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Archive for the ‘invariant basis number’ Category

Invariant basis number (3)

Posted: August 18, 2022 in invariant basis number, Noncommutative Ring Theory Notes
Tags: Artinian rings, Dedekind-finite, group algebra, invariant basis number, matrix rings, Noetherian ring

See the first two parts of this post here and here.

All rings in this post are assumed to have $1$ and all modules are left modules. Also, $M_n(R)$ denotes the ring of $n \times n$ matrices with entries from a ring $R,$ and $I_n$ denotes the $n \times n$ identity matrix.

We now give two more (clever) corollaries of Proposition 3 that we proved in the second part of this post.

Corollary 2. Let $R$ be a ring and suppose that $M_n(R)$ is Dedekind-finite for all positive integers $n.$ Then $R$ satisfies IBN.

Proof. Suppose, on the contrary, that $R$ does not satisfy IBN, and let $m,n,A,B,$ be as defined in Proposition 3 in the second part of this post. We may assume that $m < n.$ Now, let $A', B' \in M_n(R)$ be matrices obtained from $A,B,$ respectively, by adding $n-m$ zero-rows to $A$ and $n-m$ zero-columns to $B.$ Then see that $AB=I_m, \ BA=I_n$ gives $B'A'=I_n$ and $A'B' \ne I_n.$ Thus $M_n(R)$ is not Dedekind-finite, contradiction. $\ \Box$

A ring $R$ for which $M_n(R)$ is Dedekind-finite for all $n$ is called weakly-finite.

We can now show that (left or right) Noetherian rings satisfy IBN. But first, a little reminder.

Fact. If $R$ is left Noetherian (respectively, Artinian), then so is $M_n(R)$ for all positive integers $n.$

Proof. Note that $M_n(R)$ contains $\{rI_n: \ r \in R\} \cong R$ as a subring. Also, $M_n(R),$ as an $R$ -module, is generated by $n^2$ matrix units (matrices with one entry $1$ and all other entries $0$ ). Thus $M_n(R)$ is a finitely generated module over a left Noetherian (respectively, Artinian) subring, and hence it is a left Noetherian (respectively, Artinian) ring. $\ \Box$

In part (1) of this post, I gave four examples of rings that satisfy IBN. Now we can add another one, which is an important one.

Example 5. Every left Noetherian or Artinian ring satisfies IBN.

Proof. Let $R$ be a left Noetherian (respectively, Artinian) ring. Then, by the above Fact, $M_n(R)$ is also left Noetherian (respectively, Artinian) ring, and hence Dedekind-finite, by Examples 2,3 in this post. Thus $R$ satisfies IBN, by Corollary 2. $\ \Box$

Corollary 3. Let $R$ be a ring and suppose that there exist an abelian group $(G,+)$ and a group homomorphism $f: (R,+) \to (G,+)$ such that $f(ab)=f(ba), \ \forall a,b \in R,$ and the order of $f(1)$ is infinite. Then $R$ satisfies IBN.

Proof. Suppose, to the contrary, that $R$ does not satisfy IBN, and let $m,n,A=(a_{ij}),B=(b_{ij})$ be as defined in Proposition 3 in the second part of this post. Since $AB=I_m, \ BA=I_n,$ we have

$\displaystyle \sum_{j=1}^na_{ij}b_{ji}=1, \ \ 1 \le i \le m, \ \ \ \ \ \ \sum_{i=1}^mb_{ji}a_{ij}=1, \ \ 1 \le j \le n,$

and so

$\displaystyle mf(1)=\sum_{i=1}^mf(1)=\sum_{i=1}^mf \left( \sum_{j=1}^na_{ij}b_{ji}\right)=\sum_{i=1}^m\sum_{j=1}^nf(a_{ij}b_{ji})=\sum_{j=1}^n\sum_{i=1}^mf(a_{ij}b_{ji})$

$\displaystyle =\sum_{j=1}^n\sum_{i=1}^mf(b_{ji}a_{ij})=\sum_{j=1}^nf\left(\sum_{i=1}^mb_{ji}a_{ij}\right)=\sum_{j=1}^nf(1)=nf(1).$

Thus $(n-m)f(1)=0$ and so, since $f(1)$ has infinite order, $n=m,$ contradiction. $\ \Box$

We finish this post with an interesting application of Corollary 3.

Example 6. Let $C$ be a commutative ring of characteristic zero, and let $G$ be any group. Then the group algebra $R:=C[G]$ satisfies IBN.

Proof. Define the map $f: (R,+) \to (C,+)$ by $f(\sum c_g g)=c_1, \ c_g \in C, \ g \in G,$ where $c_1$ is the coefficient of $g=1_G.$ It is clear that $f$ is a group homomorphism and if $a=\sum c_g g, \ b=\sum c'_g g \in R,$ then

$\displaystyle f(ab)=\sum_{gh=1_G} c_gc'_h=\sum_{hg=1_G}c'_hc_g=f(ba).$

Finally, since the characteristic of $C$ is zero, $f(1)=1$ has infinite order and therefore, by Corollary 3, $R$ satisfies IBN. $\ \Box$

Think about this: does every Dedekind-finite ring satisfy IBN ?

Invariant basis number (2)

Posted: August 18, 2022 in invariant basis number, Noncommutative Ring Theory Notes
Tags: invariant basis number

See the first part of this post here.

In this post, we give a matrix view of rings that satisfy IBN, which turns out to be quite efficient.

Proposition 3. A ring $R$ does not satisfy IBN if and only if there exist distinct positive integers $m,n,$ an $m \times n$ matrix $A,$ and an $n \times m$ matrix $B,$ each with entries from $R,$ such that $AB=I_m, \ BA=I_n.$

Proof. Suppose first that $R$ does not satisfy IBN. So there exist distinct positive integers $m,n$ and a finitely generated free $R$ -module $M$ with two bases $\{x_1, \ldots , x_m\}, \ \{y_1, \ldots , y_n\}.$ So there exist $a_{ij}, b_{ij} \in R$ such that

$\displaystyle x_i=\sum_{j=1}^na_{ij}y_j, \ \ \ \ y_j=\sum_{k=1}^mb_{jk}x_k, \ \ \ \ \ 1 \le i \le m, \ 1 \le j \le n,$

and so

$\displaystyle x_i=\sum_{j=1}^na_{ij}\sum_{k=1}^mb_{jk}x_k=\sum_{k=1}^m\left(\sum_{j=1}^na_{ij}b_{jk}\right)x_k,$

$\displaystyle y_j=\sum_{k=1}^mb_{jk}\sum_{\ell=1}^na_{k\ell}y_{\ell}=\sum_{\ell=1}^n\left(\sum_{k=1}^mb_{jk}a_{k\ell}\right)y_{\ell},$

Thus, since $\{x_1, \ldots , x_n\}$ and $\{y_1, \ldots , y_n\}$ are $R$ -linearly independent, we must have

$\displaystyle \sum_{j=1}^na_{ij}b_{jk}=\begin{cases}1 & \text{if} \ k=i, \ \ 1 \le i \le m \\ 0 & \text{if} \ k \ne i, \ \ 1 \le i \le m,\end{cases} \ \ \ \ \ \ \ \ \ \ (1)$

$\displaystyle \sum_{k=1}^mb_{jk}a_{k\ell}=\begin{cases}1 & \text{if} \ \ell=j, \ \ 1 \le j \le n \\ 0 & \text{if} \ \ell \ne j, \ \ 1 \le j \le n.\end{cases} \ \ \ \ \ \ \ \ \ \ (2)$

which means that $AB=I_m, \ BA=I_n,$ where $A=(a_{ij}), \ B=(b_{ij}).$
Conversely, suppose that there exist distinct positive integers $m,n,$ an $m \times n$ matrix $A=(a_{ij})$ and an $n \times m$ matrix $B=(b_{ij})$ such that $AB=I_m, \ BA=I_n.$ We use Proposition 2 in the first part of this post to show that $R$ does not satisfy IBN. Let $\{x_1, \ldots , x_m\}$ and $\{y_1, \ldots , y_n\}$ be bases for $R^m, R^n,$ respectively. Define the $R$ -linear maps $f: R^m \to R^n, \ g: R^n \to R^m$ by

$f(x_i)=\sum_{j=1}^na_{ij}y_j, \ \ \ \ \ g(y_j)=\sum_{k=1}^mb_{jk}x_k,$

for all $i,j.$ Then the conditions $AB=I_m, \ BA=I_n,$ give $(1), (2),$ and so $fg(y_j)=y_i, \ gf(x_i)=x_i$ for all $i,j.$ Thus both $fg$ and $gf$ are identity maps and hence $f$ is an $R$ -module isomorphism. So, since $m \ne n, \ R$ does not satisfy IBN. $\ \Box$

The above Proposition has important corollaries.

Corollary 1. Let $R,S$ be rings, and let $f: R \to S$ be a ring homomorphism with $f(1)=1.$ If $S$ satisfies IBN, then $R$ satisfies IBN too.

Proof. Suppose, to the contrary, that $R$ does not satisfy IBN, and let $m,n, A,B$ be as defined in Proposition 3. Let $A=(a_{ij}), \ B=(b_{ij}),$ and define $A'=(f(a_{ij})), \ B'=(f(b_{ij})).$ Then since $f$ is a ring homomorphism, $f(1)=1,$ and $AB=I_m, \ BA=I_n,$ we have $A'B'=I_m, \ B'A'=I_n.$ Thus, by Proposition 3, $S$ does not satisfy IBN, contradiction. $\ \Box$

Remarks. i) If, in Corollary 1, we choose $S=R/I,$ where $I$ is any proper ideal of $R,$ then we’ll get Proposition 1 in part (1) of this post.
ii) If a ring $S$ satisfies IBN, then every subring $R$ of $S$ with $1_R=1_S$ satisfies IBN. This follows from Corollary 1 by choosing $f$ to be the inclusion map.

There are two more corollaries of Proposition 3 that we will discuss in the third part of this post, and then we will give important examples of rings that satisfy IBN.

Invariant basis number (1)

Posted: August 17, 2022 in invariant basis number, Noncommutative Ring Theory Notes
Tags: commutative ring, free modules, invariant basis number, vector space

All rings in this post are assumed to have $1$ and all modules are left modules.

Note. The main reference for this post is volume I, chapter 1, of the book Ring Theory, by Louis H. Rowen. As always, I have modified the proofs and added some comments and examples.

One of the first important facts that we learn in linear algebra is that every finite-dimensional vector space $V$ over a field $F$ has a basis and, more importantly, every two bases of $V$ have the same number of elements called the dimension of $V.$ Now, it makes sense to try to extend the concept of “dimension” by considering a finitely generated free module $M$ over any ring $R.$ So we are replacing $V$ with $M$ and $F$ with $R.$ Saying that $M$ is a finitely generated free $R$ -module means that there exists a finite set $\mathfrak{B}=\{x_1, \ldots , x_n\} \subseteq M$ such that every element of $x \in M$ can be written uniquely as $x=r_!x_1+ \ldots + r_nx_n, \ r_i \in R.$ Just like vector spaces, we call $\mathfrak{B}$ a basis or an $R$ -basis for $M.$ What comes next is naturally this question: will still be true that every two (finite) bases of $M$ have the same number of elements? The short answer is: not always, but most of the time, or at least if $R$ is “nice” enough. Let’s begin with giving those “nice” rings a name.

Definition. A ring $R$ is said to have the invariant basis number property if every two finite bases of a finitely generated free $R$ -module have the same number of elements. If $R$ has the invariant basis number property, then we say that $R$ satisfies IBN.

Example 1. By what we discussed at the beginning of this post, every field satisfies IBN.

We can now prove that every commutative ring satisfies IBN. But first, a more general fact.

Proposition 1. Let $R$ be a ring, and let $I$ be a proper ideal of $R.$ If $R/I$ satisfies IBN, then $R$ satisfies IBN too.

Proof. Let $M$ be a finitely generated free $R$ -module, and let

$\overline{R}:=R/I, \ \ \ \ \ \overline{M}:=M/IM.$

Since $I\overline{M}=(0),$ the second part of the Lemma in this post implies that $\overline{M}$ is an $\overline{R}$ -module where the scalar multiplication is defined as follows: if $\overline{r}=r+I \in \overline{R}, \ r \in R,$ and $\overline{x}=x+IM \in \overline{M}, \ x \in M,$ then $\overline{r} \ \overline{x}=rx+IM \in \overline{M}.$

Claim. If $\{x_1, \ldots , x_n\}$ is an $R$ -basis for $M,$ then $\{x_1+IM, \ldots , x_n+IM\}$ is an $\overline{R}$ -basis for $\overline{M}.$

Proof. Let $\overline{x}=x+IM \in \overline{M}, \ x \in M.$ Then $x=\sum_{i=1}^nr_ix_i$ for some (unique) $r_i \in R,$ and hence

$\displaystyle \overline{x}=\sum_{i=1}^nr_ix_i+IM=\sum_{i=1}^n(r_ix_i+IM)=\sum_{i=1}^n(r_i+I)(x_i+IM),$

which proves that the set $\mathfrak{B}:=\{x_1+IM, \ldots , x_n+IM\}$ generates $\overline{M}$ as an $\overline{R}$ -module. So we just need to show that $\mathfrak{B}$ is $\overline{R}$ -linearly independent, suppose that $\sum_{i=1}^n(r_i+I)(x_i+IM)=0,$ for some $r_i \in R.$ So $\sum_{i=1}^nr_ix_i+IM=0$ and hence $\sum_{i=1}^nr_ix_i \in IM.$ Thus there exist $a_1, \ldots , a_n \in I$ such that $\sum_{i=1}^nr_ix_i=\sum_{i=1}^na_ix_i,$ which gives $r_i=a_i \in I$ because $\{x_1, \ldots , x_n\}$ is an $R$ -basis for $M.$ Hence $r_i+I=0_{\overline{R}},$ for all $i,$ and that completes the proof of the Claim.

So, by the Claim, the number of elements of an $R$ -basis for $M$ is equal to the number of an $\overline{R}$ -basis for $\overline{M}.$ Since we’ve assumed that $\overline{R}$ satisfies IBN, that number is a constant and so all $R$ -bases for $M$ have the same number of elements, i.e. $R$ satisfies IBN. $\ \Box$

Example 2. Every commutative ring satisfies IBN.

Proof. Let $R$ be a commutative ring and let $\mathfrak{m}$ be any maximal ideal of $R.$ Then $R/\mathfrak{m}$ is a field hence satisfies IBN, by Example 1. Thus $R$ satisfies IBN, by Proposition 1. $\ \Box$

Example 3. Let $R:=C \langle x_1, \ldots , x_n\rangle,$ the ring of polynomials in noncommuting indeterminates $x_1, \ldots ,x_n$ and with coefficients from a commutative ring $C.$ Then $R$ satisfies IBN.

Proof. Let $I$ be the ideal of $R$ generated by $x_1, \ldots , x_n.$ Then $R/I \cong C$ and the result follows from Example 2 and Proposition 1. Note that the result still holds if the number of indeterminates is infinite. $\ \Box$

Example 4. Let $R$ be the ring of $n \times n$ upper triangular matrices with entries from some commutative ring $C.$ Then $R$ satisfies IBN.

Proof. Let $I$ be the set of all the elements of $R$ whose diagonal entries are all zero. See that $I$ is a proper ideal of $R$ and

$R/I \cong \underbrace{C \oplus \ldots \oplus C}_{\text{n times}},$

which is a commutative ring. The result now follows from Example 2 and Proposition 1. $\ \Box$

Proposition 2. For a positive integer $n$ and a ring $R,$ let

$R^n:= \underbrace{R \oplus \ldots \oplus R}_{\text{n times}},$

considered as an $R$ -module.

i) $M$ is a finitely generated free $R$ -module if and only if $M \cong R^n$ for some integer $n \ge 1.$

ii) $R$ satisfies IBN if and only if $R^n \cong R^m$ implies $m=n.$

Proof. Consider the following elements of $R^n$

$e_1:=(1,0, \ldots , 0), \ e_2:=(0,1,0, \ldots ,0), \ldots , e_n:=(0,0, \ldots ,0, 1),$

and let $r=(r_1,r_2, \ldots , r_n) \in R.$ Then $r=r_1e_1+r_2e_2+ \ldots + r_ne_n$ and this representation of $r$ is clearly unique. So $R^n$ is a finitely generated free $R$ -module and $\{e_1, \ldots , e_n\}$ is an $R$ -basis for $R^n.$ Now, let $M$ be any finitely generated free $R$ -module with basis $\{x_1, \ldots , x_n\}.$ Define the map $f: M \to R^n$ by $f(r_1x_1+ \ldots + r_nx_n)=r_1e_1+ \ldots + r_ne_n.$ Clearly $f$ is an $R$ -module isomorphism.

ii) Suppose first that $R$ satisfies IBN, and let $R^n \cong R^m.$ Let $f: R^n \to R^m$ be an $R$ -module isomorphism, and let $\mathfrak{B}_1, \mathfrak{B}_2$ be the bases of $R^n, R^m,$ as defined in i). Then $\mathfrak{B_1}, f^{-1}(\mathfrak{B}_2)$ are two bases for $R^n$ and hence $n=|\mathfrak{B}_1|=|f^{-1}(\mathfrak{B}_2)|=m.$ Conversely, let $M$ be a finitely generated free $R$ -module with two bases $B_1,B_2$ and let $n=|B_1|, m=|B_2|.$ Then, by i), $M \cong R^n, \ M \cong R^m,$ and so $R^n \cong R^m,$ giving $n=m.$ Thus $R$ satisfies IBN. $\ \Box$