Degree of Product of Polynomials over Ring/Corollary 2
< Degree of Product of Polynomials over Ring(Redirected from Degree of Product of Polynomials over Integral Domain)
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Theorem
Let $\struct {D, +, \circ}$ be an integral domain whose zero is $0_D$.
Let $D \sqbrk X$ be the ring of polynomials over $D$ in the indeterminate $X$.
For $f \in D \sqbrk X$ let $\map \deg f$ denote the degree of $f$.
Then:
- $\forall f, g \in D \sqbrk X: \map \deg {f g} = \map \deg f + \map \deg g$
Proof
An integral domain is a commutative and unitary ring with no proper zero divisors.
The result follows from Degree of Product of Polynomials over Ring: Corollary 1.
$\blacksquare$
Sources
- 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $6$: Polynomials and Euclidean Rings: $\S 25$. Polynomials: Theorem $48$
- 1970: B. Hartley and T.O. Hawkes: Rings, Modules and Linear Algebra ... (previous) ... (next): $\S 3.2$: Polynomial rings: Lemma $3.7 \ \text{(iii)}$