# Linear transformation, show surjection and ker=0.

• grimster
In summary, the conversation discusses a linear map from a vector space V to a quotient ring $K[X_{1},...,X_{n}]/I$, where I is an ideal generated by elements of the form $X_{i}^{q}-X_{i}$ and V consists of polynomials with degrees less than q in the variables $x_{1},...,x_{n}$. The goal is to prove that this linear map is both surjective and has a trivial kernel. This is done by showing that the elements in the quotient ring that map to zero are only multiples of the ideal I, and by using polynomial division to show that every element in the quotient ring has a preimage in V.
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.

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?

"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.

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

Last edited:
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?

edit...

Last edited:
matt grime said:
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.

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

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

matt grime said:
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 $x_{1}<q$

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

## 1. What is a linear transformation?

A linear transformation is a function that maps one vector space to another in a linear manner. This means that the output of the function is a linear combination of the inputs.

## 2. How can you show that a linear transformation is a surjection?

To show that a linear transformation is a surjection, we need to prove that every element in the output vector space has at least one corresponding input element. This can be done by showing that the rank of the transformation is equal to the dimension of the output vector space.

## 3. What does it mean for ker=0 in a linear transformation?

Ker=0 in a linear transformation means that the only element in the input vector space that maps to the zero vector in the output vector space is the zero vector itself. In other words, the null space of the transformation is only the zero vector.

## 4. How can you show that ker=0 in a linear transformation?

To show that ker=0 in a linear transformation, we need to prove that the only solution to the equation T(x) = 0 is x=0. This can be done by showing that the transformation is injective, meaning that no two input vectors map to the same output vector.

## 5. Can a linear transformation have both ker=0 and be a surjection?

Yes, it is possible for a linear transformation to have both ker=0 and be a surjection. This means that the transformation is both injective and surjective, and is therefore an isomorphism between the input and output vector spaces.

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