In any integral domain, prime implies irreducible. In any Unique Factorization Domain, irreducibles implies prime. So _a_ place to look are integral domains which aren't unique factorization domains. The go-to example of this is Z[√(−5)]. In this ring, all elements are of the form a+b√(−5), where a and b are integers. In this ring, 2 is an irreducible element which is not a prime. First, let's prove 2 is irreducible. Suppose 2 = (a+b√(−5))(c+d√(−5)), and we want to show that one of these factors is a unit. Perhaps the best way to do this is to use the distance formula in the complex plane and to note that distance is multiplicative. This means |2| = |a+b√(−5)|*|c+d√(−5)| = sqrt(a^2+5b^2)*sqrt(c^2+5d^2). Or better yet, looking at _squares of distances._ So we have 4 = (a^2+5b^2)(c^2+5d^2) Notice this is a factorization of 4 into _integers._ So the only possibilities are: a^2+5b^2 = 2 and c^2+5d^2 = 2 a^2+5b^2 = 4 and c^2+5d^2 = 1 (or vice versa, but the labels here are arbitrary, so we don't need to consider the other cases). Consider the first case (a^2+5b^2 = 2 and c^2+5d^2 = 2). This is impossible unless b = d = 0. But then a and c have to be irrational, not integers, which is a contradiction. So the first case is not possible. Consider the second case (a^2+5b^2 = 4 and c^2+5d^2 = 1). Here, this is impossible unless d = 0. So we get c^2 = 1, which means c is 1 or -1. Hence, we have c+d√(−5) = 1 or -1, a unit. Therefore, the only way to have 2 as a product of two elements in Z[√(−5)] is if one them is a unit. Hence, 2 is irreducible in Z[√(−5)]. But 2 is not prime in Z[√(−5)]. This is because 2*3 = 6, so 2 divides 6. However, 6 = (1+√(−5))(1−√(−5)), but 2 doesn't divide either 1+√(−5) or 1−√(−5). So 2 divides a product without dividing either of the factors. So 2 can't be prime.
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What is an example of a ring where the irreducible elements and the primes are not the same?
In any integral domain, prime implies irreducible. In any Unique Factorization Domain, irreducibles implies prime.
So _a_ place to look are integral domains which aren't unique factorization domains. The go-to example of this is Z[√(−5)]. In this ring, all elements are of the form a+b√(−5), where a and b are integers.
In this ring, 2 is an irreducible element which is not a prime.
First, let's prove 2 is irreducible. Suppose 2 = (a+b√(−5))(c+d√(−5)), and we want to show that one of these factors is a unit. Perhaps the best way to do this is to use the distance formula in the complex plane and to note that distance is multiplicative. This means |2| = |a+b√(−5)|*|c+d√(−5)| = sqrt(a^2+5b^2)*sqrt(c^2+5d^2). Or better yet, looking at _squares of distances._
So we have 4 = (a^2+5b^2)(c^2+5d^2)
Notice this is a factorization of 4 into _integers._ So the only possibilities are:
a^2+5b^2 = 2 and c^2+5d^2 = 2
a^2+5b^2 = 4 and c^2+5d^2 = 1 (or vice versa, but the labels here are arbitrary, so we don't need to consider the other cases).
Consider the first case (a^2+5b^2 = 2 and c^2+5d^2 = 2). This is impossible unless b = d = 0. But then a and c have to be irrational, not integers, which is a contradiction. So the first case is not possible.
Consider the second case (a^2+5b^2 = 4 and c^2+5d^2 = 1). Here, this is impossible unless d = 0. So we get c^2 = 1, which means c is 1 or -1. Hence, we have c+d√(−5) = 1 or -1, a unit. Therefore, the only way to have 2 as a product of two elements in Z[√(−5)] is if one them is a unit. Hence, 2 is irreducible in Z[√(−5)].
But 2 is not prime in Z[√(−5)]. This is because 2*3 = 6, so 2 divides 6. However, 6 = (1+√(−5))(1−√(−5)), but 2 doesn't divide either 1+√(−5) or 1−√(−5). So 2 divides a product without dividing either of the factors. So 2 can't be prime.
Thank you!