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Assume $m_i$ are maximal ideals in a ring $R$.

Then I have $m_1 \cdot \dots m_{k}$ is an ideal in $m_1 \cdot \dots m_{k-1}$ hence I can quotient to get a factor ring $m_1 \cdot \dots m_{k-1} / m_1 \cdot \dots m_{k}$.

This ring is of course also an abelian group. Now I'd like to turn this into a vector space. How do I see that the field of scalars I am looking for is $R/m_k$?

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I'm not sure what you mean by "the" field of scalars you are looking for; in general, you can define a vector space structure over many fields; if $R/m_k$ has subfields (e.g., if it is isomorphic to $\mathbb{R}$), then anything which has a natural vector space structure over $R$ also has a natural vector space structure over any subfield of $R/m_k$. –  Arturo Magidin Jul 22 '12 at 20:00
    
@ArturoMagidin Thank you. I meant: how do I see that I get a valid field of scalars if I choose $R/m_k$? –  Rudy the Reindeer Jul 22 '12 at 20:02
    
Also, your quotient of ideals is not a ring, most likely. –  Dylan Moreland Jul 22 '12 at 20:06
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Do ideals have identity elements, for example? I guess it depends on what you mean by "ring". –  Dylan Moreland Jul 22 '12 at 20:09
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@ClarkKent: For "$R$" read "$R/m_k$". If $F$ is a field, and $V$ has a "natural" (whatever that means) vector space structure over $F$, then there is a natural vector space structure for $V$ over any subfield of $F$ by restriction. –  Arturo Magidin Jul 22 '12 at 20:37

2 Answers 2

up vote 2 down vote accepted

If $M$ is an $R$-module and $I$ is an ideal of $R$, then $IM$ is a submodule of $M$, and there is a natural $R/I$ module structure on $M/IM$: namely, given $r+I\in R/I$ and $m+IM$ in $M/IM$, define $(r+I)(m+IM) = rm+IM$.

This is well-defined: if $r-s\in I$ then $(r-s)m\in IM$, so $rm+IM = sm+IM$. And if $n-m\in IM$, then we can express $n-m$ as a sum $$n-m = a_1m_1+\cdots +a_km_k,\qquad a_i\in I, m_j\in M$$ so $$r(n-m) = ra_1m_1+\cdots + ra_km_k\in IM$$ since $ra_i\in I$ for all $i$.

Here, you have $M=m_1\cdots m_{k-1}$, $I=m_k$, so there is a natural $R/m_k$ module structure on $M/IM = m_1\cdots m_{k-1}/m_1\cdots m_k$. Since $m_k$ is maximal, $R/m_k$ is a field, so this is actually a vector space structure. So this is a natural field over which to give $M/IM$ a vector space structure, but by no means the only one.

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The way I like to think about this is that an $R$-module is an abelian group $M$ together with a ring homomorphism $R \to \operatorname{End}(M)$. The kernel of this map is called the annihilator of $M$, and from this perspective it's easy to see that $M$ is an $R/I$-module for any ideal $I$ contained in the annihilator.

Another way of viewing your quotient is that you took the $R$-module $\mathfrak m_1 \cdots \mathfrak m_{k-1}$ and tensored with $R/\mathfrak m_k$. Remember that in general for an $R$-module $M$ and $R$-algebra $S$ the base change $M \otimes_R S$ is an $S$-module, and that for an ideal $I$ of $R$, $M \otimes_R (R/I) \simeq M/(IM)$ by right exactness of the tensor product.

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Regarding your first paragraph: I just verified that the action of $R/I$ on an $R$-module $M$ is well-defined if and only if $I$ is contained in the annihilator of $M$. Now I asked myself a related question: Is $R$ an $R/I$-module for an arbitrary ideal $I$? And I think the answer is yes. Is the following correct: –  Rudy the Reindeer Jul 27 '12 at 8:00
    
(i) We have an inclusion $R/I \hookrightarrow R$ hence $R/I$ is a subring of $R$ and hence closed with respect to addition. –  Rudy the Reindeer Jul 27 '12 at 8:01
    
(ii) $R/I$ is closed with respect to multiplication by $R$ and is hence an ideal of $R$. Hence it is an $R/I$-module. –  Rudy the Reindeer Jul 27 '12 at 8:01

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