Linear transformation, show surjection and ker=0.

[grimster]

I have a linear map from $ V\rightarrow K[X_{1},...,X_{n}]\rightarrow K[X_{1},...,X_{n}]/I.$

how do i prove that a linear map from $ V=\{$polynomials with $\deg _{x_{i}}f\prec q\}$ to $ K[X_{1},..X_{n}]/I.$ where I is the ideal generated by the elements $ X_{i}^{q}-X_{i},1\leq i\leq n.,$ is both surjective and that the kernel is zero. V is a vector space over K. Have $ dim_{k}V=\{$the number of different monomials\}= $ q^{n}.$ and $ \mid V\mid =q^{q^{n}}.$ K is a field with q elements.

 

[matt grime]

An element in the quotient is zero if and only if what? Do any of the elements equivalent to zero lie in the image of the natural inclusion? Given any element in the quotient, can you think of some element in the preimage that maps to it?

That is given some element f in the ring k[x_1,..,X_n] can you think of a polynomial g with degree less than q such that f and g are equivalent in the qoutient?

 

[grimster]

"An element in the quotient is zero if and only if what?"
-i'm not sure what you mean. the zero map? f(a)=0 for all a?

"Do any of the elements equivalent to zero lie in the image of the natural inclusion?"
-what do you mean by natural inclusion? i don't think I've heard that expression before.


"preimage"
-is that the same as inverse image?

"That is given some element f in the ring k[x_1,..,X_n] can you think of a polynomial g with degree less than q such that f and g are equivalent in the qoutient?"
-how does this help show that the linear map from $ V=\{$polynomials with $\deg _{x_{i}}f\prec q\}$ to $ K[X_{1},..X_{n}]/I.$ where I is the ideal generated by the elements $ X_{i}^{q}-X_{i},1\leq i\leq n.,$ is both surjective and that the kernel is zero.

 

[matt grime]

I think you need to go back to basics.

Let R be the polynomial ring, and let R/I be the quotient.

You're trying to show the map V to R to R/I is an isomorphism.

So show that the map to R/I is injective. It is injective iff the only element mapping to 0 in the quotient is 0 in V. To do this you need to describe the elements in R that map to 0 in the qoutient. Do it. Now, does the map V to R send any element other than 0 to something tha maps to zero? If you followed my last post you'd know the answer.

Now, to show surjectivity you need to show that everything in the quotient has some preimage in V, ie some element in the equivlance class in the quotient is in the image of the map V to R. So do it: hint polynomial division.


If it helps why not consider special case.

n=1, p=2.


A basis of R/I is 1,x, since ever polynomial in k[x] is equal to one of the form

ax+b+ P(x)(x^2-x)

by the polynomial divisoin algorithm,

thus the only things that get sent to 0 are multiples of x^2 - x.

so obviously V here is isomorphic to R/I

 

[grimster]

from R to R/I the only elements that are mapped to 0, are the ones with <Xi^q -Xi> as a factor? or the ideal I? is that right?

 

[grimster]

edit...

 

[grimster]

I think you need to go back to basics.

Let R be the polynomial ring, and let R/I be the quotient.

You're trying to show the map V to R to R/I is an isomorphism.

So show that the map to R/I is injective. It is injective iff the only element mapping to 0 in the quotient is 0 in V. To do this you need to describe the elements in R that map to 0 in the qoutient. Do it. Now, does the map V to R send any element other than 0 to something tha maps to zero? If you followed my last post you'd know the answer.

Now, to show surjectivity you need to show that everything in the quotient has some preimage in V, ie some element in the equivlance class in the quotient is in the image of the map V to R. So do it: hint polynomial division.


If it helps why not consider special case.

n=1, p=2.


A basis of R/I is 1,x, since ever polynomial in k[x] is equal to one of the form

ax+b+ P(x)(x^2-x)

by the polynomial divisoin algorithm,

thus the only things that get sent to 0 are multiples of x^2 - x.

so obviously V here is isomorphic to R/I

ok, this is what i have so far.

$\ker \left( \varphi :V\longrightarrow k[X_{1},...,X_{n}]\right) =0$
per definition since V is defined as a subset of $ k[X_{1},...,X_{n}]$.

then we have:
$\ker \left( \psi :V\longrightarrow k[X_{1},...,X_{n}]/I\right) =\left\{ x\in V:\psi (x)=0\right\} =\left\{ x\in V:\varphi (x)\in I\right\} =0.$

so that proves that the kernel is trivial. then all i have to do is show that it is surjective.

 

[matt grime]

It now suffices to count dimension, or as I said eariler pick a nice basis of the quotient.

 

[grimster]

so the "kernel thing" is correct? all i have to show now is surjection?

 

[grimster]

It now suffices to count dimension, or as I said eariler pick a nice basis of the quotient.

ok, I've tried to do this, but i can't figure it out.

this was what i tried:



let $ f(x_{1},...,x_{n})$ be a polynomial in R

by euclidian division we get

$ f(x_{1}..x_{n})=Q_{1}(x_{1},,x_{n})\ast (x_{1}^{q}-x_{1})+R_{1}(x_{1},,x_{n})$

$ Q_{1}$ and $ R_{1}$ r in R and deg $ R_{1}$ in $ x_{1}&lt;q$


in R/I, P and R$ _{1}$ have the same class

divide now $ R_{1}$ by $ x_{2}^{q}-x_{2}$ ...


at the end u get $ R_{n}$ in R with degree \TEXTsymbol{<}q in $ x_{1}...x_{n}$

P and $ R_{n}$ have the same class