We will use the fact that every prime ideal in a P.I.D. is a maximal ideal, which we have proved here.

Prove that if R[x] is a P.I.D. and R commutative then R is a field

Proof: given R[x] is a principal ideal domain. We have proven here R is an integral domain since it is a subring of R[x]. Further, we have that R[x]/(x) \cong R and since R is an integral domain, we do know that (x) is a prime ideal. But every prime ideal is a maximal ideal and therefore R[x]/(x) must be a field, which implicates that R is a field.

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The reader should know that only proving that the subring has no zero divisors is not enough. A subring should also contain the identity element.

Prove that a subring R of the PID R[x] is an integral domain

Proof: we take the principal ideal domain R[x] and let R be its subring. Let’s assume that we have the non-zero elements a,b \in R such that ab = 0, i.e., a and b are zero divisors. Since R \subset R[x], we have that ab = 0, but that is a contradiction since R[x] has no zero divisors since R[x] is an integral domain.

Now we need to figure out if R contains the identity element. But, R[x] contains the identity element since it is a principal ideal domain, and therefore, R too (so 1_R \neq 0_R).

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In order to prove that statement, let’s recall what actually a least common multiple is. Let a,b \in R nonzero elements, where R is a P.I.D. A least common multiple of a and b is an element e of R such that:

1. a \mid e and b \mid e
2. if a \mid e' and b \mid e', then e \mid e'

Prove that any two nonzero elements of a principal ideal domain have a least common multiple

Proof: let (a) \cap (b) = (e). Then (e) \subseteq (a) and (e) \subseteq (b). This means that there exists r_1,r_2 \in R such that a=r_1e and b=r_2e. This implies that a \mid e and b \mid e, which proves the first statement.

If a \mid e' and b \mid e', then (e') \subseteq (a) and (e') \subseteq (b). Since (a) \cap (b) = (e), we must have that e' \in (e), which means there exists r \in R such that er = e', and therefore, e \mid e'.

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Prove that any two ideal (a) and (b) in PID are comaximal iff gcd(a,b)=1

“\Rightarrow“: let d be the generator for the principal ideal generated by a and b, i.e.,

\begin{equation*}
(d) = (a,b) = \{ax+by \ | \ x,y\in R\}.
\end{equation*}

“\Leftarrow“: reverse the previous proof.

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The best way to tackle this problem is by using the fact every prime ideal is a maximal ideal in a P.I.D.

Prove that a quotient of a P.I.D. by a prime ideal is again a P.I.D.

Proof: let P be a prime ideal of the principal ideal domain R. Every prime ideal is a maximal ideal in R, which we have proven here. So, we get that R/P is a field, again, which we have seen earlier here. Since R/P, it is definitely a P.I.D. too, which completes this proof.

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The easiest way to prove this is to use the definitions of a prime ideal, maximal ideal, and P.I.D. In the proof below, we want to show that P is either a maximal ideal or a P.I.D.

Prove that every prime ideal in P.I.D. is a maximal ideal

Proof: Let P = (p) and M = (m) be a prime ideal and a maximal ideal of the principal ideal domain R, respectively. Take for the assumption that P is contained in M. Then there exists an element r \in R such that p = rm. Since P is a prime ideal, we have that either r \in P or m \in P. If m \in P, then P is the maximal ideal. If r \in P, then r = px for some x \in R. Then p = rm = pxm which means that xm = 1, so this implies that m is a unit. So P = R.

Therefore, we have that P = M) or P = R).

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