Show that the direct sum of two integral domains (E and H) is never an integral domain.
To show that the direct sum of two integral domains E and H is never an integral domain, we need to prove that it does not satisfy the defining property of an integral domain.
The direct sum of two integral domains E and H, denoted as E⊕H, is defined as the set of all ordered pairs (e, h) where e ∈ E and h ∈ H. The operations of addition and multiplication are defined component-wise:
(a, b) + (c, d) = (a + c, b + d) for all (a, b), (c, d) ∈ E⊕H
(a, b) · (c, d) = (a · c, b · d) for all (a, b), (c, d) ∈ E⊕H
To prove that E⊕H is not an integral domain, we need to find a non-zero element in E⊕H that has a zero divisor. A zero divisor is an element in a ring such that its product with a non-zero element results in zero.
Consider the following elements in E⊕H:
(e, 0) and (0, h), where e ≠ 0 and h ≠ 0
We can observe that:
(e, 0) · (0, h) = (e · 0, 0 · h) = (0, 0)
Since the product of (e, 0) and (0, h) is equal to (0, 0), which is the additive identity of E⊕H, we have found a zero divisor. This implies that E⊕H is not an integral domain.
In summary, E⊕H is not an integral domain because it contains zero divisors.