Verified answer

21 = 3 * 7 = (1 + 2√(-5))(1 - 2√(-5)) is a desired counterexample.

Now, we need to show that each of the four factors above are irreducible over Z[√(-5)].

I'll demonstrate this with one of the four factors; you can fill in the remaining details.

Suppose that 3 = (a + b√(-5)) (c + d√(-5)) for some integers a, b, c, d, where neither

(a + b√(-5)) nor (c + d√(-5)) are units (that is -1, or 1) in Z[√(-5)].

Taking conjugates:

3 = (a - b√(-5)) (c - d√(-5)).

Multiply the last two equations together:

3^2 = (a^2 + 5b^2)(c^2 + 5d^2), an equation in integers!

Since we want (a - b√(-5)) and (c - d√(-5)) to both not be units, we need

a^2 + 5b^2 = 3 = c^2 + 5d^2; it's easy to check that these have no solutions in Z.

Hence, 3 is irreducible in Z[√(-5)].

I hope this helps!

A simple counter example.

21=7.3 = (4 + √5 i ) (4 - √5 i ) ( for i = √-1 ) . So I assume that you know the stuff behind the ideal class and class number. Ideal class represents the failure that you have asked. For this case the ideal class is no more trivial .

Thank you.