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Artin defines the adjoining of an element $\alpha$ to ring $R$ satisfying polynomial $f \in R[x]$ by $$R[\alpha] = R[x]/(f).$$

If $f$ is monic with degree $n$, then we get some nice properties, such as a unique representation of any element of $R[x]/(f)$ as a $n-1$-degree polynomial in $\alpha$. $$\sum_{k=0}^{n-1} c_k \alpha^k$$ This follows from division with remainder by a monic polynomial.

However, what happens when $f$ is not monic? I am having trouble figuring out what still holds. Here's my guess: the elements of $R[x]/(f)$ are still polynomials in $\alpha$, but the representation is no longer unique, and we may need polynomials of arbitrarily large degree (in contrast with the $n-1$-degree polynomial above).

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  • $\begingroup$ You assume $f$ to be prime, right? $\endgroup$ – k.stm Feb 14 '14 at 16:18
  • $\begingroup$ It is irrelevant if f is prime or not. $\endgroup$ – Martin Brandenburg Feb 14 '14 at 16:20
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I think you are right. First of all, the case that the leading coefficient is a unit can be reduced to the monic case. But when the leading coefficient is not a unit, strange things can happen. For example, $Z/4[x]/(2x+1)$ is zero! But $Z/4[x]/(2x)$ is not, it has $8$ elements. More generally one can classify invertible polynomials as those which have an invertible constant coefficient and all the other coefficients are nilpotent. I think these examples suggest that there won't be any general normal form for the elements of a quotient ring.

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  • $\begingroup$ What do you intend to generalize when you write "more generally..."? $\endgroup$ – Bill Dubuque Feb 14 '14 at 16:40
  • $\begingroup$ This was a hint to the first quotient: 2x+1 is invertible over Z/4. $\endgroup$ – Martin Brandenburg Feb 14 '14 at 21:04

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