Question about a paper, polynomials preserving congruence In the paper, Interpolation Domains (Here), the beginning of the paper says:

Let $K$ be a field ... The same does not hold for a domain $D$ (which is not a field), as polynomials in $D[X]$ preserve congruences modulo every ideal of $D$.

I am just wondering what is meant by polynomials in $D[X]$ preserving congruence modulo every ideal of $D$. I know the usual congruence modulo an ideal in that if $R$ is a ring and $I$ is an ideal then $a$ is left congruence to $b$ modulo $I$ if $a-b \in I$. Could someone explain what they mean here?
 A: The author is stressing that in domains, unlike fields, modular reductions imply obstructions on (Lagrange) interpolation. For example $\  f(0) = 0,\ f(2) = 1 $ has no solution for $\,f \in \Bbb Z[x]\,$ since, viewed modulo $2,\,$ it implies that $\, 0 \equiv f(0) \equiv 1.$  Thus "polynomials in $D[X]$ preserve congruences modulo every ideal of D" means that for a polynomial $\,f\in D[X]\,$ with coefficients in $D$ (vs. its fraction field $K$), one can always reduce it mod any ideal $I$ of $D$ to obtain possible obstructions to interpolation. This need not be possible for polynomials whose coefficients are "fractions" $(\in K)$ since the denominator might be $\,0\,$ or a zero-divisor modulo $I,\,$ so the polynomial cannot be reduced mod $I$, preventing such a modular obstruction argument. In particular, this is why the obstruction goes away for the ring of integer-valued polynomials discussed there. Indeed, in this ring the above problem is solvable: $\,f(x) = x(x-1)/\color{#c00}2,\,$ so the modular argument breaks down because it is not possible to reduce $\,f\,$ modulo $\,\color{#c00}2.$
