Bases

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

For any set \(X\), we observed that the free \(A\)-module defined by \(X\) is given by the expression

\[F(X)=\bigoplus_{x\in X} A\]

([Algebraic Structures] §Direct Products, Direct Sums, and Tensor Products of Modules, ⁋Proposition 3](/en/math/algebraic_structures/operations_of_modules#prop3)). In this post, we examine the properties of free \(A\)-modules in more detail.

Basis

Now let an arbitrary \(A\)-module \(M\) be given, and let a family \((x_i)_{i\in I}\) of elements of \(M\) be given. If we define the function \(e:I \rightarrow M\) by \(e(i)=x_i\), then by the adjunction \(F\dashv U\) there exists a unique \(A\)-linear map \(\varepsilon:F(I) \rightarrow M\). If \((x_i)_{i\in I}\) was a generating set of \(M\), then \(\varepsilon\) must be surjective, and the converse also holds. In a similar context, we make the following definition.

Definition 1 Let an arbitrary \(A\)-module \(M\) and a family \((x_i)_{i\in I}\) of elements of \(M\) be given. For the \(A\)-linear map \(\varepsilon:F(I) \rightarrow M\) defined above, we define the following.

1. The family \((x_i)_{i\in I}\) is a free family if \(\varepsilon\) is injective.
2. The family \((x_i)_{i\in I}\) generates \(M\) if \(\varepsilon\) is surjective.
3. The family \((x_i)_{i\in I}\) is a basis of \(M\) if \(\varepsilon\) is bijective.

A family that is not a free family is called a related family.

A free family generalizes the notion of linear independence in vector spaces. That is, if \(A\) is a field and \(M\) is a vector space over \(A\), then a family \((x_i)_{i\in I}\) of elements of \(M\) being a free family is equivalent to the \(x_i\) being linearly independent. ([Linear Algebra] §Bases of Vector Spaces, ⁋Definition 5](/en/math/linear_algebra/basis#def5)) From this perspective, the elements of a related family are said to be linearly dependent.

On the other hand, any \(A\)-module \(M\) always has a spanning set. This is because at the very least, collecting all elements of \(M\) generates \(M\). From this we obtain the following.

Proposition 2 Any \(A\)-module \(M\) is isomorphic to a quotient of a suitable free \(A\)-module.

Proof

For any \(A\)-module \(M\), let \(X\) be a spanning set of \(M\). Then there exists a surjective \(A\)-linear map \(\varepsilon:F(X) \rightarrow M\). In this case, since the kernel of \(\varepsilon\) is a submodule of \(F(X)\), we have \(M\cong F(X)/\ker\varepsilon\).

That \(M\) is a finitely generated \(A\)-module is equivalent to being able to choose such a family to be finite, and in this case the free \(A\)-module in the above proof can also be chosen to have a finite basis. As a more special case, we define the following.

Definition 3 An \(A\)-module \(M\) is called monogenous if \(M\) is generated by a single element \(x\) as an \(A\)-module.

The point to note is that \(x\) need not be a free element. That is, there may exist some \(\alpha\neq 0\) such that \(\alpha x=0\), and this is a point of difference from what was treated in Linear Algebra.

Invariant Basis Number

First we prove the following general proposition.

Proposition 4 Let \(M=\bigoplus_{i\in I} N_i\), let \(I\) be an infinite set, and let \(N_i\neq 0\). Then for any generating set \(X\) of \(M\), we have \(\card X\geq \card I\).

Proof

For each \(x\in X\), let \(S(x)\subseteq I\) be the set of indices of the nonzero components when \(x\) is expressed as an element of the direct sum. By the definition of direct sum, each \(S(x)\) is a finite set.

First we show that \(I=\bigcup_{x\in X}S(x)\). If some \(i_0\in I\) does not belong to any \(S(x)\), then all elements of \(X\) have \(0\) as their \(i_0\)-th component, and hence any linear combination of elements of \(X\) also has this property. However, since \(N_{i_0}\neq 0\), there exists an element of \(M\) whose \(i_0\)-th component is not \(0\), which contradicts the assumption that \(X\) generates \(M\).

Next, \(X\) must be an infinite set. Otherwise, \(I=\bigcup_{x\in X}S(x)\) would be a finite union of finite sets and hence finite, contradicting the assumption that \(I\) is infinite.

Finally we show \(\card I\leq\card X\). For each \(i\in I\), choose one \(x(i)\in X\) such that \(i\in S(x(i))\), and for each \(x\in X\) choose an injection \(\nu_x\) from the finite set \(S(x)\) to \(\mathbb{N}\). Then the function

\[I \rightarrow X\times\mathbb{N};\qquad i\mapsto \bigl(x(i),\nu_{x(i)}(i)\bigr)\]

is an injection. Indeed, if \(\bigl(x(i),\nu_{x(i)}(i)\bigr)=\bigl(x(j),\nu_{x(j)}(j)\bigr)\), then first \(x(i)=x(j)\), and letting this be \(x\), we have \(\nu_x(i)=\nu_x(j)\), and by the injectivity of \(\nu_x\) we get \(i=j\). Therefore

\[\card I\leq \card(X\times\mathbb{N})=\card X\cdot\aleph_0\leq \card X\]

and the last inequality follows from the fact that \(X\) is an infinite set and [Set Theory] §Natural Numbers and Infinite Sets, ⁋Corollary 16](/en/math/set_theory/natural_numbers#cor16).

In the situation of Definition 3, any element of \(M\) can be written in the form \(\alpha x\) for suitable \(\alpha\in A\). Therefore in such a case we also denote \(M\) by \(Ax\). Using this notation, for any \(A\)-module \(M\) and any family \((x_i)_{i\in I}\) of its elements,

• \((x_i)_{i\in I}\) being a generating family of \(M\) is equivalent to \(M=\sum_{i\in I}Ax_i\).
• \((x_i)_{i\in I}\) being a basis of \(M\) is equivalent to the above sum \(\sum_{i\in I}Ax_i\) being a direct sum and each \(x_i\) being a free element.

Through this, if we restrict Proposition 4 to the case where each \(N_i\) is a monogenous module generated by a free element, we know that when \(A\neq 0\), all bases of a free \(A\)-module \(M\) with an infinite basis have the same cardinality. However, this does not always hold in the case where \(M\) has a finite basis.

Definition 5 For any ring \(A\), if \(A^m\cong A^n\) is always equivalent to \(m=n\), we say that \(A\) satisfies the invariant basis number property.

For example, \(A=0\) does not satisfy this property, since \(0^m\cong 0^n\) for any \(m,n\).

By [Linear Algebra] §Dimension of Vector Spaces, ⁋Lemma 2](/en/math/linear_algebra/dimension#lem2), any field has the invariant basis number property. Using this, we can show the following more general proposition.

Proposition 6 Let \(A\) be a ring, and suppose there exist a suitable field \(\mathbb{K}\) and a homomorphism \(\phi: A \rightarrow \mathbb{K}\). Then \(A\) has the IBN property.

Proof

Let an arbitrary free \(A\)-module \(M\) be given. Then there exists an isomorphism

\[M\cong \bigoplus_{i\in I} Ax_i\]

and each of the \(x_i\) is a free element. On the other hand, since \(\phi_!:\lMod{A} \rightarrow \lMod{\mathbb{K}}\) is a left adjoint, the following identity holds:

\[\phi_! M\cong\phi_!\left(\bigoplus_{i\in I} Ax_i\right)\cong \bigoplus_{i\in I}\phi_! Ax_i\]

([Algebraic Structures] §Change of Scalars, ⁋Proposition 6](/en/math/algebraic_structures/change_of_base_ring#prop6)) Also, from the fact that \(x_i\) is a free element we have \(Ax_i\cong A\), and since \(\phi_! A\cong \mathbb{K}\), we get \(\phi_! M\cong \bigoplus_{i\in I}\mathbb{K}\). Now applying [Linear Algebra] §Dimension of Vector Spaces, ⁋Lemma 2](/en/math/linear_algebra/dimension#lem2) gives the desired result.

The proof of [Linear Algebra] §Dimension of Vector Spaces, ⁋Lemma 2](/en/math/linear_algebra/dimension#lem2) uses only the fact that nonzero scalars are invertible, not the property that \(\mathbb{K}\) is commutative, so the above proposition also holds more generally if we replace \(\mathbb{K}\) with a division ring \(D\). On the other hand, any nonzero commutative ring \(A\) has a maximal ideal \(\mathfrak{m}\) by [Algebraic Structures] §Definition of Rings, ⁋Theorem 10](/en/math/algebraic_structures/rings#thm10), so there exists a homomorphism \(A \rightarrow A/\mathfrak{m}\) to a field, and hence \(A\) has the IBN property.

Definition 7 Suppose a ring \(A\) satisfies IBN. Then for any free \(A\)-module \(M\), the size of a basis of \(M\) is called the rank of \(M\).

For convenience, when a basis \((x_i)_{i\in I}\) of \(M\) is given, we denote the free module \(F(I)\) obtained from it by \(A^{\oplus I}\), and in particular if \(I\) is a finite set we also denote it by \(A^m\). These notations have notational issues when used without the guarantee that \(A\) has the IBN property, but we turn a blind eye to this slight abuse of notation.

One of the important properties of a basis is that the values of a function on the elements of a basis completely determine a linear map. To verify this, fix a free \(A\)-module \(M\) and let \((x_i)_{i\in I}\) be a basis of \(M\). That is, the function \(e_i: i\mapsto x_i\) between sets induces an \(A\)-module isomorphism \(\varepsilon:F(I)\cong M\). On the other hand, let another \(A\)-module \(N\) be given, and let a family \((y_i)_{i\in I}\) of elements of \(N\) be given; then the functions \(e_i': i\mapsto y_i\) induce \(\varepsilon': F(I) \rightarrow N\). Then there exists a unique \(A\)-linear map \(u:M \rightarrow N\) satisfying \(u(x_i)=y_i\), and explicitly this can be written as \(u=\varepsilon'\circ\varepsilon^{-1}\).

Basis of an Algebra

Now let us examine the basis of an algebra. In the Multilinear Algebra category, although our main objects of interest are \(A\)-modules, when we examine the endomorphism algebra of a fixed \(A\)-module we think of \(A\)-algebras.

Recall that when we discussed \(A\)-algebras, we always assumed that \(A\) is commutative.

Definition 8 For any \(A\)-algebra \(E\), a basis of \(E\) means a basis of \(E\) regarded as an \(A\)-module.

The point to be careful about is that while a basis of \(E\) is minimal for generating \(E\) as an \(A\)-module, a smaller set may suffice for generating \(E\) as an \(A\)-algebra. For example, to generate the polynomial algebra \(A[\x]\) as an \(A\)-module we need the elements \(1,\x,\x^2,\cdots\), but to generate it as an \(A\)-algebra, \(1\) and \(\x\) alone suffice.

Nevertheless, a basis \((e_i)_{i\in I}\) of \(E\) still contains all information about \(E\); in particular, the multiplication defined on \(E\) can be described using bases. For any \(i,j\in I\), since \(e_ie_j\) is also an element of \(E\), it can be expressed as the following sum:

\[e_ie_j=\sum_{k\in I} \gamma_{ij}^k e_k\]

Here, for each fixed \(i,j\), there are only finitely many \(k\) with \(\gamma_{ij}^k\neq 0\).

Definition 9 The family \((\gamma_{ij}^k)_{i,j,k\in I}\) defined above is called the structure constants of \(E\).

Then for any \(x,y\in E\), writing them using the basis \((e_i)\) as

\[x=\sum_{i\in I} x_i e_i,\qquad y=\sum_{j\in I} y_j e_j\]

we can write

\[xy=\sum_{i,j\in I} x_i y_j e_ie_j=\sum_{i,j,k\in I} x_i y_j \gamma_{ij}^k e_k\]

Conversely, if a family \((\gamma_{ij}^k)_{i,j,k\in I}\) satisfying the finiteness condition described above is given, we can endow the free \(A\)-module \(E\) with basis \((e_i)_{i\in I}\) with an \(A\)-algebra structure through this.

Moreover, the conditions for the multiplication of \(E\) to satisfy associativity and commutativity can also be expressed using the basis. Expressing \(x,y,z\) using the basis as above and computing \((xy)z\) and \(x(yz)\) respectively, we get

\[(xy)z=\sum_{i,j,k\in I}x_i y_jz_k(e_ie_j)e_k,\qquad x(yz)=\sum_{i,j,k\in I} x_i y_j z_k e_i(e_je_k)\]

so for associativity to hold, it suffices to verify associativity among the elements constituting the basis. For the same reason, for the multiplication of \(E\) to satisfy commutativity, it suffices to verify commutativity among the elements constituting the basis.

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References

[Bou] Bourbaki, N. Algebra I. Elements of Mathematics. Springer. 1998.