- #1
Kreizhn
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Homework Statement
Let R be a commutative ring, and let [itex] F = R^{\oplus B} [/itex] be a free R-module over R. Let m be a maximal ideal of R and take [itex] k = R/m [/itex] to be the quotient field. Show that [itex] F/mF \cong k^{\oplus B} [/itex] as k-vector spaces.
The Attempt at a Solution
If we remove the F and k notations, we essentially just want to show that
[tex] R^{\oplus B}/ m R^{\oplus B} \cong (R/m)^{\oplus B} [/tex]
and so it seems like we should use the first isomorphism theorem.
Now we define [itex] R^{\oplus B} = \{ \alpha: B \to R, \alpha(x) = 0 \text{ cofinitely in } B \} [/itex]. If [itex] \pi : R \to R/m [/itex] is the natural projection map, define [itex] \phi: R^{\oplus B} \to (R/m)^{\oplus B} [/itex] by sending [itex] \alpha \mapsto \pi \circ \alpha [/itex]. This is an R-mod homomorphism, and is surjective by the surjectivity of the projection. Hence we need only show that the kernel of this map is given by [itex] mR^{\oplus B} [/itex]
Now
[tex]\begin{align*}
\ker\phi &= \{ \alpha:B \to R, \forall x \in B \quad \pi(\alpha(x)) = 0_k \} \\
&= \{ \alpha:B \to R, \forall x \in B, \alpha(x) \in m \} \end{align*}
[/tex]
where I may have skipped a few steps in this derivation, but I think this is right. Now it's easy to show that [itex] mF \subseteq \ker \phi [/itex] since m is an ideal. However, the other inclusion is where I'm having trouble.
I guess maybe the whole question could be rephrased to avoid the baggage that comes with the question. Namely, if [itex] \alpha: B \to R [/itex] is such that [itex] \forall x \in B, \alpha(x) [/itex] lies in a maximal ideal of R, should that [itex] \alpha = r \beta [/itex] for some [itex] \beta: B \to R [/itex].
Edit: I guess I'm hoping to show they're isomorphic as R-modules, and then push that over to k-vector spaces. Maybe this is where my mistake is coming from?
Edit 2: Fixed mistake made in "without baggage" statement.
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