# How to deal with polynomial quotient rings

The question is quite general and looks to explore the properties of quotient rings of the form $$\mathbb{Z}_{m}[X] / (f(x)) \quad \text{and} \quad \mathbb{R}[X]/(f(x))$$ where $m \in \mathbb{N}$

Classic examples of how one can treat such rings is to find relationships like $$\mathbb{Z}[x]/(1-x,p) \cong \mathbb{Z_{p}}$$ for prime $p$.

However, I often struggle to intuitively understand what the elements of such rings are, and they to compute using them.

For example, what do the elements of the set $\mathbb{Z}_{2}[X] / (x^4+1)$ actually look like. Can this set be described in a clearer way to help understand the way the rings work.

Is there some general way of describing elements of such rings so that isomorphisms and computations become easier to handle?

Apologies if this question is not sufficiently clear.

• You may find my answers on the threads here and here helpful. Actually, this question might be considered a duplicate of the first thread. – Zev Chonoles May 17 '13 at 22:59
• You don't mean $\mathbb{Z}_m$ (the ring of $m$-adics), you mean $\mathbb{Z}/m\mathbb{Z}$. – Martin Brandenburg May 17 '13 at 23:03

To treat your special case $$\mathbb{Z}_{2}[X] / (x^4+1),$$ its elements are the (classes) of the remainders of Euclidean divisions by $x^4+1$, so $$\mathbb{Z}_{2}[X] / (x^4+1) = \{ [a_0 + a_1 x + a_2 x^2 + a_3 x^3] : a_i \in \mathbb{Z}_{2} \}.$$ (Here I am using brackets to denote residue classes.)

You sum the classes normally, and as to the product, you first take the product, and then take the remainder of Euclidean division by $x^4+1$. Alternatively, you take the product, and then use the relation $x^4 \equiv -1$ (which is the same as $x^4 \equiv 1$ in this case, as $1 = -1$ in $\mathbb{Z}_{2}$) to reduce the result. So it's not really different from computing in $\mathbb{Z}_{m}$, where you first take an ordinary sum or product, and then take the remainder modulo $m$.

So for instance $$[1 + x^2] \cdot [1 + x^3] = [1 + x^2 + x^3 + x^5] = [1 + x + x^2 + x^3],$$ using one of the two methods above.

The general case of $F[x] / (f(x))$, with $F$ a field, is similar, you get the classes of the remainders of Euclidean division by $f(x)$.

• This is a very good answer which helped me understand the topic very much. However, one thing I don't understand is how [1+x^2+x^3+x^5]=[1+x+x^2+x^3]. Surely, the remainder of x^5/x^4+1 is x+(-x)/(x^4+1)? So, then why has x^5 been 'simplified' to x in your calculations? (i know I'm wrong i just want to know why) – P-S.D Apr 11 '17 at 9:31
• Doesn't matter I understand now. Also, [1+x+x^3+x^5] is congruent to [1-x+x^2+x^3]. i.e. -x instead of +x. – P-S.D Apr 11 '17 at 10:15
• when you say $x^4 \equiv -1$ you mean this $x^4=-1$? – user441848 May 20 '17 at 3:59
• @AnneliseToft, I mean $x^{4} \equiv -1 \pmod{x^{4} + 1}$. – Andreas Caranti May 20 '17 at 17:06
• math.stackexchange.com/questions/3777658/… – Consider Non-Trivial Cases Aug 2 '20 at 15:20

Often you can use the universal properties (of quotients and polynomial algebras) in order to simplify the rings. For example,

$\mathbb{F}_2[x]/(x^4+1) = \mathbb{F}_2[x]/((x+1)^4) \cong \mathbb{F}_2[y]/(y^4)$

and this cannot be simplified anymore. You just "cut" the polynomials in $y$ above $y^4$. Of course $y^5$ is still there, but it equals zero. And $1+y$ is a unit, since $1+y+y^2+y^3$ is inverse to it (more generally, unit + nilpotent = unit, this is the geometric series).

Often the Chinese Remainder Theorem helps to simplify the rings. It shows that if $f \in K[x]$ decomposes as $f_1^{k_1} \cdot \dotsc \cdot f_n^{k_n}$ with irreducible $f_i$, then $K[x]/(f) \cong \prod_{i=1}^{n} K[x]/(f_i^{k_i})$.

For example, $\mathbb{R}[x]/(x^2-1) = \mathbb{R}[x]/((x+1)(x-1)) \cong \mathbb{R}[x]/(x+1) \times \mathbb{R}[x]/(x-1) \cong \mathbb{R} \times \mathbb{R}$. Another interesting example is $\mathbb{Q}[x]/(x^n-1) \cong \prod_{d|n} \mathbb{Q}(\zeta_d)$.

Since the CRT holds for arbitrary rings, we can also apply it to $\mathbb{Z}/m[x]$. But the problem is that this ring is not a PID anymore (when $m$ is not a prime), so that we have no nice factorizations and even when they exist the corresponding ideals are not coprime and therefore CRT doesn't apply. For example, I don't think that we can really simplify $\mathbb{Z}/4[x]/(2x)$. The elements have the form $a_0+a_1 x+ a_2 x^2 + \dotsc$ with integers $a_i$, where we can calculate $a_0 \bmod 4$ and $a_i \bmod 2$ for $i>0$.