Theorem. Let $R$ be a commutative ring with $1$. Any two finite bases of a free $R$-module $M$ have the same number of elements.

Proof. Let $M$ be a free module with a basis $\{x_1, x_2,\ldots, x_n\}$ and $I$ be a maximal ideal of $R$. Then $R/I = F$ is field. Since $V = M/IM$ is annihilated by $I$, $V$ is a vector space over $F$. If $[x_i] = x_i + IM$ $(1 ≤ i ≤ n)$, then $B = \{[x_1], [x_2], . . . , [x_n]\}$ is a basis of $V$ over $F$. Since any two finite bases of a vector space have the same number of elements, the theorem follows.

(See Theorem 9.6.1 of this book for detials)

I was trying to understand the above proof of the theorems but I didn't understand most of it,

  1. Why is $V$ a vector space over $F$ and under which operation?

  2. Why $V$ is annihilated by $I$?

  3. Why $B$ is a basis of $V$ over $F$ and why exactly does the invariance of cardinality of bases of $V$ imply the same for $M$?

Any help is appreciated.

  1. We have $V \cong M \otimes_R R/I = M \otimes_R F$, which is naturally a $F$-vector space. Concretely, if $[r]_I \in F$ (with $r \in R$) and $[m]_{IM} \in V$ (with $m \in M$), then $$[r]_I \cdot [m]_{IM} := [r \cdot m]_{IM}$$ provides a well-defined action of $F$ on $V$.

  2. Seen as an $R$-module, $V$ is annihilated by $I$ : pick $[m]_{IM} \in V$ (with $m \in M$) and $i \in I$. Then $i [m]_{IM} = [im]_{IM} = [0]_{IM}$.

  3. Clearly $B$ generates $V$ over $F$. We check that $B$ is linearly independent over $F$. If $\sum_j [a_j] [x_j] = 0 \in V$ with $[a_j] \in F$, then $$w := \sum_j a_j x_j \in IM.$$ Because the $x_i$'s generate $M$ over $R$, we may write $w = \sum_l i_l x_l$ for some $i_l \in I$ (since $w$ lies in $IM$). What can you conclude, using the fact that the $x_i$'s are linearly independent over $R$? Moreover, if you have another basis $\{y_1, ..., y_m\}$ of $M$ over $R$, then $\{ [y_1], ..., [y]_m \}$ gives, as before, a basis of $V$ over $F$. Therefore we must have $m=n$ since $V$ is a vector space over a field.

  • 1
    $\begingroup$ How do I prove the well-definedness of $[r]_I \cdot [m]_{IM} := [r \cdot m]_{IM}$? $\endgroup$ – user 170039 May 25 '18 at 12:12
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    $\begingroup$ @user170039 : if $r' = r+i$ and $m' = m + \sum_r i_r m_r$ then $$r' m' = (r+i)(m+ \sum_r i_r m_r) = rm + a$$ with $a \in IM$. $\endgroup$ – Watson May 25 '18 at 12:14

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