Modern Algebra: Groups, Rings, Modules, Fields

Let \(\cP\) denote the collection of all subsets \(X\) of \(V\) such that \(I\subseteq X \subseteq S\) and \(X\) is linearly independent. We make \(\cP\) into a poset by the order relation \(\subseteq\text{,}\) set containment. Note that \(I \in \cP\text{.}\)

Let \(\cT\) be any totally ordered subset of \(\cP\text{.}\) If \(\cT\) is empty, then \(\cT\) is (vacuously) bounded above by \(I\text{.}\) Assume \(\cT\) is non-empty. Let \(Z = \bigcup_{Y \in \cT} Y\text{.}\)

Given \(z_1, \dots, z_m \in Z\text{,}\) for each \(i\) we have \(z_i \in Y_i\) for some \(Y_{i} \in \cT\text{.}\) Since \(\cT\) is totally ordered, one of \(Y_1, \dots, Y_m\) contains all the others and hence it contains all the \(z_i\)’s. Since each \(Y_i\) is linearly independent, this shows \(z_1, \dots, z_m\) are linearly independent. We have shown that every finite subset of \(Z\) is linearly independent, and hence \(Z\) is linearly independent. Since \(\cT\) is non-empty, \(Z \supseteq I\text{.}\) Since each member of \(\cT\) is contained in \(S\text{,}\) \(Z \subseteq S\text{.}\) Thus, \(Z \in \cP\text{,}\) and it is an upper bound for \(\cT\text{.}\) We may thus apply Zorn’s Lemma to conclude that \(\cP\) has at least one maximal element, \(B\text{.}\)

Note that \(B\) is linearly independent and \(I \subseteq B \subseteq S\) by construction. Suppose by way of contradiction that \(B\) does not span \(V\text{.}\) Since \(S\) spans \(V\text{,}\) if \(S \subseteq \Span(B)\text{,}\) then \(\Span(B)\) would have to be all of \(V\text{.}\)

²

For note that if \(\Span(S) = V\) and \(S \subseteq \Span(B)\text{,}\) then for any \(v \in V\) we may write \(v = \sum_i c_i s_i\) for \(s_i \in S\) and \(s_i = \sum_j a_{i,j} b_{i,j}\) with \(b_{i,j} \in B\) and hence \(v= \sum_{i,j} c_i a_{i,j} b_{i,j}\text{,}\) which implies \(\Span(B) = V\text{.}\)

Since we are assuming \(\Span(B) \ne V\text{,}\) there must be at least one \(v \in S\) such that \(v \notin \Span(B)\text{.}\)

Consider \(B \cup \{v\}\text{.}\) Then \(I \subset B \cup \{v\} \subseteq S\) and, by Lemma 13.38, \(B \cup \{v\}\) is linearly independent. This shows that \(B \cup \{v\}\) is an element of \(\cP\) that is strictly bigger than \(B\text{,}\) contrary to the maximality of \(B\text{.}\) So, \(B\) must span \(V\) and hence it is a basis.

Apply Corollary 13.41 with \(B = I\) and \(S = V\text{.}\) (Since \(B\) is a basis of \(W\text{,}\) it is linearly independent, and observe that \(B\) remains linearly independent when regarded as a subset of \(V\text{.}\))

Let \(C = \{w_1, \dots, w_m\}\text{.}\) Lemma 13.38 proves the case when \(m = 1\text{.}\) The general case proceeds recursively:

Suppose that for some \(0 \leq k < m\text{,}\) we have found \(v_1, \dots, v_k \in B\) such that \(B'' := (B \setminus \{v_1, \dots, v_k\}) \cup \{w_1, \dots, w_k\}\) is a basis for some \(k < m\text{.}\) We need to show we can “swap out one more’’; that is, we need to prove there is a \(v_{k+1} \in(B \setminus \{v_1, \dots, v_k\})\) such that \((B'' \setminus \{v_{k+1}\}) \cup \{w_{k+1}\}\) is also a basis.

We will only prove this in the case of finite dimensional vector spaces, but it is indeed true in general.

Suppose \(F\) is a field and \(V\) is a finite dimensional \(F\)-vector space. Then it has a finite basis \(B\text{.}\) Let \(B'\) be any other basis, not necessarily finite. For any non-negative integer \(m\text{,}\) suppose \(C\) is any \(m\)-element subset of \(B'\text{.}\) Then \(C\) is linearly independent and so, by the Exchange Lemma, there is an \(m\)-element subset \(D\) of \(B\) such that \((B \setminus D) \cup C\) is also a basis of \(V\text{.}\) In particular, \(m \leq |B|\text{.}\) Since this holds for all \(m\text{,}\) we conclude \(|B|' \leq |B|\text{.}\) By symmetry, \(|B| \leq |B|'\) and hence \(|B|= |B'|\text{.}\)

By the Classification of Finite Dimensional Vector Spaces, \(V\) and \(V^{\prime}\) are both of the form \(V \cong F^{m}\) and \(V^{\prime} \cong F^{n}\text{,}\) while \(F^{m} \cong F^{n}\) if and only if \(m=n\text{.}\)

Pick a basis \(X\) of \(W\text{.}\) Regarded as a subset of \(V\text{,}\) \(X\) remains linearly independent and thus it may be extended to a basis of \(V\) by Corollary . Let us write this basis of \(V\) as \(X \cup Y\) with \(X \cap Y = \emptyset\text{.}\)

Let \(Z := \{y + W \mid y \in Y\} \subseteq V/W\text{.}\) I claim that \(Z\) is a basis of \(V/W\text{.}\)

Given \(v + W\) we have \(v = \sum_i a_i x_i + \sum_j b_j y_j\) for some \(x_i \in X, y_j \in Y\) and scalars \(a_i, b_j\text{.}\) Since \(x + V = 0\) for all \(x \in X\text{,}\) we have \(v + W = \sum_j b_j z_j\text{.}\) This proves \(Z\) spans. Say \(\sum_j c_j y_j + W = 0\) for some \(y_1, \dots, y_m \in Y\text{.}\) Then \(\sum_j c_j y_j \in W\) and hence \(\sum_j c_j y_j = \sum_i a_i x_i\text{,}\) whence \(\sum_j c_j y_j + \sum_i - a_i x_i = 0\text{.}\) Since \(X \cup Y\) is linearly independent, \(c_j = 0\) and \(a_i = 0\) for all \(j, i\text{.}\) This proves \(Z\) is linearly independent.