Download When divisors go bad… counterexamples with polynomial division

Math 4000: Modern Algebra and Geometry I
Spring 2014, Dr. Klipper
Extra Examples for Week 7
“When Divisors Go Bad”
In class, we have seen (or will shortly see) several important theorems that apply to
polynomials in F [x], where F is a field:
• Long division: If a(x), b(x) ∈ F [x] with a 6= 0, then there exist unique q(x), r(x) ∈ F [x]
where
b(x) = q(x)a(x) + r(x) and (r = 0 or deg(r) < deg(a))
• Root-Factor Theorem: For any f (x) ∈ F [x] and r ∈ F , we have
f (x) = q(x)(x − r) + f (r)
for some q(x) ∈ F [x]. Thus, x − r is a factor of f (x) iff r is a root of f (x).
• Euclidean algorithm: We can find d(x) = gcd(a(x), b(x)) of two polynomials by repeatedly performing (long) division, just like we did with integers, and then we can solve
for a linear combination d(x) = a(x)m(x) + b(x)n(x).
– “Linear combination” means we only multiply one polynomial (one member of
F [x]) to each of a(x) and b(x). It doesn’t mean that the polynomials m(x) and
n(x) have to be degree 1.
– Technically, the gcd should have a leading coefficient of 1, meaning it is monic.
You can always divide your equations by the leading coefficient of d(x), though,
to fix this. (I do that at the end of the process.)
However, here’s the bad news: many of these results fail in R[x] if R is not a field.
We will explore some counterexamples in this handout, which hopefully gives you a better
appreciation for fields. Zero divisors will play a major role in many of our examples, though
some examples will just work with elements that are not units.
What Goes Wrong with Zero Divisors
Multiplication of polynomials can have some strange behaviors in the ring R[x] when R
has zero divisors. We present three anomalies here.
1. A divisor may have smaller degree.
For instance, let’s use R = Z6 . In this ring, (3x + 1) | 2 because
(3x + 1) · 2 = 6x + 2 = 0x + 2 = 2
Thus, a polynomial of degree 1 can be a divisor of a polynomial of degree 0. This also
provides an excellent counterexample to the statement “deg(f g) = deg(f ) + deg(g)” which
is true for F [x].
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It’s worth noting, however, that the only way to make a counterexample like this is to use
a zero divisor for the leading coefficient. (For instance, we used 3 above, and 3 · 2 becomes
zero even though 3 and 2 are both nonzero.) If f (x) and g(x) are nonzero with f (x) | g(x),
and the leading coefficient of f (x) is not a zero divisor, then you can prove deg(f ) ≤ deg(g).
2. A polynomial may have more roots than its degree.
In your homework, you’re asked to prove that in F [x], a polynomial of degree n can have
at most n roots. This result is false in R[x] when R has zero divisors.
One example is f (x) = x2 − 1 in Z8 [x]. Before Test 1, we saw that all odd x ∈ Z satisfy
x2 ≡ 1 (mod 8), so that means 1, 3, 5, 7 are four distinct roots of f (x) in Z8 . However, f
only has degree 2. Another example is f (x) = 2x in Z4 [x]; this is a linear polynomial, but
both 0 and 2 are roots in Z4 .
This anomaly is quite related to our last one...
3. Roots of a product may not be roots of factors!
Recall that if R is an integral domain, then ab = 0 can only happen in R if a = 0 or
b = 0. This has a very important consequence for polynomial roots: if f (x)g(x) has a root
r, then f (r)g(r) = 0, and that means either f (r) = 0 or g(r) = 0 (or possibly both are zero).
This fact is used all the time when solving equations; for example, consider the following
process to solve a cubic equation in C:
z 3 + z = 0 ⇔ z(z 2 + 1) = z(z + i)(z − i) = 0 ⇔ z = 0, z + i = 0, or z − i = 0
(This gives us the three roots of z = 0, i, −i.)
However, when R is not an integral domain, f (r)g(r) = 0 could happen even when f (r)
and g(r) are nonzero! Put simply, a root of f (x)g(x) does not have to be a root of either
f (x) or of g(x). For example, consider the following examples from Z6 :
f (x) = 2x
g(x) = x(x − 1)
In both of these polynomials, 3 is a root: f (3) = g(3) = 0. However, f has two factors (2
and x), and neither has 3 as a root. Similarly, neither of g’s factors has 3 as a root. Instead,
when you substitute 3 for x, both of these polynomials give you the product 3 · 2, which
equals zero even though the factors are not zero.
This has a major impact on some of your homework problems. When you factor
a polynomial, ask yourself whether you are using an integral domain and why that matters!
When the Division Algorithm Works
The (long) division algorithm gives a step-by-step procedure for finding a quotient and
remainder when doing division in F [x], when F is a field. Can the same procedure still
work in R[x] when R is not a field? The answer is: sometimes. Here’s a simple general rule
though:
You can always perform division by a(x) if a’s leading coefficient is a unit.
Why does this work? Let’s say you’re trying to divide b(x) by a(x). At each step, you
divide the leading terms: say you’re dividing bn xn by am xm . When am is a unit, that quotient
n−m
always exists: it’s (bn a−1
.
m )x
2
−1
Example: Divide 2x + 5 by 5x + 3 in Z8 [x]. The quotient is 2 · 5 , which is 2 · 5 = 10 = 2
in Z8 :
2
5x + 3 2x + 5
−(2x + 6)
7
(In the last step, 5 − 6 = −1 = 7.) Thus, the quotient is 2 and remainder is 7, i.e.
2x + 3 = [2](5x + 3) + 7. Notice that this algorithm worked even though Z8 is not an integral
domain!
Corollary: The Root-Factor Theorem works even if R isn’t an integral domain. (This is
because you’re dividing by x − r, and the leading coefficient of x − r is 1, which is a unit.)
See Exercises 3.1 #7 and #16, for instance.
When the Division Algorithm Can Fail
We may be unable to divide a(x) when a’s leading coefficient is not a unit. For example,
consider trying to divide x + 1 by 4x + 2 in Z6 . The first step of the division algorithm would
fail: you wouldn’t be able to pick a quotient for x/4x, since 4 is not a unit in Z6 .
In fact, we can go further in this case and show that it’s impossible to get a quotient
and remainder at all, even if you don’t use the division algorithm. To prove this, suppose
for contradiction that we had polynomials q(x), r(x) ∈ Z6 [x] where either r = 0 or deg(r) <
deg(x + 1), and
x + 1 = q(x)(4x + 2) + r(x)
Because r has smaller degree than x + 1, r(x) must be constant. Thus, it doesn’t change the
linear term of the right side, so the linear term of q(x)(4x + 2) must be the same as on the
left side, namely 1x. However, neither of 4 nor 2 is a unit: no matter what you multiply to
them, you can never get 1 as a coefficient. This gives us our contradiction.
GCDs May Not Come from Linear Combinations
There’s a further issue that non-units present in R[x]. It turns out that in some cases,
you can show gcd(a(x), b(x)) exists (or at least, there’s something satisfying the main parts
of the gcd definition), and yet we cannot use the Euclidean algorithm to write it as a linear
combination! Let’s show an example in Z[x] where this can occur.
Consider a(x) = 2x and b(x) = 4x + 6. Let’s check that d(x) = 2 satisfies the most
important parts of the definition of gcd(a(x), b(x)):
(i) Is d(x) a common divisor? Yes: clearly 2x = 2(x) and 4x + 6 = 2(2x + 3).
(ii) Is d(x) a multiple of all other common divisors? Yes. Suppose e(x) divides both 2x
and 4x + 6. Therefore, e(x) divides their combination (4x + 6) − 2(2x) = 6. Since
e(x) | 6, e(x) must be a constant. Since e(x) | 2x as well, the only options for the
constant are ±1 and ±2, which all divide 2.
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However, d(x) is not monic: its leading coefficient is 2 instead of 1. And in fact, it’s
impossible to write d(x) as a linear combination in Z[x]:
2 = (2x)m(x) + (4x + 6)n(x)
Why? Suppose for contradiction such an m(x) and n(x) existed in Z[x]. The linear terms
have to cancel on the right side, so 2xm(x) = −4xn(x), showing that m(x) = −2n(x).
However, that means
2 = (2x)(−2n(x)) + (4x + 6)n(x) = −4xn(x) + 4xn(x) + 6n(x) = 6n(x)
However, 6 does not divide 2, giving us our contradiction.
This example shows why we want monic gcds and why we want to work in a field, so
that we can divide our polynomials by their leading coefficient.
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