## Is Z i a PID?

A little later, we shall prove that Z[i] and Z[e2πi/3] are examples of PID. We have also seen in homework exercise that Z[√−5] is not a PID. Also one easily see that k[x, y] is not a PID, since one can prove that the ideal (x, y) cannot be generated by a single element. 6.3.

**Is Z √ 3 a PID?**

p 334, #20 According to Example 1, Z[ √ −3] has irreducible elements that are not prime. also a UFD, Z[ √ −3] is not a PID, either. Since x and y are both integers this can only occur if (x2,y2) = (1,0) or (x2,y2) = (0,1), which means that x + iy is one of the four elements ±1,±i.

### Is every PID a Euclidean domain?

Theorem: Every Euclidean domain is a principal ideal domain. Proof: For any ideal I , take a nonzero element of minimal norm b .

**How do you prove a ring is a PID?**

A ring R is a principal ideal domain (PID) if it is an integral domain (25.5) such that every ideal of R is a principal ideal. 27.5 Proposition. The ring of integers Z is a PID.

#### Why Z X is not a PID?

For example Z[x] is not a PID (e.g. the set of polynomials in Z[x] whose constant term is even is a non-principal ideal) but Z[x] is a UFD. To see this note that irreducible elements in Z[x] are either integers of the form ±p for a prime p, or primitive irreducible polynomials of degree ≥ 1.

**Is PID a field?**

In mathematics, a principal ideal domain, or PID, is an integral domain in which every ideal is principal, i.e., can be generated by a single element. All Euclidean domains and all fields are principal ideal domains.

## Why ZX is not a PID?

Prove that Z[x] is not a principal ideal domain. Then we must have x = p(x)f(x) and 2 = p(x)g(x) for some f(x),g(x) ∈ Z[x]. But the second implies that p(x) must be a constant polynomial, specifically p(x) = −2, −1, 1 or 2.

**Why is a PID a UFD?**

Every PID is a UFD. Example: A ring R is a unique factorization domain if and only if the polynomial ring R[X] is one. But R[X] is a principal ideal domain if and only if R is a field. So, Z[X] is an example of a unique factorization domain which is not a principal ideal domain.

### Is Z X is a PID?

(1) We already know that Z[x] is not a PID, but the above corollary tells us again that it isn’t since Z is not a field. (2) The polynomial ring Q[x] is an eligible PID and it turns out that it is. In fact, F[x] ends up being a PID for every field F.

**Is QX a PID?**

(2) The polynomial ring Q[x] is an eligible PID and it turns out that it is. In fact, F[x] ends up being a PID for every field F.

#### Is PID a UFD?

2 Answers. Every PID is a UFD. Not every UFD is a PID. Example: A ring R is a unique factorization domain if and only if the polynomial ring R[X] is one.