Download *7. Polynomials

*7. Polynomials
Definition For a set F , a polynomial over F with variable x is of the form
an xn + an−1 xn−1 + an−2 xn−2 + ... + a1 x + a0 ,
where an , an−1 , ..., a1 , a0 ∈ F . The ai , 0 ≤ i ≤ n are the coefficients of the
polynomial. If xn is the largest power of x appearing in the polynomial then
n is the degree of the polynomial, an xn is the leading term and an is the
leading coefficient. The collection of all polynomials with one variable x and
with coefficients from F will be denoted by F [x] . (Note the square brackets.)
Note that 0 ∈ F [x], being ... + 0x2 + 0x + 0, but it is not said to have a
degree, though some books give it degree −1 or even −∞.
Part of the aim of this section is to show how similar are the properties
of F [x] and Z.
Examples
3x2 + 5x − 1 ∈ Z [x] ,
5
3 3
x − x2 + x ∈ Q [x] ,
7
12
x2 − π ∈ R [x] .
We could also look at polynomials in Zm [x] or C [x].
If we can add and multiply numbers in the set F then we can add and
multiply the polynomials in F [x].
Examples (i) In Z [x] the sum of 3x2 + 5x − 1 and 5x3 − 3x2 + 2x + 1 is
3x2 + 5x − 1
+ 5x3 − 3x2 + 2x + 1
= 5x3
+ 7x.
(ii) In Z [x] the difference of x2 + 1 and 9x + 7 is
x2 +
1
−
9x + 7
= x2 − 9x − 6.
Notice how subtraction is easier than for integers. For example 101 − 97
is complicated by the fact that we can subtract 7 from 1 and put −6 in the
units place, we have to instead “borrow” 10 from the next column.
1
(iii) In Z [x] the product of x2 + 2x + 3 and x2 + 4 is
x2 + 2x + 3 x2 + 4 =
3x2
+ 12
3
+ 2x
+ 8x
+x4
+ 4x2
= x4 + 2x3 + 7x2 + 8x + 12
We can do all the above again in more “interesting” sets.
(i) Addition in Z3 [x]. The sum of 2x3 + 2x2 + x + 1 and x3 + 2x2 + 2 is
2x3 + 2x2 + x + 1
+ x3 + 2x2
+2
=
x2 + x,
using 3 ≡ 0 mod 3 and 4 ≡ 1 mod 3.
(ii) Subtraction in Z5 [x]. The difference of 4x3 +3x2 +x+3 and 2x3 +3x+4
is
4x3 + 3x2 + x + 3
− 2x3
+ 3x + 4
= 2x3 + 3x2 + 3x + 4
using −2 ≡ 3 mod 5 and −1 ≡ 4 mod 5.
(iii) Multiplication in Z2 [x] . The product of x3 + x + 1 and x2 + x + 1 is
x3 + x + 1 x2 + x + 1 = x5 + x4 + x3
+ x3 + x2 + x
+ x2 + x + 1
= x5 + x4
+ 1.
using 2 ≡ 0 mod 2.
Note that in Z [x] , Q [x] , R [x] , C [x] or Zp [x], with p prime, we have
deg f g = deg f + deg g. This may not hold in Zm [x] with m composite.
Example In Z6 [x] multiply f (x) = 2x2 + 1 and g (x) = 3x5 + 1 to get
2x2 + 1 3x4 + 1 = 6x6 + 3x4 + 2x2 + 1
= 3x4 + 2x2 + 1.
Hence, in this example, deg f g < deg f + deg g.
2
Subtle point. Recall that two functions f and g are equal on a set X if,
and only if, f (x) = g (x) for all x ∈ X. Consider f (x) = x4 + 2x3 + 1 and
g (x) = (1 + x)2 over Z3 . There are only three elements in Z3 and for these,
f (0) = 1 = g (0) ,
f (1) = 1 = g (1) ,
f (2) = 0 = g (2) .
So as functions over Z3 these are equal though as polynomials they are
different.
Question If we can multiply polynomials can we factor them?
Assumption
∀α ∈ F, α 6= 0, α−1 ∈ F.
So, from now on F will be one of Q, R, C or Zp , with p prime because, to
look at factorization, or dividing, it is best to restrict to sets F in which
we can find inverses. (We can talk of dividing in Q, R or C but not in Zp .
Instead we have to remember that to divide by something is to multiply by
it’s inverse.)
(Aside: In Z6 does [3]6 have an inverse? i.e. does there exist an element
with [3]6 [x]6 = [1]6 . If there did, we could multiply both sides by [2]6 to get
[6]6 [x]6 = [2]6 , i.e. [0]6 = [2]6 , contradiction. So not every element in Z6 has
an inverse. The same argument works in general for Zm with m composite.)
The first step to answering this question about factoring is to look at long
division for polynomials.
Example In Q [x] divide x2 − x − 1 into x4 + 2x3 + 3x2 + 4x + 5.
Solution
x2 + 3x + 7
x2 − x − 1
x4 + 2x3 + 3x2 + 4x + 5
− x 4 + x3 + x2
3x3 + 4x2 + 4x
− 3x3 + 3x2 + 3x
7x2 + 7x + 5
− 7x2 + 7x + 7
14x + 12
Hence x4 + 2x3 + 3x2 + 4x + 5 = (x2 + 3x + 7) (x2 − x − 1) + (14x + 12) .
3
Always, always check by multiplying out. You should never end with
an incorrect answer.
Thus given f and g ∈ Q [x] we have found q, r ∈ Q [x] such that f = qg+r
and deg r < deg g. This is very reminiscent of the division theorem for
integers and we have exactly the same result for polynomials.
Example In Z3 [x] divide x2 + x + 1 into x4 + x2 + 1.
x2 + 2x + 1
x2 + x + 1 x4
+ x2
+1
4
3
2
x +x +x
2x3
+1
3
2
2x + 2x + 2x
x2 + x + 1
x2 + x + 1
0
(Often using −2 = 1, −1 = 2 modulo 3.)
Hence x4 + x2 + 1 = (x2 + 2x + 1) (x2 + x + 1) and so x2 + x + 1 divides
4
x + x2 + 1.
Definition If f, g ∈ F [x] we say that g (x) divides f (x) , and write g|f , if
there exists q (x) ∈ F [x] such that f (x) = g (x) q (x). We also say that f if
a multiple of g.
Note that if g|f then either g ≡ 0 (we say it is identically zero) or deg g ≤
deg f .
Careful If m, n ∈ Z, then m|n and n|m imply m = ±n. If we then demand
that both m and n are positive we get m = n. But if we have polynomials
f, g ∈ F [x] then f |g implies deg f ≤ deg g. Similarly, g|f implies deg g ≤
deg f . Hence deg f = deg g. But f |g also means f = gu for some polynomial
u. Then deg f = deg g means deg u = 0, and so u is a non-zero constant.
Hence f |g and g|f imply only that f = cg for some constant c.
Definition If f ∈ F [x] can be factored f (x) = g (x) h (x), with g, h ∈ F [x]
and
1 ≤ deg g, deg h < deg f,
then we say f is reducible. If we cannot factor f in this way we say it is
irreducible. Compare this to the definition of prime numbers. It can be hard
to check if a given polynomial is irreducible. And it can depend on the set
F , see below.
4
Theorem Division Theorem for Polynomials. Let F be as above.
Let f (x) , g (x) ∈ F [x] with g (x) 6≡ 0. Then there exist unique polynomials q (x) , r (x) ∈ F [x] such that
f (x) = q (x) g (x) + r (x)
and either r (x) ≡ 0 or r is non-zero and has a lower degree than g.
Proof Not given in course, but see appendix.
An important deduction connects the roots of a polynomial with factorization.
Corollary Let F be as above. Let p (x) ∈ F [x]. Then p (a) = 0 if, and only
if, (x − a) |p (x) .
Proof (⇒) Assume p (a) = 0. Apply the Theorem with f = p and g = x − a
to find q, r ∈ F [x] with p (x) = q (x) (x − a) + r (x) , and either r ≡ 0 or
deg r < deg g. If r ≡ 0 we are finished. Otherwise deg r < deg g = 1 and
so deg r = 0, i.e. r is a constant, c say. Thus p (x) = q (x) (x − a) + c. Put
x = a to see that c = 0. So again x − a divides p (x) .
(⇐) Assume (x − a) |p (x). But this means p (x) = q (x) (x − a) for some
q (x) ∈ F [x]. Putting x = a and we see that p (a) = 0.
Since a polynomial in F [x] of degree n can be the product of at most n
linear factors x − a, it can have at most n roots in F . But will it have exactly
n roots?
√
Example x2 − 2 ∈ Q [x] yet it has no roots in Q. Its roots are ± 2 ∈ R.
Further x2 + 1 ∈ R [x] but it has no roots in R. Its roots are ±i ∈ C.
Something different happens for C.
Theorem Fundamental Theorem of Algebra. If f ∈ C [x] has deg f ≥ 1
then f has a root in C. Thus f factorizes completely into n linear factors.
Proof Not given in this course.
So a polynomial in C [x] of degree n has exactly n roots in C. This means
that the only irreducible polynomials in C [x] are the linear polynomials,
ax + b.
The Corollary can help us factorize polynomials; to find linear factors it
suffices to find roots.
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Examples (i) Factorize p (x) = x4 + x3 − 2x2 − 6x − 4 over R.
Solution Look at p (a) for small a. So p (0) = −4, p (1) = −8, p (−1) = 0.
Thus x + 1 divides p (x).
x3 − 2x − 4
x + 1 x + x − 2x2 − 6x − 4
x4 + x 3
−2x2 − 6x − 4
−2x2 − 2x
−4x − 4
−4x − 4
0
4
3
x4 + x3 − 2x2 − 6x − 4 = (x + 1) x3 − 2x − 4 .
Writing q (x) = x3 − 2x − 4 we see that q (2) = 0 so x − 2 divides q (x) .
2
3 x + 2x + 2
x−2 x
− 2x − 4
3
x − 2x2
2x2 − 2x − 4
2x2 − 4x
2x − 4
2x − 4
0
x3 − 2x − 4 = (x − 2) x2 + 2x + 2 .
Does x2 +2x+2 factorize over R? Note that x2 +2x+2 = (x + 1)2 +1 ≥ 1
and so can never be zero for real x. Thus x2 + 2x + 2 has no linear factors.
Hence we finish with
x4 + x3 − 2x2 − 6x − 4 = (x + 1) (x − 2) x2 + 2x + 2 .
(ii) Factorize p (x) = x4 + x3 − 2x2 − 6x − 4 over Z5 .
The polynomial is the same as in (ii) so we still have
x4 + x3 − 2x2 − 6x − 4 = (x + 1) (x − 2) x2 + 2x + 2 .
But now,
x
0
1
2
3
4
x2 + 2x + 2 mod 5
2
5≡0
10 ≡ 0
17 ≡ 2
26 ≡ 1.
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Thus modulo 5, x2 + 2x + 2 has roots at 1 and 2. So, in Z5 [x] , x2 + 2x + 2
= (x − 1) (x − 2) = (x + 4) (x + 3). (Check this by multiplying out, modulo
5.)
Hence, in Z5 [x] ,
x4 + x3 − 2x2 − 6x − 4 = (x + 1) (x − 2) (x + 4) (x + 3)
= (x + 1) (x + 3)2 (x + 4) .
Aside So here we have seen that x2 +2x+2 is irreducible in R [x] yet reducible
in Z5 [x] And of course it is reducible in C [x] :
x2 + 2x + 2 = (x + 1)2 + 1 = (x + 1)2 − i2
= ((x + 1) + i) ((x + 1) − i) , “difference of squares”
= (x + 1 − i) (x + 1 + i) .
(iii) Factorize p (x) = x4 + x3 + x2 + 3x + 4 over Z5 .
Solution We need only check x = 0, 1, 2, 3 ≡ −2 or 4 ≡ −1 mod 5. In turn,
p (0) = 4, p (1) = 0, p (2) = 3 while p (−2) = 0 and p (−1) = 2, all calculation
modulo 5. Thus x − 1 and x + 2 divide p (x).
So for some q (x) we have p (x) = (x − 1) (x + 2) q (x) = (x2 + x − 2) q (x) ,
i.e.
x2
+3
2
4
3
2
x + x − 2 x + x + x + 3x + 4
x4 + x3 − 2x2
3x2 + 3x + 4
3x2 + 3x + 4
0
x4 + x3 + x2 + 3x + 4 = (x − 1) (x + 2) x2 + 3 .
You should check that x2 + 3 has no zeros. If p (x) has repeated roots
then x2 + 3 could have 1 or −2 as zeros, just as p (x) did. But
x x2 + 3 mod 5
1
4
−2
7 ≡ 2.
Hence
x4 + x3 + x2 + 3x + 4 = (x − 1) (x + 2) x2 + 3
= (x + 4) (x + 2) x2 + 3
in Z5 [x].
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The division Theorem above is very similar to a result in integers. Other
ideas can be generalized to polynomials such as
Definition Let F be one of Q, R, C or Zp , with p prime. Let f, g ∈ F [x] be
polynomials not both identically zero. Then a greatest common divisor is a
polynomial d ∈ F [x] such that
(i) d (x) divides both f (x) and g (x),
(ii) if c (x) divides both f (x) and g (x) then c (x) divides d (x).
Note that if the gcd exists it is not unique. For if d satisfies (i) and (ii) and
λ ∈ F is non-zero then λd also satisfies the two properties. For two polynomials f and g write gcd (f, g) to represent any greatest common divisor.
A particular gcd is often picked out by taking λ = c−1 where c is the
leading coefficient of d. Then c−1 d has leading coefficient 1, and is called a
monic polynomial.
We need to justify that such a gcd exists. We do not use the same
argument as used for the gcd of integers.
Theorem Let F be as above. Let f, g ∈ F [x] be polynomials not both
identically zero. Then the gcd (f, g) exists.
Proof Not given in course but see the appendix.
In any given problem we construct a gcd using repeated use of the Division
Theorem. And if we formalize this method we get
Theorem The Euclidean Algorithm for Polynomials. Let F be as above.
Let f, g ∈ F [x] be polynomials. If g divides f then g is a gcd of f and g.
Otherwise apply the Division Theorem to obtain a series of equations
f = gq1 + r1 , 0 < deg r1 < deg g,
g = r1 q2 + r2 , 0 < deg r2 < deg r1 ,
r1 = r2 q3 + r3 , 0 < deg r3 < deg r2 ,
..
.
rj−2 = rj−1 qj + rj ,
rj−1 = rj qj+1 .
0 < deg rj < deg rj−1 ,
Then rj , the last non-zero remainder, is a greatest common divisor of f
and g. Further, by working back up this list we can find p (x) , q (x) ∈ F [x]
such that
gcd (f (x) , g (x)) = p (x) f (x) + q (x) g (x) .
Proof Left to student.
8
Examples (i) Over Q [x] find a greatest common divisor of f (x) = x5 +
3x4 + 5x3 + 5x2 + 4x + 2 and g (x) = x4 + 2x3 + 4x2 + 4x + 4 and write the
answer as a linear combination of f and g.
Solution Applying the division algorithm we get
x+1
4
3
2
5
4
3
2
x + 2x + 4x + 4x + 4
x + 3x + 5x + 5x + 4x + 2
− x5 − 2x4 − 4x3 − 4x2 − 4x
x4 + x3 + x2
+2
4
3
2
− x − 2x − 4x − 4x − 4
− x3 − 3x2 − 4x − 2
− x − 3x − 4x − 2
3
2
−x+1
x + 2x + 4x + 4x + 4
− x4 − 3x3 − 4x2 − 2x
− x3
+ 2x + 4
3
2
x + 3x + 4x + 2
3x2 + 6x + 6
4
3
2
− 31 x − 31
3x2 + 6x + 6 − x3 − 3x2 − 4x − 2
x3 + 2x2 + 2x
− x2 − 2x − 2
x2 + 2x + 2
0
So we have found a zero remainder. The last non-zero remainder can be
taken as the greatest common divisor i.e. 3x2 + 6x + 6.
We can work back and write the gcd as a linear multiple of f and g. So
r2 (x) = g (x) − (−x + 1) −x3 − 3x2 − 4x − 2
= g (x) − (−x + 1) (f (x) − (x + 1) g (x))
= (1 + (−x + 1) (x + 1)) g (x) − (−x + 1) f (x)
= 2 − x2 g (x) + (x − 1) f (x) .
Always, always check your answer by multiplying out:
2 − x2 x4 + 2x3 + 4x2 + 4x + 4 + (x − 1) x5 + 3x4 + 5x3 + 5x2 + 4x + 2
= 3x2 + 6x + 6
9
(ii) Over Z3 [x] find a greatest common divisor of x3 + 2x2 + 2x + 1
and x4 + 2 and express your answer as a linear combination of the original
polynomials.
Solution Firstly
x4 + 2 = x x3 + 2x2 + 2x + 1 + x3 + x2 + 2x + 2
= (x + 1) x3 + 2x2 + 2x + 1 + 2x2 + 1.
Next
x3 + 2x2 + 2x + 1 = 2x 2x2 + 1 + 2x2 + 1
= (2x + 1) 2x2 + 1 .
So gcd (x3 + 2x2 + 2x + 1, x4 + 2) = 2x2 + 1. And, quite simply,
2x2 + 1 = x4 + 2 − (x + 1) x3 + 2x2 + 2x + 1 .
10
Appendix
1) Theorem Division Theorem for Polynomials. (H.P p. 264) Let F be as
above.
Let f (x) , g (x) ∈ F [x] with g (x) 6≡ 0. Then there exist unique polynomials q (x) , r (x) ∈ F [x] such that
f (x) = q (x) g (x) + r (x)
and either r (x) ≡ 0 or r is non-zero and has a lower degree than g.
Proof (Existence) If f (x) ≡ 0 we can write 0 = 0g (x) + 0, so we can assume
that f (x) 6≡ 0.
If g|f then f (x) = g (x) q (x) for some q (x) and the result follows with
r (x) ≡ 0. In particular if deg g = 0, i.e. g (x) ≡ c a non-zero constant, we
can take q (x) = c−1 f (x).
Hence we can assume that g - f and n = deg g ≥ 1.
Consider the set
S = {f − qg : q ∈ F [x]} .
Because g - f this set does not contain 0 so every element in this set has
a degree. This set is non-empty, since it contains f − 0g. If we look at D,
the set of degrees of the polynomials in S, we have a non-empty set of nonnegative integers so, by the well-ordering principle, it contains a minimum
element. Choose a q ∈ F [x] such that f − qg takes this minimum value and
set r = f − qg. Since g - f we have r (x) 6≡ 0 and so we have to show that
deg r < deg g.
Assume for the sake of contradiction that deg r ≥ deg g.
Let m = deg r and write r (x) = am xm + am−1 xm−1 + ... + a1 x + a0 , along
with g (x) = bn xn + bn−1 xn−1 + ... + b1 x + b0 . So m ≥ n, am 6= 0 and bn 6= 0.
Consider the polynomial
m−n
m−n
r (x) − am b−1
g (x) = (f (x) − q (x) g (x)) − am b−1
g (x)
n x
n x
−1 m−n
= f (x) − q + am bn x
g (x) ∈ S.
It is in S but the degree of this polynomial is ≤ m − 1, i.e. it is <
deg r, contradicting the choice of r as having minimum degree. Hence our
assumption is false and deg r < deg g as required.
(Uniqueness) Assume for contradiction that, for some f and g, the division is not unique. For this pair we can find distinct divisions
f = qg + r and f = q 0 g + r0
11
with either r ≡ 0 or deg r < deg g and either r0 ≡ 0 or deg r0 < deg g. Then
qg + r = q 0 g + r0 , or r − r0 = (q 0 − q) g.
Thus g divides r − r0 . This means either r − r0 ≡ 0, i.e. r = r0 , and thus
q = q 0 , contradicting the fact that we started with distinct divisions, or
deg g ≤ deg (r − r0 ) ≤ max (deg r, deg r0 ) < deg g,
again a contradiction. Hence our assumption is false and the division is
unique.
2) Theorem Let F be as above. Let f, g ∈ F [x] be polynomials not both
identically zero. Then the gcd (f, g) exists.
Proof Let
S = {f (x) K (x) + g (x) L (x) : K (x) , L (x) ∈ Q [x]} {0} .
Let D be the set of degrees of polynomials in S. W.l.o.g. assume f is
not identically zero and take K ≡ 1 and L ≡ 0 to see that f = 1 × f +
0 × g ∈ S. Thus S, and and in turn D, are non-empty. D is a set of nonnegative integers so, by the well-ordering principle it has a least element.
Take K (x) , L (x) ∈ F [x] so that d (x) = f (x) K (x) + g (x) L (x) has this
minimum degree. We have to show that (i) and (ii) of the definition of gcd
both hold for this d (x).
(i) Use the division Theorem to write f (x) = q (x) d (x) + r (x) for some
q (x) , r (x) ∈ F [x] with either r ≡ 0 or deg r < deg d. Then
r (x) = f (x) − q (x) d (x) = f (x) − q (x) (f (x) F (x) + g (x) L (x))
= f (x) (1 − q (x) K (x)) + (−q (x) L (x)) g (x)
which is either ≡ 0 or in S. If it is not identically zero then we have r (x) ∈ S
and deg r < deg d, which contradicts the minimality of deg d. Hence r ≡ 0
and f (x) = q (x) d (x), i.e. d|f .
Similarly you can show that d|g.
(ii) Assume that c (x) divides both f (x) and g (x), so f (x) = c (x) u (x)
and g (x) = c (x) v (x) for some u (x) , v (x) ∈ F [x]. Then
d (x) = f (x) K (x) + g (x) L (x) = c (x) u (x) K (x) + c (x) v (x) L (x)
= c (x) (u (x) K (x) + v (x) L (x))
and so c (x) divides d (x).
12
3) Recall, in Z we chose an m ∈ Z, and looked at the congruence classes
[a] = {n ∈ Z : m| (n − a)} .
We then defined addition and multiplication on the set of classes, Zm .
We have seen above many similarities between Z and F [x] where F is
one of Q, R, C or Zp , with p prime. So we can try to copy what we did in
constructing the congruence classes of Z. Fix a polynomial f (x) ∈ F [x] and
look at the classes
[g (x)] = {h (x) : f (x) | (h (x) − g (x))} .
Then, using the factor notation from the section on relations (since we
could define a relation h (x) ∼ g (x) if and only if f (x) | (h (x) − g (x))), we
denote the collection of classes by F [x] / (f (x)). We then define addition
and multiplication on this set of classes by
[g1 (x)] + [g2 (x)] = [g1 (x) + g2 (x)] and [g1 (x)] [g2 (x)] = [g1 (x) g2 (x)] .
remember, you should check that these are well-defined operations.
Example Let f (x) = x2 + 1 ∈ R [x]. Given any polynomial p (x) we can use
the division theorem to write p (x) = q (x) (x2 + 1) + r (x) with either r ≡ 0
or deg r (x) < deg (x2 + 2) = 2. Hence
[p (x)] = [r (x)] = [a + bx]
with a, b ∈ R. So all the classes in R [x] / (x2 + 1) can be labelled simply as
[a + bx], a, b ∈ R. For addition and multiplication we have
[a + bx] + [c + dx] = [(a + c) + (b + d) x]
[a + bx] [c + dx] = ac + bcx + adx + bdx2
= (ac − bd) + (bc + ad) x + db x2 − 1
= [(ac − bd) + (bc + ad) x] .
Note that
(a + bi) (c + di) = (ac − bd) + (bc + ad) i.
So the map R [x] / (x2 + 1) → C, [a + bx] 7→ a + bi is a map that preserves
addition and multiplication. Or we can say that R [x] / (x2 + 1) is a way of
constructing the complex numbers from the reals.
Example Take x2 + x + 2 ∈ Z3 [x]. Again the classes are of the form [a + bx],
a, b ∈ Z3 . There are 9 classes, 3 choices for a and 3 for b. They are
[0] , [1] , [2] , [x] , [1 + x] , [2 + x] , [2x] , [1 + 2x] and [2 + 2x] .
13
For the addition and multiplication we can write out the tables
table[1].pdf
For the multiplication table we need only look at non-zero classes.
×
[1]
[2]
[x]
[1 + x]
[2 + x]
[2x]
[1 + 2x]
[2 + 2x]
[1]
[1]
[2]
[x]
[1 + x]
[2 + x]
[2x]
[1 + 2x]
[2 + 2x]
[2]
[2]
[1]
[2x]
[2 + 2x]
[1 + 2x]
[x]
[2 + x]
[1 + x]
[x]
[x]
[2x]
[1 + 2x]
[1]
[1 + x]
[2 + x]
[2 + 2x]
[2]
[1 + x]
[1 + x]
[2 + 2x]
[1]
[2 + x]
[2x]
[2]
[x]
[1 + 2x]
[2 + x]
[2 + x]
[1 + 2x]
[1 + x]
[2x]
[2]
[2 + 2x]
[1]
[x]
[2x]
[2x]
[x]
[2 + x]
[2]
[2 + 2x]
[1 + 2x]
[1 + x]
[1]
[1 + 2x] [1 + 2x]
[2 + x]
[2 + 2x]
[x]
[1]
[1 + x]
[2]
[2x]
[2 + 2x] [2 + 2x]
[1 + x]
[2]
[1 + 2x]
[x]
[1]
[2x]
[2 + x]
Note that every row (and column) is simply a permutation of the classes,
none is missed and none replicated. We see no zero classes, i.e. we have no
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divisors of zero. And all classes have inverses, namely
[1]−1 = [1] , [2]−1 = [1] , [x]−1 = [1 + x] , [1 + x]−1 = [x] ,
[2 + x]−1 = [1 + 2x] , [2x]−1 = [2 + 2x] , [1 + 2x]−1 = [2 + x]
and [2 + 2x]−1 = [2x] .
Example Student to take x3 + x + 1 ∈ Z2 [x]. Draw up the tables for
(Z2 [x] / (x3 + x + 1) , +) and (Z2 [x] / (x3 + x + 1) \ {[0]} , ×). The different
classes are
[0] , [1] , [x] , [1 + x] , x2 , 1 + x2 , x + x2 and 1 + x + x2 .
×
[1]
[x]
[1+x]
[x2 ]
[1+x2 ]
[x+x2 ]
[1+x+x2 ]
[1]
[1]
[x]
[1+x]
[x2 ]
[1+x2 ]
[x+x2 ]
[1+x+x2 ]
[x]
[x]
[x2 ]
[x+x2 ]
[1+x]
[1]
[1+x+x2 ]
[1+x2 ]
[1+x]
[1+x]
[x+x2 ]
[1+x2 ]
[1+x+x2 ]
[x2 ]
[1]
[x]
[x2 ]
[x2 ]
[1+x]
[1+x+x2 ]
[x+x2 ]
[x]
[1+x2 ]
[1]
[1+x2 ]
[1+x2 ]
[1]
[x2 ]
[x]
[1+x+x2 ]
[1+x]
[x+x2 ]
[x+x2 ]
[x+x2 ]
[1+x+x2 ]
[1]
[1+x2 ]
[1+x]
[x]
[x2 ]
[1+x2 ]
[x]
[1]
[x+x2 ]
[x2 ]
[1+x]
[1+x+x2 ] [1+x+x2 ]
Note that [x]2 = [x2 ], [x]3 = [1 + x], [x]4 = [x + x2 ], [x]5 = [1 + x + x2 ] ,
[x]6 = [1 + x2 ] and [x]7 = [1].
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