## proving another isomorphism

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.

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

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 Recognitions: Gold Member Science Advisor Staff Emeritus You can use these tags to generate LaTex: [ itex ] ... [ /itex ]. (Without the excess spaces) For example, the first LaTeX expression in your post becomes $Map(n,K)\thickapprox K[X_{1},..X_{n}]/I.$