Consider Zx which is an UFD unique factorization domain True

Consider Z[x], which is an UFD (unique factorization domain). True or false: (2, x) consists of all polynomials in Z[x] whose constant terms are even. True or false: If f(x) Z[x] satisfies f(x) | 2 and f(x) | x, then f(x) = 1 or -1. Prove that the ideal (2, x) is not principal in Z[x].

Solution

Solution 1:

We need to prove whether etthe given statement is true or false.

Given, (2,x) consists of all polynomials in z[x] whose constant terms are even.

Any element of the form 2f + xg, where f and g are arbitrary polynomials from z[x].

Let us consider the constant term of the polynomial. The polynomial xg has no constant term, and so the

constant term for 2f + xg is equal to the constant term of 2f. This constant must be even. Thus, every element in

(2, x) has even constant term. But conversely, consider any polynomial in z[x] with even constant term. We can

write such a polynomial as 2n + xh, where n is an integer and h is a polynomial. Thus, (2, x) consists of all polynomials with even constant term. Hence, it is true.

Solution 2:

Since it is not principal in z[x]. It does not satisfy f(x)/2 and f(x)/x. If it were, with generator f, then all polynomials in the ideal would bemultiples f. In particular, 2, x would be multiples of f. Then, the degree must be 0.

Therefore, it is false.

Solution 3:

Suppose that the ideal I = (2, x) in Z[x] is principal. Then there exists f(x) Z[x] such that I = f(x). As 2 I there exists g(x) Z[x] such that 2 = f(x)g(x).

Hence deg f(x) + deg g(x) = deg f(x) g(x) = deg 2 = 0

so deg f(x) = deg g(x) = 0.

Thus f(x) = a, g(x) = b, a, b Z

As 2 = ab we have a = ±1 or ± 2.

Equating constant terms in these polynomials we obtain 1 = 2 r(0) so that r(0) = 1/2.

contradicting r(0) Z.

If a = ±2 then (2, x) = 2. Hence there exists v(x) Z[xX] such that x = 2v(x).

Then v(x) = (1/2)X, contradicting v(x) Z[x]. This proves that (2, x) is non principal.

Hence, the ideal (2, x) is not principal in z[x].