $x_1x_2=-x_2x_1.$ Now, an elements of $\Lambda_2(k)$ is a polynomial in $x_1,x_2$ and so it’s a $k$ -linear combination of $1$ and monomials in $x_1,x_2.$ But a monomial in $x_1,x_2$ is in the form $z=y_1y_2 \ldots y_m$ where $m$ is any positive integer and $y_i \in \{x_1,x_2\}$ for all $i.$ Thus reordering $y_1, \ldots, y_m$ in $z$ if necessarily, using the relations $x_1x_2=-x_2x_1,$ and the fact that $x_i^2=0$ for all $i,$ we see that $z \in \{0,x_1,x_2,x_1x_2\}. \ \Box$

Following the above example, you should be able to easily see now that if $\text{char}(k) \ne 2,$ then, for example, $\Lambda_3(k)=k+kx_1+kx_2+kx_3+kx_1x_2+kx_2x_3+kx_1x_3+kx_1x_2x_3.$ Can you now see how $\Lambda_n(k)$ looks like in general? The following theorem gives you the view and some basic properties of $\Lambda_n(k).$

Theorem. Suppose that $\text{char}(k) \ne 2$ and let $\Lambda_n:=\Lambda_n(k).$ Then

i) $\Lambda_n$ is a finite dimensional $k$ -algebra with the $k$ -basis consisting of $1$ and all the monomials in the form $x_{i_1}x_{i_2} \ldots x_{i_m},$ where $1 \le m \le n, \ \ 1 \le i_1 < i_2 < \ldots < i_m \le n,$

ii) $\dim_k \Lambda_n=2^n,$

iii) $x_i\Lambda_nx_i=\{0\}$ for all $1 \le i \le n,$

iv) if $z_1, \ldots ,z_{n+1}$ are any elements of $\Lambda_n$ with constant terms $0,$ then $z_1z_2 \ldots z_{n+1}=0,$

v) $\Lambda_n$ is both Artinian and Noetherian,

vi) every element of $\Lambda_n$ is either a unit or a zero-divisor,

vii) an element of $\Lambda_n$ is a zero-divisor if and only if its constant term is zero,

viii) an element of $\Lambda_n$ is nilpotent if and only if it’s a zero-divisor,

ix) the set $\mathfrak{m}$ of non-units of $\Lambda_n$ is an ideal and $\mathfrak{m}^{n+1}=(0).$

Proof. i) If we choose $i=j,$ then from the relations $x_ix_j+x_jx_i=0$ we get $2x_i^2=0$ and so $x_i^2=0,$ for all $i,$ because $\text{char}(k) \ne 2.$ Thus, again using the relations $x_ix_j=-x_jx_i,$ every monomial is either $0$ or is in the form $x_{i_1}x_{i_2} \ldots x_{i_m}$ for some positive integer $m \le n$ and $1 \le i_1 < i_2 < \ldots i_m \le n.$

ii) By i), $\dim_k \Lambda_n=1+\sum_{m=1}^n \binom{n}{m}=2^n.$

iii) As shown in i), $x_i^2=0$ for all $i,$ and so, using the relation $x_ix_j=-x_jx_i$ we have

$x_i x_{i_1}x_{i_2} \ldots x_{i_m}x_i=\pm x_i^2x_{i_1}x_{i_2} \ldots x_{i_m}=0,$

for all $m, i_1, \ldots , i_m.$

iv) Let $z:=z_1z_2 \ldots z_{n+1}.$ Since the constant term of each $z_i$ is $0,$ every monomial appearing in $z$ is a product of at least $n+1$ elements of the set $\{x_1, \ldots, x_n\},$ and hence it’s $0,$ by iii).

v) That’s a property of every finite dimensional algebra $A$ over a field. To see that, first note that a (left/right) ideal of $A$ is clearly a $k$ -vector subspace of $A.$ So if $I_1 \subseteq I_2 \subseteq \ldots$ and $J_1 \supseteq J_2 \supseteq \ldots$ are chains of (left/right) ideals of $A,$ then $\dim_k I_1 \le \dim_kI_2 \le \ldots$ and $\dim_k J_1 \ge \dim_k J_2 \ge \ldots$ and that cannot go on forever because the dimensions are finite.

vi) By ii), $\Lambda_n$ is finite dimensional hence algebraic over $k.$ The result now follows from this post.

vii) If the constant term of $z \in A$ is $0,$ then $z^{n+1}=0,$ by iv), and so $z$ is nilpotent hence a zero-divisor. Conversely, if the constant term of $z$ is not $0,$ then we can write $z=c+y$ for some $0 \ne c \in k$ and some $y \in \Lambda_n$ whose constant term is $0.$ By iv), $y$ is nilpotent and so $c^{-1}y$ is nilpotent. Hence $1+c^{-1}y$ is a unit. Thus $z=c(1+c^{-1}y)$ is a unit, and so it can’t be a zero-divisor.

viii) If $z \in \Lambda_n$ is nilpotent, it’s clearly a zero-divisor too. Conversely, if $z$ is a zero-divisor, then, by vii), the constant term of $z$ is $0$ and so, by iv), $z^{n+1}=0$ hence $z$ is nilpotent.

ix) Clear by iv), vi), and vii). $\ \Box$

An Application. Using the exterior algebra, Shmuel Rosset gave a clever proof of the celebrated Amitsur-Levitzki theorem: for any commutative ring $C$ and any $2n$ matrices $A_1, \cdots , A_{2n} \in M_n(C),$

$\displaystyle \sum_{\sigma \in S_{2n}}\text{sgn}(\sigma)A_{\sigma(1)}A_{\sigma(2)} \cdots A_{\sigma(2n)}=0,$

where $S_{2n}$ is the symmetric group on the set $\{1,2, \ldots , 2n\}$ and $\text{sgn}(\sigma)$ is the sign of $\sigma \in S_{2n}.$

Exercise 1. Show that if $\text{char}(k) \ne 2,$ then for every monomial $0 \ne z \in \Lambda_n(k)$ there exist $r_i \in \{0,1\}$ such that $z=x_1^{r_1}x_2^{r_2} \ldots x_n^{r_n}.$ This gives another proof of $\dim_k \Lambda_n(k)=2^n.$

Exercise 2. Show that $x_1x_2 \ldots x_n$ is in the center of $\Lambda_n(k).$ Assuming $\text{char}(k) \ne 2,$ can you find the center of $\Lambda_n(k) ?$

Exercise 3. Show that $(z_1z_2-z_2z_1)z_3=z_3(z_1z_2-z_2z_1)$ for all $z_1,z_2,z_3 \in \Lambda_n(k).$ In other words, for every $z_1,z_2 \in \Lambda_n(k),$ the commutator $z_1z_2-z_2z_1$ is central.

Reference

Hopfian algebras (2)

Posted: December 19, 2021 in Noncommutative Ring Theory Notes, Other Rings & Algebras
Tags: GK-dimension, Hopfian algebra, Hopfian ring, matrix ring, ring of endomorphisms

See the first part of the post here.

Let $k$ be a field, and let $V$ be a $k$ -vector space. If $\dim_k V=n$ is finite, then $\text{End}_k(V) \cong M_n(k)$ is a simple (Noetherian) ring hence Hopfian, by Example 1. As the next example shows, this result still holds even when $\dim_k V$ is infinite, countably.

Example 5. Let $k$ be a field, and let $V$ be a $k$ -vector space. Let $R:=\text{End}_k(V),$ the ring of $k$ -linear maps $V \to V.$ If $\dim_k V$ is countably infinite, then $R$ is Hopfian.

Proof. We showed here that $R$ has only one non-trivial two sided ideal, say $I.$ So $R$ is not a simple ring but $R/I$ is, and hence the rings $R/I$ and $R$ cannot be isomorphic. Thus $R$ is Hopfian, by Remark 2. $\Box$

Example 6. The matrix ring $M_n(R)$ over a commutative ring $R$ is Hopfian if and only if $R$ is Hopfian.

Proof. Recall that every two-sided ideal of $M_n(R)$ is in the form $M_n(I)$ for some ideal $I$ of $R.$ Now, If $I$ is an ideal of $R,$ and $A=[r_{ij}] \in M_n(R),$ then $\overline{A}:=[a_{ij}+I] \in M_n(R/I)$ and so we have a map

$f: M_n(R) \to M_n(R/I), \ \ \ \ \ f(A)=\overline{A}.$

It is easy to see that $f$ is a surjective ring homomorphism and $\ker f=M_n(I).$ Thus we have the following ring isomorphism

$M_n(R)/M_n(I) \cong M_n(R/I). \ \ \ \ \ \ \ \ \ \ (*)$

Now, suppose that $R$ is Hopfian, but $M_n(R)$ is not. So, by the above Remark, $M_n(R)/M_n(I) \cong M_n(R)$ for some non-zero ideal $I$ of $R.$ Thus, by $(*), \ M_n(R) \cong M_n(R/I),$ and so taking the center of both sides, we get $R \cong R/I,$ which is impossible because $R$ is Hopfian.
Conversely, suppose that $M_n(R)$ is Hopfian, but $R$ is not. Then $R$ has an ideal $I \ne (0)$ such that $R \cong R/I.$ But then, by $(*), \ M_n(R) \cong M_n(R/I) \cong M_n(R)/M_n(I),$ which is impossible because $M_n(R)$ is Hopfian and $M_n(I) \ne (0). \ \Box$

Example 7. Let $k$ be a field, and let $R$ be a $k$ -algebra of finite GK-dimension. If $R$ is a domain, then $R$ is Hopfian, as a $k$ -algebra.

Proof. Let $I \ne (0)$ be an ideal of $R,$ and choose $0 \ne x \in I.$ Since $R$ is a domain,

$\text{l.ann}_R(x):=\{r \in R: \ rx=0\}=(0)$

and hence, by Fact 6 in this post, $\text{GKdim}(R) \ne \text{GKdim}(R/I).$ Thus, as $k$ -algebras, $R$ and $R/I$ cannot be isomorphic, and so $R$ is Hopfian. $\Box$

Notes
$\bullet$ Let $k$ be a field. In this paper, the following two examples of Hopfian $k$ -algebras are given.
i) Every finitely generated algebra which is both prime and PI (Corollary 2.3).
ii) Every finitely generated free algebra (Theorem 2.7).
$\bullet$ See this paper for more examples of Hopfian algebras.

Hopfian algebras (1)

Posted: December 18, 2021 in Noncommutative Ring Theory Notes, Other Rings & Algebras
Tags: Finite Rings, finitely generated commutative algebra, Hopfian algebra, Hopfian ring, Noetherian ring, simple ring

All rings in this post are assumed to have $1,$ the multiplicative identity, and all ring homomorphisms map $1$ to $1.$ Also, $C$ is always a commutative ring.

Let $R$ be a $C$ -algebra. Recall that a ring homomorphism $f: R \to R$ is a $C$ -algebra homomorphism if $f(cr)=cf(r)$ for all $c \in C, r \in R.$

Definition 1. Let $R$ be a $C$ -algebra. We say that $R$ is Hopfian, as a $C$ -algebra, if every surjective $C$ -algebra homomorphism $R \to R$ is injective.

Remark 1. Any ring $R$ is a $\mathbb{Z}$ -algebra, and a map $f: R \to R$ is clearly a $\mathbb{Z}$ -algebra homomorphism if and only if $f$ is a ring homomorphism. So Definition 1 gives us the following definition.

Definition 2. A ring $R$ is said to be Hopfian, as a ring, if it is Hopfian as a $\mathbb{Z}$ -algebra, i.e. if every surjective ring homomorphism $R \to R$ is injective.

Note that if a $C$ -algebra is Hopfian as a ring, it is also Hopfian as a $C$ -algebra, but the converse is not necessarily true.

A $C$ -algebra $R$ is called co-Hopfian, as a $C$ -algebra, if every injective $C$ -algebra homomorphism $R \to R$ is surjective. Hopfian and co-Hopfian groups and modules are defined analogously. In this post, we only consider Hopfian algebras. We begin with a simple yet important remark.

Remark 2. Let $R$ be a $C$ -algebra. Then $R$ is not Hopfian, as a $C$ -algebra, if and only if there exists a two-sided ideal $I \ne (0)$ of $R$ such that, as $C$ -algebras, $R/I \cong R.$

Proof. If $R$ is not Hopfian, there exists a surjective $C$ -algebra homomorphism $g: R \to R$ such that $I:=\ker f \ne (0),$ and so $R/I \cong R.$ Conversely, suppose that there exists a $C$ -algebra isomorphism $f: R/I \to R$ for some two-sided ideal $I \ne (0)$ of $R.$ Then the $C$ -algebra homomorphism $g: R \to R$ defined by $g(r)=f(r+I), \ r \in R,$ is surjective but not injective, because $\ker g=I \ne (0),$ and hence $R$ is not Hopfian. $\Box$

Example 1. Let $R$ be a ring. If $R$ is simple, or finite, or $R=\mathbb{Z},$ then $R$ is Hopfian.

Proof. If $R$ is simple, then $R$ has no non-zero proper two-sided ideal of $R$ and hence, by Remark 2, $R$ is a Hopfian ring. If $R$ is finite, then, by Problem 1 in this post, every surjective map $R \to R$ is injective, and hence $R$ is Hopfian. If $R=\mathbb{Z},$ and $I \ne (0)$ is an ideal of $\mathbb{Z},$ then $I=n\mathbb{Z}$ for some integer $n > 0$ and $R/I \cong \mathbb{Z}_n,$ which is a finite ring hence not isomorphic to $\mathbb{Z}.$ So $\mathbb{Z}$ is Hopfian, by Remark 2. $\Box$

Finite rings and $\mathbb{Z}$ are just basic examples of Noetherian rings. We now show that in fact every Noetherian ring is Hopfian.

Example 2. Every Noetherian ring is Hopfian.

Proof. Let $R$ be a (left or right) Noetherian ring, and let $f: R \to R$ be a surjective ring homomorphism. We need to show that $\ker f=(0).$ We have the following ascending chain of two-sided ideals

$\ker f \subseteq \ker f^2 \subseteq \ker f^3 \subseteq \cdots ,$

and so, since $R$ is Noetherian, $\ker f^n=\ker f^{n+1}$ for some integer $n \ge 1.$ Now let $r \in \ker f.$ Since $f$ is surjective, $f^n$ is surjective too, and hence $f^n(x)=r$ for some $x \in R.$ Thus $0=f(r)=f^{n+1}(x)$ and therefore $x \in \ker f^{n+1}=\ker f^n,$ which gives $r=f^n(x)=0.$ So we have shown that $\ker f=(0),$ i.e. $f$ is injective, and hence $R$ is Hopfian. $\Box$

If $C$ is a field or $C=\mathbb{Z},$ then the commutative polynomial ring $C[x_1, \cdots ,x_n]$ is Noetherian, by the Hilbert basis theorem, and hence Hopfian, by Example 2. But if the number of variables is not finite, then polynomial rings are not Hopfian, as the next example shows.

Example 3. The commutative polynomial ring $R:=C[x_1,x_2, \cdots ]$ in infinitely many variables $x_1, x_2, \cdots$ is not Hopfian as a $C$ -algebra (hence not Hopfian as a ring).

Proof. We have the $C$ -algebra isomorphism $R \cong C[x_2,x_3, \cdots ],$ where the map is $x_i \to x_{i+1}, \ c \to c,$ for all $i$ and all $c \in C.$ So $R/\langle x_1 \rangle \cong C[x_2, x_3, \cdots] \cong R,$ and hence $R$ is not Hopfian, by Remark 2. $\Box$

Remark 3. Let $T:=C[x_1, \cdots,x_n],$ the commutative polynomial ring in the variables $x_1, \cdots , x_n,$ and let $R:=C[r_1, \cdots, r_n],$ a finitely generated commutative $C$ -algebra. We have a surjective $C$ -algebra homomorphism $f: T \to R$ defined by $x_i \to r_i, \ c \to c,$ for all $i$ and all $c \in C.$ So $R$ is a homomorphic image of $T.$ If $C$ is Noetherian, then $T$ is Noetherian, by the Hilbert basis theorem, and hence $R$ is Noetherian too. Therefore, by Example 2, $R$ is Hopfian, as a ring, provided that $C$ is Noetherian. But what if $C$ is not Noetherian? The next example shows that $R$ will still be Hopfian but as a $C$ -algebra not necessarily as a ring. But a first a little remark.

Remark 4. Let $R$ be a ring, and let $S:=\{m1_R: \ m \in \mathbb{Z}\}.$ Then $S$ is a Noetherian subring of $R.$

Proof. Let $\ell$ be the characteristic of $R.$ If $\ell=0,$ then $S \cong \mathbb{Z},$ and if $\ell > 0,$ then $S \cong \mathbb{Z}_{\ell}.$ Either way, $S$ is clearly Noetherian. $\Box$

Example 4 (Orzech & Ribes, 1970). Every finitely generated commutative algebra $R=C[r_1, \cdots, r_n]$ is Hopfian, as a $C$ -algebra.

Proof. Let $f: R \to R$ be a surjective $C$ -algebra homomorphism. Note that $f(c)=cf(1)=c$ for all $c \in C.$ Let $r \in \ker f.$ We are done if we show that $r=0.$ Since $f$ is surjective, for each $1 \le i \le n$ there exists $r_i' \in R$ such that $f(r'_i)=r_i.$ Note that every element of $R$ is a polynomial in the elements $r_1, \cdots, r_n$ with coefficients from $C.$ Let $\{c_1, \cdots , c_{\ell}\}$ be the set of all elements of $C$ that appear as a coefficient in at least one of the $2n+1$ polynomials $r, r'_1, \cdots , r_n', f(r_1), \cdots , f(r_n).$ Let $S$ be the Noetherian subring of $R$ described in Remark 4, and let $C':=S[c_1, \cdots, c_{\ell}], \ R':=C'[r_1, \cdots, r_n].$ By Remark 3, $C'$ and hence $R',$ is Noetherian. Note that $f(c')=c'$ for all $c' \in C',$ and clearly $r, r'_1, \cdots , r_n', f(r_1), \cdots , f(r_n) \in R'.$ Thus $f':=f \Big\vert_{R'},$ the restriction of $f$ to $R',$ is a surjective ring homomorphism $R' \to R',$ and hence, by Example 2, $f'$ is injective. So, since $f'(r)=f(r)=0,$ we get $r=0. \ \Box$

In the second part of this post, I’ll give more examples of Hopfian algebras.

Cyclotomic polynomials

Posted: September 24, 2011 in Noncommutative Ring Theory Notes, Other Rings & Algebras
Tags: cyclotomic polynomials

Throughout this post, $\zeta_n:= e^{2 \pi i/n},$ where $n$ is a positive integer.

The complex roots of the polynomial $x^n-1,$ where $n$ is a positive integer, are $x=\zeta_n^k, \ 1 \le k \le n.$ So the set of complex roots of $x^n-1,$ which is clearly a group under multiplication, is generated by $\zeta_n.$ In fact the set of the generators of the group is $S:=\{\zeta_n^k: \ 1 \le k \le n, \ \gcd(k,n)=1\}.$ An element of $S$ is called a primitive $n$ –th root of unity. The $n$ -th cyclotomic polynomial is defined to be the monic polynomial whose zeros are exactly the elements of $S.$ Let’s make this definition official.

Definition. Given a positive integer $n,$ the $n$ -th cyclotomic polynomial $\Phi_n(x)$ is defined by

$\displaystyle \Phi_n(x) = \prod_{1 \leq k \leq n, \ \gcd(k,n)=1}(x - \zeta_n^k).$

Proposition 1. Let $n$ be a positive integer.

i) $x^n - 1 = \prod_{d \mid n} \Phi_d(x).$ In particular, $\Phi_n(x) \mid x^n -1.$

ii) $\Phi_n(x) \in \mathbb{Z}[x],$ and $\deg \Phi_n(x)=\phi(n),$ where $\phi$ is the Euler’s totient function.

iii) If $1 \leq d < n$ and $d \mid n,$ then $\displaystyle \Phi_n(x) \mid \frac{x^n - 1}{x^d - 1}.$

Proof. i) We have

$\displaystyle x^n-1 = \prod_{j=1}^n (x - \zeta_n^j)=\prod_{d \mid n} \prod_{\gcd(j,n)=d}(x-\zeta_n^j).$

But

$\displaystyle \prod_{\gcd(j,n)=d} (x - \zeta_n^j) = \prod_{\gcd(k, n/d)=1}(x - \zeta_n^{kd})$

and obviously $\zeta_n^d = \zeta_{n/d}.$ Hence

$\displaystyle x^n-1 = \prod_{d \mid n} \prod_{\gcd(k,n/d)=1}(x - \zeta_{n/d}^k)=\prod_{d \mid n} \Phi_{n/d}(x)=\prod_{d \mid n} \Phi_d(x).$

ii) It is clear that $\deg \Phi_n(x)=\phi(n)$ because, by the definition of $\phi,$ the number of elements of the set $\{1 \leq k \leq n, \ \gcd(k,n)=1\}$ is $\phi(n).$ To show that $\Phi_n(x) \in \mathbb{Z}[x],$ we use induction over $n.$ There is nothing to prove if $n=1$ because $\Phi_1(x)=x-1.$ Now let $n \geq 2$ and suppose that $\Phi_m(x) \in \mathbb{Z}[x]$ for all $m < n.$ Note that cyclotomic polynomials are all monic. Thus, by i),

$x^n-1 = g(x) \Phi_n(x),$

for some monic polynomial $g(x) \in \mathbb{Z}[x].$ Thus, since $x^n-1$ is monic too, $\Phi_n(x) \in \mathbb{Z}[x].$

iii). By i),

$\displaystyle x^n - 1 = \prod_{m \mid n} \Phi_m(x)=\Phi_n(x) \prod_{m<n, \ m \mid n} \Phi_m(x)=\Phi_n(x) \prod_{m \mid d} \Phi_m(x) \prod_{m \nmid d, \ m \mid n} \Phi_m(x).$

Therefore, again by i), $x^n-1 = \Phi_n(x) (x^d -1) f(x),$ where $f(x) =\prod_{m \nmid d, \ m \mid n} \Phi_m(x). \ \Box$

Examples. Using Proposition 1, i), we can find explicit formulas for $\Phi_n(x)$ for certain values of $n.$

i) Since $x-1=\prod_{d \mid 1}\Phi_d(x)= \Phi_1(x),$ we have $\Phi_1(x)=x-1.$

ii) Let $p$ be a prime number and $k$ a positive integer. Then

$\displaystyle x^{p^k}-1=\prod_{d \mid p^k}\Phi_d(x)=\Phi_{p^k}(x)\prod_{d \mid p^{k-1}}\Phi_d(x)=\Phi_{p^k}(x)(x^{p^{k-1}}-1)$

and so

$\displaystyle \Phi_{p^k}(x)=\frac{x^{p^k}-1}{x^{p^{k-1}}-1}=\sum_{j=0}^{p-1}x^{jp^{k-1}}.$

In particular, $\displaystyle \Phi_p(x)=1+x+ \cdots + x^{p-1}$ and $\Phi_{p^k}(x)=\Phi_p(x^{p^{k-1}}).$

iii) Let $p,q$ be two distinct prime numbers and $k$ a positive integer. Then

$\displaystyle \begin{aligned}x^{p^kq}-1=\prod_{d \mid p^kq}\Phi_d(x)=\Phi_{p^kq}(x)\Phi_{p^k}(x)\prod_{d \mid p^{k-1}q}\Phi_d(x)=\Phi_{p^kq}(x)\Phi_{p^k}(x)(x^{p^{k-1}q}-1)\end{aligned}$

and so, by ii),

$\displaystyle \Phi_{p^kq}(x)=\frac{(x^{p^kq}-1)(x^{p^{k-1}}-1)}{(x^{p^{k-1}q}-1)(x^{p^k}-1)}=\frac{\Phi_{p^k}(x^q)}{\Phi_{p^k}(x)}.$

We now prove that cyclotomic polynomials are irreducible in $\mathbb{Q}[x]$ but first a simple lemma.

Lemma. Suppose that $\Phi_n(x)=u(x)v(x)$ for some positive integer $n$ and some non-constant polynomials $u(x),v(x) \in \mathbb{Z}[x].$ Then there exist a primitive $n$ -th root of unity $\zeta$ and prime number $p$ with $p \nmid n$ such that $u(\zeta)=0, \ v(\zeta^p)=0.$

Proof. The proof is by contradiction. Let $\zeta$ be a root of $u(x).$ Clearly $\zeta$ is a primitive $n$ -th root of unity because every root of $u$ is a root of $\Phi_n.$ So, since every root of $\Phi_n$ is either a root of $u$ or a root of $v,$ we must have $u(\zeta^p)=0$ for all primes $p \nmid n.$ Since $\zeta^p$ is again a primitive $n$ -th root of unity, we must have $u(\zeta^{pq})=0$ for all prime $q \nmid n.$ Repeating this argument gives $u(\zeta^k)=0$ for all positive integers $k$ with $\gcd(k,n)=1$ and so every root of $\Phi_n$ is a root of $u,$ which implies that $v$ is a constant, which is a contradiction. $\ \Box$

Remark. Given a prime number $p$ and $q(x)=\sum a_ix^i \in \mathbb{Z}[x], \ a_i \in \mathbb{Z},$ let

$\overline{q(x)}:=\sum \overline{a}_ix^i \in \mathbb{Z}_p[x],$

where $\overline{a}_i=a_i \mod p.$ Since, by Fermat’s little theorem, $a^p \equiv a \mod p,$ for any integer $a,$ it follows that $(\overline{q(x)})^p=\overline{q(x^p)}.$

Proposition 2. The cyclotomic polynomial $\Phi_n(x)$ is irreducible in $\mathbb{Q}[x]$ for all $n.$

Proof. Suppose, to the contrary, that $\Phi_n(x)$ is reducible in $\mathbb{Q}[x]$ for some positive integer $n.$ So, since $\Phi_n(x) \in \mathbb{Z}[x]$ is monic, there exist monic non-constant polynomials $u(x),v(x) \in \mathbb{Z}[x]$ such that $\Phi_n(x)=u(x)v(x).$ Thus, by the Lemma, there exist a primitive $n$ -th root of unity $\zeta$ and prime number $p \nmid n$ such that $u(\zeta)=0$ and $v(\zeta^p)=0.$ So $\zeta$ is a root of both $u(x), v(x^p)$ and therefore the minimal polynomial of $\zeta$ over $\mathbb{Q}$ divides both $u(x), v(x^p).$ Thus there exists a monic non-constant polynomial $w(x) \in \mathbb{Z}[x]$ such that $w(x)$ divides both $u(x),v(x^p).$ Now, let’s look at this in the polynomial ring $\mathbb{Z}_p[x].$ So $\overline{w(x)}$ divides both $\overline{u(x)}$ and $\overline{v(x^p)}.$ So since, by the above Remark, $\overline{v(x^p)}=(\overline{v(x)})^p,$ it follows that $\overline{w(x)}$ divides both $\overline{u(x)}, (\overline{v(x)})^p.$ So there exists a monic non-constant polynomial $\overline{w_0(x)} \in \mathbb{Z}_p[x]$ such that $\overline{w_0(x)}$ divides both $\overline{u(x)}, \overline{v(x)}.$ Hence $(\overline{w_0(x)})^2$ divides $\overline{u(x)} \ \overline{v(x)}=\overline{\Phi_n(x)}$ and therefore, since $\Phi_n(x) \mid x^n-1,$ it follows that $(\overline{w_0(x)})^2$ divides $\overline{x^n-1}=x^n-\overline{1}.$ Thus $\overline{w_o(x)}$ divides both $x^n-\overline{1}$ and its derivative $\overline{n}x^{n-1},$ which is clearly impossible unless $\overline{n}x^{n-1}=0,$ i.e. $p \mid n,$ which is not the case. This contradiction complete the proof of the Proposition. $\ \Box$

Dedekind-finite rings

Posted: September 17, 2010 in Noncommutative Ring Theory Notes, Other Rings & Algebras
Tags: algebraic algebra, Dedekind-finite, Hopfian modules, Jacobson radical, Kaplansky theorem, Noetherian ring, PI algebra, reduced ring, reversible ring, subdirect product

All rings in this post are assumed to have $1.$

Question. Is it true that in any ring $R,$ if $ab=1$ for some $a,b \in R,$ then $ba=1 ?$

Answer. No, and here’s an example. Let $V$ be a (countably) infinite dimensional vector space over some field $F$ and let $\{v_1,v_2, \cdots \}$ be a basis for $V.$ Now consider $R:=\text{End}_F(V),$ the ring of $F$ -linear maps $V \to V.$ Define $a,b \in R$ by $a(v_1)=0, \ a(v_j)=v_{j-1}, \ j \ge 2,$ and $b(v_j)=v_{j+1}, \ j \ge 1.$ See that $ab=\text{id}_V$ but $ba \ne \text{id}_V$ because $ba(v_1)=0.$

Definition 1. A ring $R$ is called Dedekind-finite if $\forall a,b \in R: \ ab=1 \Longrightarrow ba=1.$

Some trivial examples of Dedekind-finite rings: commutative rings, any direct product of Dedekind-finite rings, any subring of a Dedekind-finite ring.

Proposition. A ring $R$ is Dedekind-finite if and only if $R,$ as a left $R$ -module, is Hopfian, i.e. every surjective $R$ -module homomorphism $R \to R$ is injective.

Proof. Suppose first that $R$ is Dedekind-finite and $f: R \to R$ is a surjective $R$ -module homomorphism. So there exists $a \in R$ such that $1=f(a)=af(1)$ and hence $f(1)a=1.$ Now, suppose that $f(x)=0.$ Then $0=f(x)a=xf(1)a=x$ and so $\ker f=(0),$ i.e. $f$ is injective. Conversely, suppose that every surjective $R$ -module homomorphism $R \to R$ is injective, and $ab=1$ for some $a,b \in R.$ Consider the $R$ -module homomorphism $f: R \to R$ defined by $f(r)=rb,$ which is surjective because $f(ra)=rab=r$ for all $r \in R.$ So $f$ must be injective. Thus, since $b=f(1)=f(ba),$ we get that $ba=1$ and so $R$ is Dedekind-finite. $\ \Box$

Definition 2. A ring $R$ is called reversible if $\forall a,b \in R : \ ab = 0 \Longrightarrow ba = 0.$

Example 1. Every reversible ring $R$ is Dedekind-finite. In particular, reduced rings are Dedekind-finite.

Proof. Suppose that $ab=1$ for some $a,b \in R.$ Then $(ba-1)b=b(ab)-b=0$ and thus $b(ba-1)=0.$ So $b^2a=b$ and hence $ab^2a=ab=1.$ It follows that $ba=(ab^2a)ba=(ab^2)(ab)a=ab^2a=1.$ So $R$ is Dedekind-finite. Finally, note that every reduced ring is reversible because if $ab=0,$ for some $a,b \in R,$ then $(ba)^2=b(ab)a=0$ and thus $ba=0. \Box$

Example 2. Every (left or right) Noetherian ring $R$ is Dedekind-finite.

Proof. We will assume that $R$ is left Noetherian, the proof for right Noetherian rings is similar. Suppose that $ab=1$ for some $a,b \in R.$ Define the map $f: R \longrightarrow R$ by $f(r)=rb.$ Clearly $f$ is an $R$ -module homomorphism and $f$ is onto because $f(ra)=(ra)b=r(ab)=r,$ for all $r \in R.$ Now we have an ascending chain of left ideals of $R$

$\ker f \subseteq \ker f^2 \subseteq \cdots.$

Since $R$ is left Noetherian, this chain stabilizes at some point, i.e. there exists some $n$ such that $\ker f^n = \ker f^{n+1}.$ Clearly $f^n$ is onto because $f$ is onto. Thus $f^n(c)=ba-1$ for some $c \in R.$ Then

$f^{n+1}(c)=f(ba-1)=(ba-1)b=b(ab)-b=0.$

Hence $c \in \ker f^{n+1}=\ker f^n$ and therefore $ba-1=f^n (c) = 0. \Box$

Example 3. Every (left or right) Artinian ring $R$ is Dedekind-finite.

First Proof. Artinian rings are Noetherian, by Hopkins-Levitzki theorem, and hence Dedekind-finite, by Example 2.

Second Proof. Suppose that $ab=1$ for some $a,b \in R.$ First note that since $ab=1,$ we have

$a^nb^n=a^{n-1}(ab)b^{n-1}=a^{n-1}b^{n-1}= \cdots =ab=1,$

for all integers $n \ge 1.$ Now consider the descending chain of left ideals $Ra \supseteq Ra^2 \supseteq Ra^3 \supseteq \cdots .$ Since $R$ is left Artinian, there exists an integer $n \ge 1$ such that $Ra^n=Ra^{n+1}.$ Hence $a^n=ra^{n+1}$ for some $r \in R$ and so $(1-ra)a^n=0.$ Thus $1-ra=(1-ra)a^nb^n=0,$ i.e. $ra=1,$ and therefore

$ba=(ra)ba=r(ab)a=ra=1,$

proving that $R$ is Dedekind-finite. The proof for right Artinian rings is similar; the only difference is that you’ll need to consider the descending chain of right ideals $bR \supseteq b^2R \supseteq b^3R \supseteq \cdots. \ \Box$

Example 4. Finite rings are obviously Noetherian and so Dedekind-finite, by Example 2. More generally:

Example 5. If the number of nilpotent elements of a ring is finite, then the ring is Dedekind-finite. See here.

Note that Example 5 implies that every reduced ring is Dedekind-finite; a fact that we proved in Example 1.

Example 6. Let $k$ be a field and let $R$ be a finite dimensional $k$ -algebra. Then $R$ is Dedekind-finite.

Proof. Every left ideal of $R$ is clearly a $k$ -vector subspace of $R$ and thus, since $\dim_k R < \infty,$ any ascending chain of left ideals of $R$ will stop at some point. So $R$ is left Noetherian and thus, by Example 2, Dedekind-finite. $\Box$

Remark 1. Two important cases of Example 6 are $M_n(k),$ the ring of $n \times n$ matrices over a field $k,$ and, in general, semisimple rings. So $M_n(R)$ is Dedekind-finite for any commutative domain $R$ because $M_n(R)$ is a subring of $M_n(Q(R))$ , where $Q(R)$ is the quotient field of $R.$ In fact, $M_n(R)$ is Dedekind-finite for any commutative ring $R$ (see Example 7, and Example 10 for a generalization).
So the ring of $n \times n$ matrices, where $n \geq 2,$ over a field is an example of a Dedekind-finite ring which is not reversible, i.e. the converse of Example 1 is not true. Now let $R_i = \mathbb{Z}, \ i \geq 1.$ Then $R= \prod_{i=1}^{\infty} R_i$ is clearly Dedekind-finite but not Noetherian. So the converse of Example 2 is not true.

Example 7. $M_n(R)$ is Dedekind-finite for any commutative ring $R.$

Proof. There are several ways to prove that, here is one. Suppose that $AB=I$ for some $A,B \in M_n(R),$ where $I$ is the identity matrix. So $\det A \det B=\det I=1$ and hence $\det A$ is an invertible element of $R.$ Thus, since $\text{adj}(A)A=A\text{adj}(A)=\det(A)I,$ we get that $A$ is invertible and $A^{-1}=(\det A)^{-1}\text{adj}(A).$ Thus $AB=I$ gives $B=A^{-1}$ and so $BA=I. \ \Box$

Examples 8 and 10 are two generalizations of Example 6.

Example 8. Every algebraic algebra $R$ over a field $k$ is Dedekind-finite.

Proof. Suppose that $ab=1$ for some $a,b \in R.$ Since $R$ is algebraic over $k,$ there exist integers $n \geq m \geq 0$ and some $\alpha_i \in k$ with $\alpha_n \alpha_m \neq 0$ such that $\sum_{i=m}^n \alpha_i b^i = 0.$ We will assume that $n$ is as small as possible. Suppose that $m \geq 1.$ Then, since $ab=1,$ we have

$\displaystyle \sum_{i=m}^n \alpha_i b^{i-1}=a \sum_{i=m}^n \alpha_i b^i = 0,$

which contradicts the minimality of $n.$ So $m = 0.$ Let $c = -\alpha_0^{-1}\sum_{i=1}^n \alpha_i b^{i-1}$ and see that $bc=cb=1.$ But then $a=a(bc)=(ab)c=c$ and therefore $ba=bc=1. \ \Box$

Remark 2. Regarding Examples 6 and 8, note that although any finite dimensional $k$ -algebra $R$ is algebraic over $k,$ but $R$ being algebraic over $k$ does not necessarily imply that $R$ is finite dimensional over $k.$ For example, if $\overline{\mathbb{Q}}$ is the algebraic closure of $\mathbb{Q}$ in $\mathbb{C},$ then it is easily seen that $\dim_{\mathbb{Q}} \overline{\mathbb{Q}}=\infty.$ Thus the matrix ring $R = M_n(\overline{\mathbb{Q}})$ is an algebraic $\mathbb{Q}$ -algebra which is not finite dimensional over $\mathbb{Q}.$ So $R$ is Dedekind-finite by Example 8 (or 7) not Example 6.

Example 9. For a ring $R,$ let $J(R)$ be the Jacobson radical of $R.$ If $S:=R/J(R)$ is Dedekind-finite, then $R$ is Dedekind-finite too. In particular, every semilocal ring is Dedekind-finite.

Proof. Suppose that $ab = 1$ for some $a,b \in R$ and let $c,d$ be the image of $a,b$ in $S$ respectively. Clearly $cd=1_S$ and so $dc=1_S.$ Thus $1-ba \in J(R)$ and so $ba=1-(1-ba)$ is invertible. Hence there exists $e \in R$ such that $e(ba)=1.$ But then $eb=(eb)ab=e(ba)b=b$ and hence $ba=(eb)a=e(ba)=1. \Box$

Example 10. Every PI-algebra $R$ is Dedekind-finite.

Proof. If $J(R)=(0),$ then $R$ is a subdirect product of primitive algebras $R/P_i,$ where $P_i$ are the primitive ideals of $R.$ Since $R$ is PI, each $R/P_i$ is PI too and thus, by Kaplansky’s theorem, $R/P_i$ is a matrix ring over some division algebra and thus Dedekind-finite by Example 2. Thus $\prod R/P_i$ is Dedekind-finite and so $R,$ which is a subalgebra of $\prod R/P_i,$ is also Dedekind-finite. For the general case, let $S=R/J(R).$ Now, $S$ is PI, because $R$ is PI, and $J(S)=\{0\}.$ Therefore, by what we just proved, $S$ is Dedekind-finite and so $R$ is Dedekind-finite, by Example 9. $\ \Box$