Describe a residue class R/pR

Question

I must describe a residue class $R/pR$ with the euclidean ring $R$ and prime element $p$ for: $R=\mathbb{F_3}[T]$ $p=T^2+1$ I don't know what I have to do exactly

Answer 1

Given the polynomial ring over the field $\Bbb F_3$ in the single indeterminate $T$,

$R = \Bbb F_3[T], \tag 1$

and the quadratic polynomial

$p(T) = T^2 + 1 \in R = \Bbb F_3[T], \tag 2$

we may describe the residue classes of

$R/pR = R/(p) = \Bbb F_3[T]/(T^2 + 1) \tag 3$

by first observing that $p(T)$ is irreducible in $\Bbb F_3[T]$, which is easy to see since

$\deg p(T) = \deg (T^2 + 1) = 2; \tag 4$

thus to show $p(T)$ irreducible we only need verify it has no roots in $\Bbb F_3$; but this is easy to do by direct calculation since

$\Bbb F_3 = \{0, 1, 2\} \tag 5$

has only three elements; we have:

$p(0) = 1, \tag 6$

$p(1) = 2, \tag 7$

$p(2) = 5 \equiv 2 \mod 3, \tag 8$

and so we find that $p(T)$ is indeed irreducible. This being the case, and since $\Bbb F_3[T]$ is a principal ideal domain, it follows that $(T^2 + 1)$ is a maximal ideal in $\Bbb F_3[T]$, and hence that $\Bbb F_3[T]/(T^2 + 1)$ is a field; we wish to explicitly express the field operations in $\Bbb F_3[T]/(T^2 + 1)$. To this end we note that

$\dim_{\Bbb F_3} \Bbb F_3[T]/(T^2 + 1) = 2, \tag 9$

and therefore we may seek a vector-space basis with two elements; denoting the cosets of $(T^2 + 1)$ by

$\bar x = x + (T^2 + 1), \tag{10}$

we take one basis vector to be

$\bar 1 = 1 + (T^2 + 1), \tag{11}$

and for the other

$\bar T = T + (T^2 + 1); \tag{12}$

it is easy to see that $\bar 1$ and $\bar T$ are linearly independent over $\Bbb F_3$; otherwise,

$\exists a, b \in \Bbb F_3, \; a \bar 1 + b \bar T = \bar 0; \tag{13}$

this translates to

$(a + bT) + (T^2 + 1) = (T^2 + 1)\Longrightarrow a + bT \in (T^2 + 1); \tag{14}$

however, the inclusion on the right of (14) is impossible unless $a = b = 0$, since every non-zero polynomial in $(T^2 + 1)$ is of degree at least $2$; we see that $\bar 1$ and $\bar T$ are linearly independent over $\Bbb F_3$, and by virtue of (9),

$\Bbb F_3[T]/(T^2 + 1) = \text{span} \{\bar 1, \bar T \}; \tag{15}$

therefore any element of $\Bbb F_3[T]/(T^2 + 1)$ may be written in the form

$a \bar 1 + b \bar T \in \Bbb F_3[T]/(T^2 + 1); \tag{16}$

now if $c \bar 1 + d \bar T \in \Bbb F_3[T]/(T^2 + 1)$ as well, the the rules of vector space addition imply that

$(a \bar 1 + b \bar T) + (c \bar 1 + d \bar T) = (a + c) \bar 1 + (b + d) \bar T, \tag{17}$

which shows how elements of $\Bbb F_3[T]/(T^2 + 1)$ add; as for multiplication, we first use the ordinary ring axioms, which hold in $\Bbb F_3[T]/(T^2 + 1)$, and compute

$(a \bar 1 + b \bar T)(c \bar 1 + d \bar T) = ac \bar 1 + ad \bar T + bc \bar T + bd \bar T^2, \tag{18}$

which suffers further reduction since, via (11) and (12),

$\bar T^2 + \bar 1 = T^2 + 1 + (T^2 + 1) = \bar 0 + (T^2 + 1), \tag{19}$

since $T^2 + 1 \in (T^2 + 1)$; thus,

$\bar T^2 = -\bar 1, \tag{20}$

and thus (18) becomes

$(a \bar 1 + b \bar T)(c \bar 1 + d \bar T) = ac \bar 1 + ad \bar T + bc \bar T - bd \bar 1 = (ac - bd)\bar 1 + (ad + bc)\bar T, \tag{21}$

which shows how elements of $\Bbb F_3[T]/(T^2 + 1)$ multiply.

We see that the pattern expressed in (21) shows a strong resemblance to multiplication of ordinary complex numbers

$(a + bi)(c + di) = (ac - bd) + (ad + bc)i, \; a, b, c, s \in \Bbb R, \; i \in \Bbb C, \; i^2 = -1, \tag{22}$

which comes as no surprise since the field operations are the same in either case.

In fact, there is no need to restrict ouselves to either $\Bbb F_3$ or $\Bbb R$; we need merely ensure that the quadratic $T^2 + 1$ has no roots--that is, does not split--in the base field $\Bbb F$; so for example $\Bbb Z_5$ may be ruled out, since

$2^2 + 1 \equiv 0 \mod 5, \tag{23}$

but $\Bbb F_7$ is acceptable, as is shown by easy calculation, as is $\Bbb F_{11}$; $\Bbb F_{13}$ fails the test, however, since

$5^2 + 1 \equiv 8^2 + 1 \equiv 0 \mod{13}. \tag{24}$

other finite fields may be so checked provided one is willing, and has the time, to do the necessary arithmetic.