# BenjaLim

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 May11 comment How does the differential $df$ act on an element of $T_pM$?@Hurkyl Is it true that the space of all derivations at a point $p$ on a manifold $M$ has basis $\frac{\partial}{\partial x^i}|_p$, $1\leq i \leq n$? May11 comment How does the differential $df$ act on an element of $T_pM$?@Hurkyl Thanks. May11 revised How does the differential $df$ act on an element of $T_pM$?added 35 characters in body May11 comment How does the differential $df$ act on an element of $T_pM$?Dear Kofi, your $X^i$ are real numbers yes? May11 comment How does the differential $df$ act on an element of $T_pM$?Sorry Mariano, I realised that what was more meaningful to ask is the edited version of my question. I'm sorry; could you help me with this edited version? Thanks. May11 revised How does the differential $df$ act on an element of $T_pM$?added 59 characters in body May11 comment How does the differential $df$ act on an element of $T_pM$?Thanks for your answer. I forgot that the differential can also be thought of a kind of map between tangent spaces. However if now I think of $v$ as a vector on $S^2$, how do I compute this? For example, if $p$ is the south pole, then I can consider the vector $(1,1,0)$ based at $p$. How can I know what $df(v)$ is? May11 asked How does the differential $df$ act on an element of $T_pM$? May10 revised The operator in Tensor algebra.deleted 1 characters in body May8 comment Proving that $\mathbb{Z}[x]$ and $\mathbb{Q}[x]$ are not isomorphic.Biggest thermonuclear weapon.... May8 awarded Caucus May8 comment Proving that $\mathbb{Z}[x]$ and $\mathbb{Q}[x]$ are not isomorphic.@MartinBrandenburg It's ok don't worry. May6 comment Proving that $\mathbb{Z}[x]$ and $\mathbb{Q}[x]$ are not isomorphic.@KCd I have edited my answer. May6 revised Proving that $\mathbb{Z}[x]$ and $\mathbb{Q}[x]$ are not isomorphic.added 138 characters in body May6 answered Proving that $\mathbb{Z}[x]$ and $\mathbb{Q}[x]$ are not isomorphic. May5 revised Exercise 2.4 Fulton's Algebraic Curvesdeleted 12 characters in body May5 comment Exercise 2.4 Fulton's Algebraic Curves@Sebastian Dear Sebastian, your claim is certainly true. In fact it's true not just for affine varieties but for more general algebraic subsets of $\Bbb{A}^n$. May5 comment Exercise 2.4 Fulton's Algebraic Curves@Sebastian Dear Sebastian, no problems! I should say in your question above you are spot on in saying that $\Bbb{C}[x]/((x-1)(x-2))$ is finite dimensional over $\Bbb{C}$: By the Chinese remainder theorem, your ring is isomorphic to $\Bbb{C}[x]/(x-1) \times \Bbb{C}[x]/(x-2) \cong \Bbb{C} \times \Bbb{C}$ that is two dimensional over $\Bbb{C}$. Also you are right when you write that $I(\{1,2\}) = (x-1)(x-2)$. We have that $I(\{1,2\}) = I(\{1\}) \cap I(\{2\}) = (x-1) \cap (x-2) = (x-1)(x-2)$. The last equality comes from the fact that $(x-1)$ and $(x-2)$ are coprime. Regards, May5 revised Exercise 2.4 Fulton's Algebraic Curvesadded 508 characters in body May5 answered Exercise 2.4 Fulton's Algebraic Curves