# Irreducible polynomial $f$ as quotient - effect on $\mathbb{Z}_5[x]$

I want to get a better understanding of quotient rings so I have two questions.

Let $f(x) = x^2 + 2$

Let $R = \mathbb{Z}_{5}/(f(x))$

Now as $f$ is irreducible in $\mathbb{Z}_{5}$ we have that $R$ is a field with elements being all polynomials in $\mathbb{Z}_{5}$ with degree less than $2$, i.e.

$\{0, 1, 2, 3, 4,$
$x, x + 1, \dots, x + 4,$
$,\dots,$
$4x, 4x + 1, ..., 4x + 4 \}$

• First question - But how can, say, the element $x \in R$ be a unit?

Also now if we let the quotient be a reducible polynomial, the ring is not supposed to be a field? I.e.

Let $g(x) = x^2 + 1$

Let $S = \mathbb{Z}_{5}/(g(x))$

• Second question - It seems to me that $S$ will have exactly the same elements as $R$ and hence it will also be a field?
-

In $R$: $x^2 + 2 = 0$, so $(2x)x = 2 x^2 = -4 = 1$, i.e., $x^{-1} = 2x$.
For the second question, $R$ does not have the same elements as $S$; $R$ and $S$ only happen to have the same number of elements and they can be represented by the same elements of ${\mathbb Z}_5[x]$, but that's it. Now because $x^2 + 1$ is reducible over ${\mathbb Z}_5$ ($x^2 + 1 = (x + 2)(x - 2)$), the ring $S$ is not an integral domain ($(x + 2)(x-2) = 0$ in $S$) and therefore not a field.
It may be worthwhile to stress that elements of $R$ (and of $S$) are not polynomials over ${\mathbb Z}_5$ with degree less than 2, they can merely be represented by those polynomials. The elements are residue classes; the residue class of $h(x) \in {\mathbb{Z}_5[x]}$ is $\{ h(x) + a(x) f(x) \;\mid\; a(x) \in {\mathbb Z}_5[x] \}$ (and letting $h(x)$ range over the polynomials of degree less than 2, you get every residue class exactly once). Writing down these residue classes all the time is not really enlightening, which is why most of the time you just write $h(x)$ (saying that computations are "in $R$" or "modulo $(f(x))$") or maybe $\overline{h(x)}$.
How did you spot that $x^{-1} = 2x$ without checking it against all the $ax + b$ in $R$? – sonicboom Nov 17 '13 at 19:26
I started with $x^2 = -2 = 3$ and then multiplied by the inverse of $3$ in $\mathbb Z_5$. – Magdiragdag Nov 17 '13 at 19:28
Is it possible to have some $f(x) \in \mathbb{Z_5[x]}$ with $deg(f) > 1$ that is irreducible? – sonicboom Nov 17 '13 at 19:37
What do you mean? The polynomial you started with, $x^2 + 2 \in {\mathbb Z}_5[x]$, is irreducible. – Magdiragdag Nov 17 '13 at 19:54
In general, in $k[x]/(f(x))$ with $f$ irreducble, you could find the inverse of a $\overline g$ by computing the gcd of $f$ and $g$ (which is 1) with the extended Euclidean algorithm. This gives you polynomials $a$ and $b$ with $af + bg = 1$, so ${\bar b} {\bar g} = \bar 1$. – Magdiragdag Nov 17 '13 at 20:52