MATH 57091 - Algebra for High School Teachers
Polynomials & Degrees
Professor Donald L. White
Department of Mathematical Sciences
Kent State University
D.L. White (Kent State University)
1/7
Review & Definitions
Recall our previous setup for polynomial rings:
R is a commutative ring.
R[x] is the set of all polynomials in x with coefficients from R.
Addition and multiplication in R[x]
are the usual polynomial operations
(see text §3.1 for formal definitions).
R[x] is a commutative ring.
If R is a ring with 1, then R[x] has a 1.
By viewing r ∈ R as a constant polynomial,
we consider R to be contained in R[x].
D.L. White (Kent State University)
2/7
Review & Definitions
Definition
Let f (x) = an x n + · · · + a1 x + a0 be a polynomial in R[x],
with n > 0, ai ∈ R, and an 6= 0 (so f (x) is not the zero polynomial).
We define:
an x n is the leading term of f ;
an is the leading coefficient of f ;
the degree of f is n (we write deg f (x) = n);
if an = 1, we say f is a monic polynomial.
NOTES:
If f (x) ∈ R[x] is a non-zero constant polynomial, then deg f (x) = 0.
The degree of the zero polynomial is undefined.
Alternatively, we could define deg 0 = −∞, and all of our results
on degrees would hold, even including the zero polynomial.
(This is not the case if we attempt to define the degree as 0.)
D.L. White (Kent State University)
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Properties of Degrees
Theorem 1
If p(x), q(x) are non-zero polynomials in R[x] such that p(x) + q(x) 6= 0,
then deg[p(x) + q(x)] 6 max{deg p(x), deg q(x)}.
Proof: Let n = max{deg p(x), deg q(x)}, so that
p(x) = an x n + · · · + a1 x + a0 and q(x) = bn x n + · · · + b1 x + b0
(and one of an or bn may be 0). We then have
p(x) + q(x) = (an + bn )x n + · · · + (a1 + b1 )x + (a0 + b0 ).
Therefore, deg[p(x) + q(x)] 6 n = max{deg p(x), deg q(x)}.
(Note that an + bn could be 0, and so the degree could be less than n.) EXAMPLE: If p(x) = 2x 2 + 3x + 4 and q(x) = 3x 2 + x + 2 in Z5 [x],
so that max{deg p(x), deg q(x)} = 2,
then p(x) + q(x) = 4x + 1 and deg[p(x) + q(x)] = 1.
D.L. White (Kent State University)
4/7
Properties of Degrees
Theorem 2
If p(x), q(x) are non-zero polynomials in R[x] such that p(x)q(x) 6= 0,
then
i
deg[p(x)q(x)] 6 deg p(x) + deg q(x) and
ii
if R is an integral domain, then deg[p(x)q(x)] = deg p(x) + deg q(x).
Proof: Let
p(x) = am x m + · · · + a1 x + a0 and q(x) = bn x n + · · · + b1 x + b0 ,
where am 6= 0 and bn 6= 0, so that deg p(x) = m and deg q(x) = n.
The highest possible degree term in p(x)q(x) is am x m bn x n = am bn x m+n .
If am bn 6= 0, then deg[p(x)q(x)] = m + n = deg p(x) + deg q(x).
If am bn = 0, then deg[p(x)q(x)] < m + n = deg p(x) + deg q(x).
Therefore, (i) holds.
If R is an integral domain, then am 6= 0 and bn 6= 0 imply am bn 6= 0,
hence deg[p(x)q(x)] = m + n = deg p(x) + deg q(x) and (ii) holds.
D.L. White (Kent State University)
5/7
Properties of Degrees
EXAMPLE: Let p(x) = 2x 3 + 3 and q(x) = 3x 2 + 2.
1
If p(x) and q(x) are in Z[x], then
p(x)q(x) = 6x 5 + 4x 3 + 9x 2 + 6,
and
deg p(x)q(x) = 5 = deg p(x) + deg q(x).
2
If p(x) and q(x) are in Z6 [x], then
p(x)q(x) = 4x 3 + 3x 2 ,
and
deg p(x)q(x) = 3 < 3 + 2 = deg p(x) + deg q(x).
D.L. White (Kent State University)
6/7
Units
Theorem 3
Let R be an integral domain.
A polynomial p(x) ∈ R[x] is a unit in R[x] if and only if
p(x) is a constant polynomial r , where r is a unit in R.
Proof: If r is a unit in R, so rs = 1R for some s ∈ R, then r , s ∈ R[x]
and 1R[x] = 1R , and so r is a unit in R[x].
Conversely, suppose p(x) is a unit in R[x], so that p(x)q(x) = 1
for some q(x) ∈ R[x]. In particular, p(x) and q(x) are non-zero,
and since R is an integral domain, Theorem 2 implies
deg p(x) + deg q(x) = deg[p(x)q(x)] = deg 1 = 0.
Since deg p(x) and deg q(x) are both non-negative, this implies
deg p(x) = deg q(x) = 0.
Thus, p(x) = r and q(x) = s are both constant polynomials, so are in R.
Since rs = 1, we have that r = p(x) is a unit in R.
D.L. White (Kent State University)
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