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grimster

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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 $K[X_{1},...,X_{n}]/I\thickapprox map(n,K).$