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We known that "Every function can represent as a polynomial function" in Finite field .what about in finite ring. when we consider the polynomial $x^2-1\in\mathbb Z_8[x]$ has $4$ roots $(1,3,5,7) $in $\mathbb Z_8$. By fundamental theorem of algebra , degree of polynomial and number of roots of the polynomial must be same(at most). here degree two polynomial have 4 roots. this violates fundamental theorem. which means polynomial$\nRightarrow$ function in Finite ring. is its converse true over finite ring?

Anitha Gandhi
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    If the ring is of finite characteristic then you can express any function as a polynomial – MAS Mar 15 '22 at 06:41
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    @learner Really? If $F$ is an infinite field of characteristic $2$, there are $|F|^{|F|}$ functions, but only $|F|$ polynomials. Finiteness is a necessary condition. – egreg Mar 15 '22 at 08:28
  • @egreg how can we speak finite characteristic to infinite field ? – Anitha Gandhi Mar 15 '22 at 10:54
  • @egreg yeah! if $|F|=n$ there are $n^n $ functions can defined includes constant function. can we treat these constant function as constant polynomial? if yes, how it reduce $|F|$ alone polynomials? – Anitha Gandhi Mar 15 '22 at 11:00
  • @AnithaGandhi Take any finite field $K$ and consider the field of quotients $F$ of $K[X]$. This is an infinite field having the same characteristic as $F$. – egreg Mar 15 '22 at 11:41
  • @egreg, thanks for pointing the error. – MAS Mar 16 '22 at 01:14

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This fails for all $\Bbb{Z}_m$, $m$ not a prime. Consider any polynomial $f(x)\in\Bbb{Z}_4[x]$. We easily see that $f(0)\equiv f(2)\pmod 2$ as well as $f(1)\equiv f(3)\pmod2$. Therefore it is impossible to get the function $g:\Bbb{Z}_4\to\Bbb{Z}_4, 0\mapsto 1, 1\mapsto0,2\mapsto2, 3\mapsto3$ by evaluating any polynomial.

Jyrki Lahtonen
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  • See 1,2 for more. – Jyrki Lahtonen Mar 15 '22 at 09:36
  • Take $f(x)=x^3+2x+1 \in \mathbb{Z}_4[x]$. Then $f(1)=4$ and $f(3)=34$. So $f(1) \not\equiv f(3)$ (mod $4$). Do you mean mod $2$ ? – MAS Mar 16 '22 at 06:15
  • Oops. Typo in the second congruence. Thanks @learner. – Jyrki Lahtonen Mar 16 '22 at 07:23
  • For my above example, $f(0)=1$ and $f(1)=0$ in $\mathbb{Z}_4$. But I didn't understand the maps $2 \mapsto 2$ and $3 \mapsto 3$. Can you please explain it ? – MAS Mar 16 '22 at 08:33
  • If $f(x)\in\Bbb{Z}_4[x]$, then automatically $f(0)\equiv f(2)\pmod2$. So any function $g$ such that $g(0)=1, g(2)=2$ cannot be gotten by evaluating a polynomial. I decided to make the example a permutation, but that is not necessary. $g0)=1$m $g(k)=0$ for all $k\in\Bbb{Z}_4,k\neq0$ is also impossible for a polynomial $g$. – Jyrki Lahtonen Mar 16 '22 at 08:38