Let \[ R = \operatorname{End}_{D}(V) \] denote the ring of all \( D \)-linear endomorphisms of \( V \). Then \( R \) is a ring under addition and composition of linear maps.
Consider \( R \) as a left module over itself. Since \[ R = R \cdot 1_R, \] it follows that \( R \) is a free \( R \)-module with basis \( \{1_R\} \), and hence \( R \) has a basis consisting of a single element.
We shall prove that for every positive integer \( n \), there exists an \( R \)-basis \[ B_n = \{f_1, f_2, \dots, f_n\} \] of \( R \) having exactly \( n \) elements. This will show that the free module \( R \) admits bases of different cardinalities.
Since \( V \) has countably infinite dimension over \( D \), there exists a basis \[ B = \{e_k\}_{k=1}^{\infty} \] of \( V \) over \( D \).
Using this basis, we define endomorphisms \[ f_1, f_2, \dots, f_n \in R \] by prescribing their values on the basis vectors \( e_k \) as follows:
For each integer \( k \geq 0 \) and \( 1 \leq i \leq n \), define \[ f_i(e_{kn+i}) = e_{k+1}, \] and set \[ f_i(e_j) = 0 \quad \text{for all other indices } j. \]
The action of these maps is summarized in the following table:
| f₁ | f₂ | f₃ | ⋯ | fₙ | |
|---|---|---|---|---|---|
| e₁ | e₁ | 0 | 0 | ⋯ | 0 |
| e₂ | 0 | e₁ | 0 | ⋯ | 0 |
| ⋮ | ⋱ | ||||
| eₙ | 0 | 0 | 0 | ⋯ | e₁ |
| eₙ₊₁ | e₂ | 0 | 0 | ⋯ | 0 |
| ⋮ | ⋱ |
Thus, on each block \( (e_{kn+1}, \dots, e_{(k+1)n}) \), exactly one vector is sent to \( e_{k+1} \), and all others are sent to zero.
(i) Linear Independence
Suppose \[ \sum_{i=1}^{n} \alpha_i f_i = 0, \quad \alpha_i \in R. \] Evaluating this on basis vectors yields \[ \alpha_1(e_{k+1}) = \alpha_2(e_{k+1}) = \cdots = \alpha_n(e_{k+1}) = 0 \] for all \( k \geq 0 \). Hence \( \alpha_i = 0 \) for all \( i \), so \( B_n \) is linearly independent.
(ii) Spanning
Let \( f \in R \). Define endomorphisms \( \alpha_1, \dots, \alpha_n \) by \[ \alpha_i(e_{k+1}) = f(e_{kn+i}). \] Then for every basis vector \( e_{kn+i} \), \[ \left( \sum_{j=1}^{n} \alpha_j f_j \right)(e_{kn+i}) = f(e_{kn+i}), \] hence \[ f = \sum_{i=1}^{n} \alpha_i f_i. \] Therefore, \( B_n \) spans \( R \).
Combining (i) and (ii), we conclude that \( B_n \) is an \( R \)-basis of \( R \). Hence, for each positive integer \( n \), the free module \( R \) admits a basis consisting of \( n \) elements.
| α1 | α2 | α3 | ⋯ | αn | |
|---|---|---|---|---|---|
| e1 | f(e1) | f(e2) | f(e3) | ⋯ | f(en) |
| e2 | f(en+1) | f(en+2) | f(en+3) | ⋯ | f(e2n) |
| e3 | f(e2n+1) | f(e2n+2) | f(e2n+3) | ⋯ | f(e3n) |
| ⋮ | ⋮ | ⋮ | ⋮ | ⋱ | ⋮ |
| ek+1 | f(ekn+1) | f(ekn+2) | f(ekn+3) | ⋯ | f(e(k+1)n) |
| ⋮ | ⋮ | ⋮ | ⋮ | ⋱ | ⋮ |