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Let $I = \left < f_1, \dots, f_n \right > \subset R$ be an ideal generated by homogeneous elements where $\deg(f_i) = d_i$ and $\phi$ be the graded $R$-mod homomorphism $$\phi: R(-d_1) \oplus \cdots \oplus R(-d_s) \to I$$ where $e_i \to f_i$. So it maps generators to $f_i$. If one treats $g = g_1e_1 + \dots + g_ne_n \in R(-d_1) \oplus \cdots \oplus R(-d_s) $ as a column vector $g = \begin{bmatrix} g_1\\ \vdots\\ g_n \end{bmatrix}$ with $g_i \in R(-d_i)$, then why can we view $\phi$ as $$\phi(g) = \begin{bmatrix} f_1 & \cdots &f_n \end{bmatrix} \begin{bmatrix} g_1\\ \vdots\\ g_n\end{bmatrix}.$$

I am not sure what it is asking. The map $\phi$ is a homomorphism, and it maps from vector space to vector space and is a linear map so it has a matrix representation. Is it just asking us to expand the matrix operation on the RHS of the given map ($f$ with $g$?)

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  • $\begingroup$ It seems that a lot of information about $R$ is missing from this question. What are the $e_i$ for example? $\endgroup$ – Tobias Kildetoft Mar 29 '16 at 7:47
  • $\begingroup$ @TobiasKildetoft, $e_i$ are the generators of $R(-d_i)$ and $R$ is a graded $R-$Module $\endgroup$ – jacob smith Mar 29 '16 at 7:48
  • $\begingroup$ What do you mean by "the" generators? $\endgroup$ – Tobias Kildetoft Mar 29 '16 at 7:49
  • $\begingroup$ I just mean for notation. $\endgroup$ – jacob smith Mar 29 '16 at 7:55
  • $\begingroup$ That still does not make any sense to me. An $R$-module does not have any distinguished generator usually, so you need to specify one. $\endgroup$ – Tobias Kildetoft Mar 29 '16 at 7:57
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You're correct, there's basically nothing to check here.

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  • $\begingroup$ This looks better as a comment. $\endgroup$ – user26857 Mar 29 '16 at 8:37
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    $\begingroup$ I've never really liked the idea of answering a question in a comment for stylistic reasons, because then the system doesn't know that the question has been answered. $\endgroup$ – Daniel McLaury Mar 29 '16 at 16:02

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