2
$\begingroup$

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.

$\endgroup$
3
$\begingroup$
  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.

$\endgroup$
  • 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
  • 2
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.