Given:
Four statements are given in which we have to find out which one is not a Euclidean domain.
Concept Used:
Euclidean domain: Let R be an integral domain i.e., let R be a commutative ring without zero divisors. Then R is said to be a Euclidean ring if to every non-zero element a ∈ R we can assign a non-negative integer d(a) such that:
- For all a, b ∈ R, both non-zero, d(ab) ≥ d(a).
- For any a, b ∈ R and b = 0, there exist q, r ∈ R such that a = qb + r where either r = 0 or d(r) < d(b).
Solution:
Option 1: Z[i] is a ring of Gaussian integers is a Euclidean domain..
Let (G, +, .) be the ring of Gaussian integers where G = {x + iy : x, y ∈ Z}
Let the d function on the non-zero elements of G be defined as d(x + iy) = x2 + y2 ∀ 0 + i0 = x + iy ∈ G.
Now if x + iy is a non-zero elements of G be defined non-negative integer. Thus we have assigned a non-negative integer to every non-zero element of G.
If x + iy and m + in are two non-zero elements of G, then d[ (x + iy) (m +in)] = d [(xm - ny) + i(my + xn)]
⇒ (xm - ny)2 + (my + xn)2
⇒ x2 m2 + n2 y2 + m2 y2 + x2 n2
⇒ (x2 + y2) (m2 + n2)
⇒ ≥ x2 + y2 {since m2 + n2 ≥ 1}
Thus d[(x + iy) (m + in)] ≥ d (x + iy).
Now to show the existence of division algorithm in G.
Let α ∈ G and let β be a non-zero element of G. Let α = x + iy and β = m + in. Define a complex number γ by the equation
γ = βα = m+inx+iy = m2+n2(x+iy)(m−in) = p + iq, where p, q are rational numbers.
Here γ is not necessarily a Gaussian integer.
Also division by β is possible since β = 0.
Let p' and q' be the nearest integers to p and q respectively. Then obviously | p - p' | ≤ 21, | q - q' | ≤ 21.
Let γ' = p' + iq'. Then γ' is a gaussian integer.
Now γ = βα ⇒ α = γ(\beta\) ⇒ α = γ' β + γ(\beta\) - γ' β.
Thus α = γ′β+(γ−γ′)β..........................(1)
Hence (γ−γ′)β is also a Gaussian integer.
Now if p and q are both integers then p = p', q = q'.
So γ−γ′ = (p - p') + i(q - q') = 0 + i0. Thus (γ−γ′)β = 0 + i0
If p and q are not both integers, then (γ−γ′)β is a non-zero Gaussian integer and we have
d[(γ−γ′)β] = d [{(p - p') + i(q - q')} (m + in)]
⇒ [(p - p')2 + (q - q')2] (m2 + n2) = [(p - p')2 + (q - q')2] d(β)
⇒ ≤ (1/4 + 1/4) d(β)
⇒ 1/2d(β) < d(β).
Thus α = γ′β+(γ−γ′)β where γ' and (γ−γ′)β are Gaussian integers and either (γ−γ′)β = 0 or d[(γ−γ′)β] < d(β).
Hence the gaussian integers is a Euclidean domain.
Option 2: Every field F with the norm N as N(a) = 1 is a Euclidean Domain.
Since every field F with the norm N as N(a) = 1 will satisfy both the conditions of Euclidean domain so it is a Euclidean domain.
Option 3: The ring of integers is a Euclidean domain.
Let (z, +, .) be the ring of integers where Z = {........., - 3, - 2, - 1, 0, 1, 2, 3,.....}.
Let the d function on the non-zero elements of Z be defined as d(a) = |a| ∀ 0 = a ∈ Z.
Now if 0 = a ∈ Z, then |a| is a non-zero integer. Thus we have assigned a non-zero integer to every non-zero element
a ∈ Z.
[d (- 5)] = | - 5 | = 5, d(- 1) = | - 1 | = 1.
Further if a, b ∈ Z and are both non zero, then
|ab| = |a| |b|
⇒ |ab| ≥ |a| {since |b| ≥ 1 if 0 = b ∈ Z}
⇒ d(ab) ≥ d(a).
Finally, we know that if a ∈ Z and 0 = b then there exist two integers q and r such that
a = qb + r where 0 ≤ r < |b|
either r = 0 or 1 ≤ r < |b|
⇒ either r = 0 or d(r) < d(b).
Hence the ring of integers is a Euclidean domain.
Option 4: The field of rational numbers under absolute value given by N(a) = |a|.
Since the field of rational number is not an integral domain because it does not have zero divisor, so it is not a Euclidean domain.
∴ Option 4 is correct.
