Recent content by mobe

There are other factors: a suitable advisor, a right area,...but in my opinion, the most important factor is hard work.

Thanks for all the notes mathmonk! They are quite useful.

Hi Mathwonk, I am a first year MA student in Math at a small university. I am interested in studying algebra/algebraic geometry. I have noticed that your research area is algebraic geometry. My question for you is: what are some good universities to study algebraic geometry? I have looked at...

Also, is there an age limit for postdocs?

Same here :)

Is it true that the F[\lambda] module determined by a linear transformation T is cyclic iff the characteristic polynomial of T equals the minimum polynomial of T?

I can see now how to do the first one, the other direction is proved using the fact that the intersection of N and K is trivial. Thanks! For part (2): how do I show the existence of K? How does the assumption that D is a P.I.D change the problem?

I could not fix my post and so I posted it again. Can someone delete one post for me? Thanks!

Homework Statement Suppose M is a D_module and N is a submodule. N is called pure iff for any y \in N and a \in D ax = y is solvable in N iff it is solvable in M. N is a direct summand of M iff there is a submodule K with M = N \oplus K. Prove: (1) If N is a direct summand, then N is pure. (2)...

Homework Statement Suppose M is a D_module and N is a submodule. N is called pure iff for any y \in N and a \in D ax = y is solvable in N iff it is solvable in M. N is a direct summand of M iff there is a submodule K with M = N \oplus K. Prove: (1) If N is a direct summand, then N is pure. (2)...

Just ordered :)

I think g:\mathbb{R}^2 \rightarrow \mathbb{R} \text{ and } Jg \in M_{1\times 2} (\mathbb{R} ) is right. I will use your suggestion and see where it gets me. Thanks! ( I may post more question).

Just to move it down here.

Yes, I believe that it is the dot product.

Thanks for the reply! I got up to g = x - f'(x)Click to see the LaTeX code for this image.f(x) in term of J_f and J_f^-1. I want to compute J_g but continue doing the same thing ( taking partial derivatives) would make J_g look awfully ugly. I am curious if there is another way to get J_g? I got...