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 1d answered Question about field extension notation 2d comment Proj of some ring. This should have dimension 10, since the ring has dimension 10 if we invert the variables. That said, I believe this is just $\mathbb{P}^{10}$. We have an embedding $\mathbb{P}^{10}\to \mathbb{P}^{2^5}$ given by taking the generators of the subalgebra, and this is an isomorphism onto the image. The proof that this corrosponds to the desired subset follows from an aurgument similar to that of the Veronese mapping. Apr15 revised If $P(A) < P(A \cup B)$, does that mean that $A\subsetneq (A\cup B)$? deleted 111 characters in body Apr15 comment If $P(A) < P(A \cup B)$, does that mean that $A\subsetneq (A\cup B)$? Ah yes, I did. It should be fixed now. Thank you. Apr15 revised If $P(A) < P(A \cup B)$, does that mean that $A\subsetneq (A\cup B)$? added 66 characters in body Apr15 revised If $P(A) < P(A \cup B)$, does that mean that $A\subsetneq (A\cup B)$? added 6 characters in body Apr15 answered Representing Matrix Subgroups Apr15 answered If $P(A) < P(A \cup B)$, does that mean that $A\subsetneq (A\cup B)$? Apr13 comment Necessary Conditions for Saddle Value point What your looking for is an understanding of he Morse Lemma. It states, in two dimensions, that if we have a point with zero gradient and nonzero Hessian, that the function is asymmtopically of the form $\pm x^2\pm y^2$. We have also, that the Hessian being zero implies that we can write it in the form $x^2-y^2$, which is what you want. A good refrence is Milnor's Book on Morse theory. Feb9 comment how to compute the Euler characters of a Grassmannian? Unless I am mistaken, for $\chi(Gr(1, 2m+1))=\chi(\mathbb{R}P^{2m+1})=0$ despite your claim. I think the problem is that the compact manifolds are not orientable, which is needed to apply the theorem. But the general result is obtained easily from apply the same argument to the orientable double cover (which is just oriented $n$-dimensional subspaces). Nov4 comment Prove isomorphism for a commutative ring Well observe that we have a map $\phi:Hom_{R}(R, M)\to M$ given by $\phi(\psi)=\psi(1)$. Nov4 comment The index of a smooth vector field is well-defined Well, what is your definition of the index of a vector field? Nov2 awarded Yearling Aug10 answered Is the sphere $S^n$ always arcwise connected? Jul5 comment Axioms of associative algebra? Oh, also, lay off lang, get Artin, it's far better. Jul5 comment Axioms of associative algebra? No, and in general it is not. It is best to think of such a structure as a vecotorspace (and if your more inclined, a module) that just happens to have a notion of multiplication. Then the first axiom just expresses that the actions respect each other. An example you could look at are of-course function algebras, fields over their characteristic, or one of the classics, the quaternions (which are not commutative!). Algebras are actually very nice, for example, Commutative Rings are just $Z$-algebras (you can define algebras independent of rings, so this is actually meaning-ful). Hope this help Jul2 awarded Curious Jul1 comment Holomorphically simply connected implies simply connected The idea would be that if the set is not simply connected, then there is a loop that cannot be contracted. Not being contractable means there is a hole in the inside of the loop, so that we can use the standard $1/z$ trick to get a holomorphic (on our set) function that has nonzero integral, and thus isn't simply connected. Notably, this is a proof by contradiction. I don't know of a constructive proof or method, and I would be surprised if one existed. Jul1 comment How to solve this simple linear equation Felt like this was too short for an answer. We have that $2x+(5+3)=2x+8=46+3x$ so that subtracting $2x$, $2x+8-2x=46+3x-2x$ which is the same as $2x-2x+8=(3-2)x+46$ or $8=x+46$ or thus subtracting $46$, $x=-38$. Jul1 comment Is set of even integers an integral domain? I would say no, but then again, that is just a matter of taste and what you tend to use rings for.