An integral domain is a commutative ring with unity that has no zero divisors. This means that if a and b are nonzero elements of the ring, then their product ab is also nonzero. In this article, we will discuss how to prove that a given ring is an integral domain.

Prove that a Given Ring is an Integral Domain

Steps

To prove that a given ring is an integral domain, we need to show that it satisfies the three defining properties of an integral domain: commutativity, associativity, and the absence of zero divisors. Here are the steps to follow:

1. Show that the ring is commutative: To prove that the ring is commutative, we need to show that for any two elements a and b in the ring, we have a*b = b*a. This can be done by using the properties of the ring and the distributive law of multiplication over addition.
2. Show that the ring is associative: To prove that the ring is associative, we need to show that for any three elements a, b, and c in the ring, we have (a*b)*c = a*(b*c). This can be done by using the properties of the ring and the associative law of multiplication.
3. Show that the ring has no zero divisors: To prove that the ring has no zero divisors, we need to show that for any two nonzero elements a and b in the ring, we have a*b ≠ 0. This can be done by assuming that a*b = 0 and then showing that either a or b must be zero, which contradicts our assumption.

Example

Consider the ring R = {a + b*sqrt(2) | a, b ∈ ℤ} with the usual addition and multiplication of real numbers. To prove that R is an integral domain, we need to show that it satisfies the three properties of an integral domain.

1. Commutativity: Let a, b ∈ R. Then a = x + y*sqrt(2) and b = w + z*sqrt(2) for some integers x, y, w, and z. We have a*b = (x + y*sqrt(2))(w + z*sqrt(2)) = (xw + 2yz) + (xz + yw)*sqrt(2) and b*a = (w + z*sqrt(2))(x + y*sqrt(2)) = (xw + 2yz) + (xz + yw)*sqrt(2). Therefore, a*b = b*a, and R is commutative.
2. Associativity: Let a, b, and c ∈ R. Then a = x + y*sqrt(2), b = w + z*sqrt(2), and c = u + v*sqrt(2) for some integers x, y, w, z, u, and v. We have (a*b)*c = ((xw + 2yz) + (xz + yw)*sqrt(2))(u + v*sqrt(2)) = [(xw + 2yz)u + 2(xz + yw)v] + [(xz + yw)u + (xw + 2yz)v]*sqrt(2) and a*(b*c) = (x + y*sqrt(2))[(wu + 2(zu + vw)*sqrt(2)] = [(xwu + 2xyzu + 2yzwu + 4yzvw) + (xzu + xvw + ywu + yvw)*sqrt(2)]. Therefore, (a*b)*c = a*(b*c), and R is associative.
3. Absence of zero divisors: Let a, b ∈ R such that a*b = 0. Then we have (x + y*sqrt(2))(w + z*sqrt(2)) = 0 for some integers x, y, w, and z. This implies that xw + 2yz + (xz + yw)*sqrt(2) = 0. If xz + yw ≠ 0, then we have yw = -xz and zw = -xy, which implies that z² = 2y², a contradiction since √2 is irrational. Therefore, xz + yw = 0, and we have xw = -2yz. Since x, y, w, and z are integers, this implies that x = 0 or w = 0. If x = 0, then yw = 0, which implies that y = 0 or w = 0. Similarly, if w = 0, then xz = 0, which implies that z = 0 or x = 0. Therefore, either a = 0 or b = 0, and R has no zero divisors.

Therefore, we have shown that R is commutative, associative, and has no zero divisors, which implies that R is an integral domain.

Conclusion

To prove that a given ring is an integral domain, we need to show that it satisfies the three defining properties of an integral domain: commutativity, associativity, and the absence of zero divisors. By following the steps outlined in this article, we can show that a given ring is an integral domain.