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Chapter 10: Q10.2.4E (page 341)

Is a field a UFD?

Short Answer

Expert verified

Yes, every field is a UFD.

Step by step solution

01

Defining field and UFD

Field

A field is a set of elements that satisfy all field axioms related to both addition and multiplication and is a commutative division algebra.

UFD (Unique Factorization Domain)

It is an integral domain in which each non-zero and non-invertible element has a unique factorization.

02

Proving that every field is a UFD

According to the definition of UFD, in UFD we always consider the factorization of non-zero and non-invertible elements.

As we know that a field is a unit, therefore, we can conclude that every field is a UFD.

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Most popular questions from this chapter

LetR be an integral domain of characteristic 0 (see Exercises 41-43 in Section 3.2).

(a) Prove thatR has a subring isomorphic to $ℤ$ [Hint: Consider $\{n 1_{R} | n \in ℤ\}$ .]

(b) Prove that a field of characteristic 0 contains a subfield isomorphic to $ℚ$ . [Hint: Theorem 10.31.]

If $a = b c$ with $a \neq 0$ and b and cnon units, show that a is not an associate of b.

If the statement is true, prove it; if it is false, give a counterexample:

If a | b and c | d in R, then .
If a | b c | d in R, then ( a + c ) | ( b + d )

Let R be a Euclidean domain. If the function $δ$ is a constant function, prove that R is a field.

Let R be a PID. If (C) is a nonzero ideal in R, then show that there are only finitely many ideals in R that contain (C) . Consider the divisors of c

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