Can Polynomial Maps and Ideals Demonstrate a Ring Isomorphism in Finite Fields?

[grimster]

ok, I've pasting some of the stuff I've done in scientific workplace 3.0. should be easier to read than in plain text. hope some of you can help me... just ask if there is something you don't get.


I am supposed to prove that $Map(n,K)\thickapprox K[X_{1},..X_{n}]/I.$ where I is the ideal generated by the elements $X_{i}^{q}-X_{i},1\leq i\leq n.$ Map(n,K) is the ring of polynomial mappings. here is a link that explains what a polynomial map is:

http://mathworld.wolfram.com/PolynomialMap.html

K is a field of q elements.

I start with this:

have a homomorphism $\phi :(K[X_{1},..X_{n}])\rightarrow GMAP(K^{n},K)$ where GMAP is the group of all mappings from $K^{n}\rightarrow K.$

this is an evaluation homomorphism. $\phi (f):K^{n}\rightarrow K$ and $\phi (f)(a_{1},...a_{n}):=f(a_{1},...,a_{n}).$ $I=\ker \phi .J=<X_{i}^{q}-X_{i}>.$

here is my plan. i want to use the first isomorphism theorem.

if we have a homomorphism $f:G\rightarrow H,$ then we have that $G/\ker f\thickapprox \func{Im}f.$ In my case G is $K[X_{1},..X_{n}]$ and H is $GMap(K^{n},K)$

to do this i have to prove first that I=J. This can be done by showing that $I\subseteq J$ and that $J\subseteq I.$ i must also prove that $\func{Im}f=Map(K^{n},K)=K[X]/I.$

we have that every X$^{q}$ can be replaced by X. given a polynomial $f=\sum_{i_{1},...,i_{n}}a_{i_{n},...,i_{n}}X_{1}^{i_{1}}X_{2}^{i_{2}}...X_{n}^{i_{n}}. $ What i want to do is reduce all parts, so that all exponents are $\leq q.$


given $f=\sum_{i_{1},...,i_{n}}a_{i_{1}},...,_{i_{n}}X_{1}^{i_{1}}X_{2}^{i_{2}}...X_{n}^{i_{n}}\in I. $ if i.e. i$_{1}\geq q,$ then i would i have $f^{|}=f-(X_{1}^{q}-X_{i})\cdot a_{i_{1}},...,_{i_{n}}X_{1}^{i-q}X_{2}^{i_{2}}...X_{n}^{i_{n}}.$ This is done for all i, so that every exponent is $\leq q.$ Then $f^{|}\in I.$ An element in both I and J. The monomial $a_{i_{1}},...,_{i_{n}}X_{1}^{i_{1}}X_{2}^{i_{2}}...X_{n}^{i_{n}}$ ''becomes'' $a_{i_{1}},...,_{i_{n}}X_{1}^{i_{1}-q+1}X_{2}^{i_{2}}...X_{n}^{i_{n}}.$

so $f^{|}\in I.f-f^{|}\in J.$ $\deg f^{|}\prec ($less than) $q.$ then a corrollary from lang(p.177 c.1.8) says:let k be a finite field with q elements. let f be a polynomial in n variables over k such that the degree of f in each variable is less than q. if f induces the zero function on k$^{(n)}$ then f=0.

so considering this, $f\in J.$ Does all of this make any sense or am i waaaay off here? how do i show the oter way around? $J\subseteq I?$

i know it's a lot to read, but bare with me here:

so when one has shown that I=J, i must prove the other part.

$V=\{$polynomials with $\deg x_{i}f\prec q\}.$ a vector space over K. $dim_{k}V=\{$the number of different monomials\}= q$^{n}.$ then $|V|=q^{q^{n}} $.

we have linear mappings $V\rightarrow K[X_{1},...,X_{n}]\rightarrow K[]/I.$ So if i show that ker = 0 and that this is surjective(from V to K[]/I) then we have an isomorphism from V to K[]/I. $\func{Im}\phi =Map.$ So then from the isomorphism theorem, G/$\ker f\thickapprox \func{Im}f.$ That should give $K[X_{1},...,X_{n}]/I\thickapprox map(n,K).$

 

[Hurkyl]

You can use these tags to generate LaTex: [ itex ] ... [ /itex ]. (Without the excess spaces)

For example, the first LaTeX expression in your post becomes Map(n,K)\thickapprox K[X_{1},..X_{n}]/I.

 

[M98Ranger]

Firstly, I would like to commend you for the thorough and detailed work you have done so far. It is clear that you have a good understanding of the concepts and have put in a lot of effort to prove this isomorphism. However, there are a few areas where I believe you may have gone off track or could improve upon.

Firstly, in your plan to use the first isomorphism theorem, you mention that you need to prove that $I=J$ in order to apply the theorem. However, this is not entirely necessary. The first isomorphism theorem states that if we have a homomorphism $f:G\rightarrow H,$ then $G/\ker f\thickapprox \func{Im}f.$ In your case, you have already defined the homomorphism $\phi: K[X_{1},...,X_{n}]\rightarrow \func{Map}(K^{n},K)$ and have shown that $I=\ker \phi.$ Therefore, you can directly apply the first isomorphism theorem without needing to prove that $I=J.$

Additionally, in your proof that $I\subseteq J,$ you have shown that any polynomial $f\in I$ can be written as $f^{|}\in I.$ However, this is not sufficient to prove that $I\subseteq J.$ In order to show this, you must also show that $f^{|}$ can be written as a linear combination of elements in $J.$ This is necessary because $J$ is defined as the ideal generated by the elements $X_{i}^{q}-X_{i},$ so any element in $J$ must be a linear combination of these elements. Similarly, in your proof that $J\subseteq I,$ you need to show that any element in $J$ can be written as a linear combination of elements in $I.$ This will require some more work, but it is possible to show that $J\subseteq I$ in a similar manner to how you showed $I\subseteq J.$

Finally, in your proof that $\func{Im}\phi= \func{Map}(K^{n},K),$ you mention that you have linear mappings $V\rightarrow K[X_{1},...,X_{n}]\rightarrow K[]/I.$ However, it is not clear how you have defined these linear mappings. It seems that you are trying to use the evaluation homomorphism