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>> No.11169688 [View]
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11169688

What multiplication?????

Also, if R^r is isomorphic to the reals, isn't Hom(R^{r-q},R^q) extremely simple?

like won't the only homomorphism between them have to send every basis element of R^{r-q} to 1, and the other homomorphism is sending them all to zero?

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