Mark each of the following statements True or False. If a statement is False, gi

Question

Mark each of the following statements True or False. If a statement is False, give an example that shows it is false. (a) Let F be a field. Then F[x] is a commutative ring with identity. (b) Let F be a field and let f(x) be a polynomial in F[x]. If f(x)1S irreducible, then f(x) has no roots in F. (c) Let F be a field and let f(x) be a polynomial in F[x]. If f(x) has no roots in F then f(x) is irreducible. (d) Let F be a field and let f(x) be a nonzero polynomial in F[x]. If the degree of f(x) is less than or equal to n, then f(x) has at most n roots in F. (e) Let F be a field and let f(x) be a nonzero polynomial in F(x). If f(x) has n roots in F, then the degree of f(x) is n.

Explanation / Answer

a) A eld F is a commutative ring with identity in which every non-zero element is a unit. That is, any a F with a 6= 0 has a multiplicative inverse. In a non-trivial eld, the set F = F {0} is an Abelian group under multiplication, with identity 1.

So statement a is True.

b) LEt F is field and f(x) is nonconstant polynomial in F then f(x) is irreducible if f(x) has no roots in F.

This is True because

If F is a field, a non-constant polynomial is irreducible over F if its coefficients belong to F and it cannot be factored into the product of two non-constant polynomials with coefficients in F.

A polynomial with integer coefficients, or, more generally, with coefficients in a unique factorization domain R, is sometimes said to be irreducible (or irreducible over R) if it is an irreducible element of the polynomial ring, that is, it is not invertible, not zero, and cannot be factored into the product of two non-invertible polynomials with coefficients in R.

c) Let F is field and f(x) be nonconstant polynomial in F[x]. If f(x) has no roots in F then f(x) is irreducible.

It is false

because if it dont has roots in F but we are not sure coefficients are in F or not so we cant say it is irreducible or not. So it is false

d) Let F be field and let f(x) be nonzero polynomial in F. If the degree of f(x) is less than or equal to n then f(x) has at most n roots in F.

It is true.

Let p(x) be a nonzero polynomial in F[x], F a eld, of degree d. Then p(x) has at most d distinct roots in F.

proof:

The proof proceeds by induction on d. The result is clearly true for d = 0,1. Assume now that d > 1 and that the proposition is true for all polynomials of degree less than d. Consider a polynomial p(x) of degree d. If p(x) has no roots in F, then the proposition clearly holds for p(x) (as 0 < d). Thus we may assume that p(x) has at least one root, say. Then p(x) = (x)q(x), for some q(x) of degree d1 (by Proposition A.2.1). Any root of p(x) other than must also be a root of q(x). However by induction q(x) has at most d1 roots. Therefore p(x) has at most 1+(d1) = d roots, as claimed. This completes the induction.

e) Let F be field and let f(x) be nonzero polynomial in F. If f(x) has n roots in F then the degree of f(x) is less than or equal to n .

It is false. Because f(x) has n roots but we are not sure those are distinct or not and we are not sure those are the maximum number of roots or not. because to be n as degree it has to have n at most distinct roots. So if we only know that there are n roots but we dont know they are maximum or not and distinct or not. So it is false.

Related Questions

Navigate

• Browse (All)
• Browse M
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.