If R is an integral domain with unit element, prove that any unit in R[x] must already be a unit in R.

Related BrainMass Solutions

• If R is an integral domain with unit element, prove that any unit in R[x] is a unit in R and any unit in R is also a unit in R[x].

  127645 If R is an integral domain with unit element, prove that any unit in R[x] is a unit in R and any unit in R is also a unit in R[x].

• If R is an Integral Domain, then so is R[x].

  138887 If R is an Integral Domain, then so is R[x]. If R is an integral domain, then so is R[x]. Prove that if R is an integral domain, then R[x] is also an integral domain. See attached file for full problem description.

• Finite ring proofs

  Suppose that R has identity, prove that if x E R is not a zero divisor, then it is a unit Proof: (a) Since R is finite, assume |R| = n, then for any non-zero element x in R, we have x^n = x.

• Ring proofs

  Prove that the polynomial f(x) is a unit in R[x] if and only if f(x) is a nonzero constant which is a unit in R. b.

• If R is an integral domain, then so is R[x_1, x_2, ... , x_n] . Or, Prove that if R is an integral domain, then R[x_1, x_2, ... , x_n] is also an integral domain.

  Or, Prove that if R is an integral domain, then R[x_1, x_2, ... , x_n] is also an integral domain.

• Integral Domains and Fields : Embedding Theorem

  Proof: R is an integral domain, it means R is a commutative ring with the multiplicative unit e. Suppose R C F, where F is a field. Now we construct a map f: Q->F. For any element x=a/b in Q, where a,b are integers, we define f(x)=f(a/b)=ae/be.

• Ring and fields

  If F is a field, then the field of fractions Q of F consists of elements of the form ab^{-1} with a in F and b nonzero in F. If x is a nonzero element of F, then x = (xb)b^{-1}, for any nonzero b in F.

• Local ring proof

  Because if M is contained by another ideal I, then I contains some element x that does not belong to M, then x must be a unit. So we can find some y in R, such that yx = 1. Since I is an ideal, then yx = 1 is in I, thus I = R.

• Ideal proof

  226937 Ideals Let R be a commutative ring with 1. Prove that the principal ideal generated by x in the polynomial ring R[x] is a prime ideal if and only if R is an integral domain. Prove that (x) is a maximal ideal if and only f R is a field.