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Hi Darij.

$Ind_H^G X = R[G] \otimes_{R[H]} X$ and hence there are two canonical maps:

$R[G]\otimes V \otimes W \to R[G] \otimes V\otimes R[G] \otimes W, x\otimes v\otimes w\mapsto x\otimes v\otimes 1 \otimes w$ and

$R[G]\otimes V \otimes R[G] \otimes W \to R[G] \otimes V \otimes W, x\otimes v\otimes y \times otimes w\mapsto xy \otimes v \otimes w$.

Obviously the second is a right inverse to the first. Hence the first one is injective and the second is surjective. If $R$ is a field, then there cannot be injective (surjective) maps in the other direction because the dimensions don't agree.

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Hi Darij.

$Ind_H^G X = R[G] \otimes_{R[H]} X$ and hence there are two canonical maps:

$R[G]\otimes V \otimes W \to R[G] \otimes V\otimes R[G] \otimes W, x\otimes v\otimes w\mapsto x\otimes v\otimes 1 \otimes w$ and

$R[G]\otimes V \otimes R[G] \otimes W \to R[G] \otimes V \otimes W, x\otimes v\otimes y \times w\mapsto xy \otimes v \otimes w$.

Obviously the second is a right inverse to the first. Hence the first one is injective and the second is surjective. If $R$ is a field, then there cannot be injective (surjective) maps in the other direction because the dimensions don't agree.