Exercise 3.1.4 (Polynomials and functions)

Question

Show that whenever $R$ is a finite nontrivial ring, it is possible to find distinct polynomials over $R$ that induce the same function $R 	o R$. (Hint: are there finitely or infinitely many polynomials over $R$? Functions $R 	o R$?)

Richard Ganaye · Accepted Answer

\begin{proof} Since $R$ is not trivial, $0 
e 1$. Therefore the polynomials
$$1 = (1,0,0,\ldots), \quad t = (0,1,0,0,\ldots),\quad  t^3 =(0,0,1,0,\ldots),\ldots,$$
are distinct, so there are infinitely formal polynomials over $R$.

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But le set $R^R$ of all functions $u : R 	o R$ is finite, with cardinality $|R|^{|R|}$. A fortiori, the cardinality of the set $\mathscr{P}(R) \subset R^R$ of polynomial functions is finite.

Therefore the ring homomorphism
$$
\chi 
\left\{
\begin{array}{ccl}
R[t] & \mapsto & \mathscr{P}(R) \
p = \sum\limits_{i=0}^d a_i t^i & \mapsto &
\chi(p)
\left\{
\begin{array}{ccl}
R & 	o & R\
x & \mapsto & p(x) = \sum\limits_{i=0}^d a_i x^i
\end{array}
ight.
\end{array}
ight.
$$
cannot be injective (one to one).

This means that we can find distinct polynomials over $R$ that induce the same function $R 	o R$.
\end{proof}

Exercise 3.1.4 (Polynomials and functions)

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Exercise 3.1.4 (Polynomials and functions)

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