What's the relationship between domain and the degree of a polynomial?

Question

Since d(x)|2, and since Z[x] is a domain, the degree of d(x) cannot exceed that of 2.

(Provided d(x) ∈ Z[x])

Why d(x) | 2 and Z[x] is a domain implies that the degree of d(x) cannot exceed 2?

Answer 1

You've misread the English: it is saying

The degree of $d(x)$ cannot exceed the degree of $2$.

Maybe it helps to rename things. If we define $f$ to be the polynomial $f(x) = 2$, then the quoted passage intends to say

The degree of $d(x)$ cannot exceed the degree of $f(x)$.

Answer 2

If you are working with polynomials over a ring that admits zero divisors, then it is not always true that $\deg(fg) = \deg(f) + \deg(g)$. For instance, if the coefficient ring is $\mathbb Z_4$, then $(2x+1)(2x+1) = 1$.

But if there are no zero-divisors in the coefficient ring (i.e. it is a domain), then the highest-degree term of the product will always come from the highest-degree terms of the factors.

This is why it makes sense to mention that $\mathbb Z$ is a domain before concluding that the degree of a polynomial does not exceed the degree of a multiple of that polynomial.

Answer 3

In general, if $R$ is a domain, then for any two poynomials $f,g \in R[x]$, it is true that

$$f | g \implies \deg(f) \le \deg(g)$$

The statement you are asking about is just the special case of this in which $g$ is the constant polynomial $g=2$. It asserts that $$f| 2 \implies \deg(f) \le \deg(2)$$ (The phrase "less than that of 2" is a shorter way of saying "less than the degree of the polynomial 2".) In particular, the sentence is not saying that $\deg(f) \le 2$; in fact it's saying something quite a bit stronger than that, namely that $\deg(f) = 0$, i.e. $f$ is a constant.