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Why do we prefer to let the coefficients of our polynomial be members of a field instead of just members of a ring? For instance, e.g. instead of considering polynomials over GF(p) what "nice" properties do we lose by just letting the coefficients be members of $\mathbb{Z}/12\mathbb{Z}$.

One thing I can think of in the case of finite fields GP(p) is that it allows us to construct polynomials with degree up to $p-1$ (using a general residue ring of order d might result in powers higher than $\phi(d)$ being redundant). But this doesn't seem like good enough motivation, since it would seem like many of the things such as factoring etc. can still be done in this context.

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    Why? Because $K[X]$ is a Euclidean ring, hence a PID for a field $K$. This is not longer true in general, e.g., it is false for $\Bbb Z[X]$. If your ring has zero divisors, we have even more problems. – Dietrich Burde May 02 '19 at 19:01
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    We also work with polynomials over rings. But of course fields are much "nicer" than rings since we can do a lot more things there. An example of an important property that we lose is that over rings the number of roots of a non zero polynomial might be bigger than its degree. But over fields it is impossible. – Mark May 02 '19 at 19:03
  • There is nothing stopping you from talking about polynomials over rings. It is a standard construction. And as mentioned already, they can be much different from something like $F[x]$. – rschwieb May 02 '19 at 19:07
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    For example, see here for how the notion of UFD bifurcates into various inequivalent notions in commutative rings with zero divisors. – Bill Dubuque May 02 '19 at 19:19
  • The polynomial ring over a finite field is not equal to the ring of polynomial functions on the field. In particular higher powers are not redundant, and there's no simplification to lower degree, even though there is a lower degree polynomial that takes the same values. – Matt Samuel May 02 '19 at 19:30

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