relationship between Chinese remainder theorem and roots of polynomials

Question

According to https://eprint.iacr.org/2020/1481.pdf on page 9:

What is $\mathbb{Z}_p[\eta]$? I mean, $\eta := [X \mod F_1(X)]$. Does this mean a polynomial evaluated at $\eta$? If so, which polynomial? Or does it mean a space of polynomials with variable at $\eta$?

Answer 1

If $R$ is a commutative ring, $A$ is an $R$-algebra, and $\alpha$ is an element of $A$, then $R[\alpha]$ is the $R$-subalgebra of $A$ generated by $\alpha$. One can prove that $R[\alpha]$ has a ‘nice’ expression: $$ R[\alpha] = \{f(\alpha) : f \in R[x]\}, $$ where $R[x]$ is the ring of polynomials in $x$ (the indeterminate) with coefficients in $R$.

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Now, let $k$ be a field, and $p \in k[x]$ an irreducible polynomial. Then $(p)$ is a maximal ideal of $k[x]$ (because $k[x]$ is a PID), and so the ring $$ K := k[x]/(p) $$ is, in fact, a field. Moreover, the ring homomorphism $$ k \to K, \quad a \mapsto a \bmod p $$ (that is, the composition of the inclusion $k \to k[x]$ with the quotient map $k[x] \to K$) allows us to see $K$ as a $k$-algebra. Less abstractly, we can use the $k$-algebra structure of $k[x]$ to give the scalar multiplication on $K$: If $a \in k$ and $g \in k[x]$, $$ a(g \bmod p) := (ag) \bmod p. \tag{$*$} $$

We claim that if $\eta$ is the element $x \bmod p$ of $K$, then $$ k[\eta] = K. $$ Indeed, given $f \in k[x]$, say $f = \sum_{i\geq0} a_ix^i$, \begin{align} f \bmod p &= \sum_{i\geq0} ((a_ix^i) \bmod p) \quad (\textrm{by the sum def. on $K$}) \\ &= \sum_{i\geq0} a_i(x^i \bmod p) \quad (\textrm{by $(*)$}) \\ &= \sum_{i\geq0} a_i(x \bmod p)^i \quad (\textrm{by the product def. on $K$}) \\ &= \sum_{i\geq0} a_i\eta^i = f(\eta). \end{align} This proves $k[\eta] = K$. Moreover, the above equality says also that $$ p(\eta) = (p \bmod p) = (0 \bmod p) =: 0 $$ i.e. $\eta$ is a root of $p$.

Answer 2

$\eta$ is a primitive m-th root of unity of an irreducible polynomial. $\mathbb{Z}_p[\eta]$ is the finite field generated by the irreducible polynomial. One may choose any irreducible polynomial of degree $d$ and all fields generated are isomorphic to each other. That is why the choice of the irreducible polynomial is arbitrary.