$Tensor Product$

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Intro to Tensor Products exercise.

Let $R$ be an integral domain, $F$ its field of fractions and $M$ a left $R$-module.
Look at $F$ as an $F$-$R$-bimodule using the operations of $F$ so that $F?_RM$ has a structure of $F$-vector space.

Show that if $X$ is the generating set of $M$ as $R$-module, then the set $\{1?x|x?X\}$ generates $F?_RM$ as $F$-vector space.

Abstract Algebra

Ssvnormandysr1

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We must show that every element of $F\otimes M$ is an $F$-linear combination of elements of $\{1\otimes x \}$. But elements of $F\otimes M$ are of the form $\sum_{i\le k} a_i\otimes m_i$ for some $a_i \in F$ and $m_i \in M$. So it suffices to show that each $a\otimes m$ is an $F$-linear combination of elements of $\{1\otimes x \}$. But $X$ generates $M$ so we have $m=\sum_{i\le n} r_i x_i$ for some $r_i \in R$ and $x_i \in X$. Now $$ a\otimes m=a\otimes (\sum r_i x_i )= \sum a\otimes r_i x_i =\sum ar_i \otimes x_i =\sum ar_i (1\otimes x_i )$$Note that $ar_i \in F$. So each $a\otimes m$ is an $F$-linear combination of elements of $\{1\otimes x \}$, and thus $$ \sum_{i\le k} a_i\otimes m_i=\sum_{i\le k}\sum _{j\le n_i} a_{ij} (1\otimes x_{ij})$$is also an $F$-linear combination of elements of $\{1\otimes x \}$.

Martin

Ssvnormandysr1

Hey, are the a_i, m_i, r_i, x_i just arbitrary elements? Didn't quite figure that out.
Martin

yes they are arbitrary elements, although they may depend on the other elements introduced before. For example in the expansion of the arbitrary element m the r_i and x_i depend on m

$Tensor Product$

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