If F is a field, then it has no proper ideal. T OF
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- Since this section presents a method for constructing a field of quotients for an arbitrary integral domain D, we might ask what happens if D is already a field. As an example, consider the situation when D=5. a. With D=5, write out all the elements of S, sort these elements according to the relation , and then list all the distinct elements of Q. b. Exhibit an isomorphism from D to Q.Let R be a commutative ring with unity whose only ideals are {0} and R Prove that R is a field.(Hint: See Exercise 30.)Suppose S is a subset of an field F that contains at least two elements and satisfies both of the following conditions: xS and yS imply xyS, and xS and y0S imply xy1S. Prove that S is a field. This S is called a subfield of F. [Type here][Type here]
- Prove that if F is an ordered field with F+ as its set of positive elements, then F+nen+, where e denotes the multiplicative identity in F. (Hint: See Theorem 5.34 and its proof.) Theorem 5.34: Well-Ordered D+ If D is an ordered integral domain in which the set D+ of positive elements is well-ordered, then e is the least element of D+ and D+=nen+.Let ab in a field F. Show that x+a and x+b are relatively prime in F[x].15. Prove that if is an ideal in a commutative ring with unity, then is an ideal in .