Question

Prove that an ideal (P) in a PID is maximal if and only if p is irreducible.

Short answer

It is proved that ideal (P) in a PID is maximal if and only if p is irreducible.

Step-by-step solution

Defining principal ideal domain (PID)

PID

A principal ideal domain (PID) is an integral domain in which every ideal is principal.

Proving that p is irreducible

Let’s assume R is a PID and (P) is amaximal ideal in R.

Suppose p=ab for some ab $\in$R .

Then, a l p .

Therefore,(p)$\subset$(a) or (p) = (a).

So, p and a are associates in R.

Assume (p)⊂(a)

Since (p)⊂(a) and p is maximal in R, so (a)=R .

This means, a is unit in R.

Since p is a non-unit element in R and p=ab , which implies either of aand b is a unit of R.

Therefore, p is irreducible in R.

Proving that ideal (p)  in a PID is maximal

Now, we will assume that p is irreducible in R.

Let Ibe a proper ideal in R and it contains (p) .

Then, (p) ⊂|⊂ R

As we know R is a PID and I is principal ideal, there must be r in R such that, l=(r) . .

Since (p) ⊂| ⇒(p)⊂(r) , p=cr for some c∈R .

Since P is irreducible, therefore implies that either r is a unit or r is an associate in R.

If we take r is a unit then, (r)=R and this cannot be true because (r)⊂R .

If r is an associate of p then,(p)=(r) , again it is not true because (p)⊂ (r) .

Therefore,(p)⊂|⊂ R implies either(p)=l or l=R .

Hence, ideal (p) in a PID is maximal.

So, from the above result, we can conclude that ideal (p) in a PID is maximal if and only if p is irreducible.