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Let $R$ be a commutative ring which is a vector space over some field $F$ ; for $r \in R$ consider the function $f:R \to R$ defined as $f(x)=rx , \forall x \in R$ , is this function a vector space homomorphism i.e. is it a linear map ? I can easily verify the additivity but having trouble with the homogeneity ; I have calculated for $\alpha \in F$ , $f(\alpha . x)= r(\alpha .x)=(\alpha . x)r$ and $\alpha . f(x)=\alpha .(rx)=\alpha . (xr)$ , but cannot see how they are equal . Please help .

  • ring multiplication is associative. – MooS Mar 15 '15 at 12:28
  • @MooS : that is completely ok , but how can I say $\alpha . (xr)=(\alpha . x)r$ ? first of all $\alpha $ I don't know whether is in $R$ or not and second of all '$.$' , the external scalar product , is different from ring multiplication . Am I making any mistakes ? –  Mar 15 '15 at 12:31
  • So you do not require $R$ to be a $F$-Algebra? – MooS Mar 15 '15 at 12:42
  • @MooS : actually I was having trouble in understanding the second answer to this question http://math.stackexchange.com/questions/63950/proof-that-an-integral-domain-that-is-a-finite-dimensional-f-vector-space-is-i .... this is my primary query –  Mar 15 '15 at 12:45
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    In this case we have $1 \in R$ since $R$ is an integral domain and we get $F \subset R$ via $F \cong F \cdot 1 \subset R$. So the scalar multiplication $F \times R \to R$ is just the restriction of the ring multiplication $R \times R \to R$. – MooS Mar 15 '15 at 12:48
  • @MooS : How is $F.1 \subset R$ and what is this "$.$" here ? –  Mar 15 '15 at 12:52
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    $F.1$ denotes the $F$-subspace spanned by the vector $1$. – MooS Mar 15 '15 at 13:02
  • @MooS: So , we define $\alpha . x := (\alpha .1)x$ ? –  Mar 15 '15 at 13:06
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    Yes, exactly this. – MooS Mar 15 '15 at 13:07
  • @MooS : THANKS a lot , that truly helped , thank you . –  Mar 15 '15 at 13:08
  • @MooS Please consider converting your comments into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. – Julian Kuelshammer Apr 12 '15 at 20:31

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