If a is prime, then a is irreducible, but not conversely test the validity of the statement.
Using the standard notation m|n to indicate that m divides n.
Suppose that a is
prime, and that a = bc. Then certainly a|bc, so by definition of prime, a|b or a|c. Let us say a|b.
So, if b = ad then b = bcd, so cd = 1 and therefore c is a unit. (Note that b cannot be 0, for
if so, a = bc = 0, which is not possible since a is prime.)
Similarly, if a|c with c = ad then
c = bcd, so bd = 1 and b is a unit.
Showing that a is product of two units. And therefore a is irreducible.
Using a counter example of an irreducible element that is not prime.
prime,
consider R = Z[ ] =
{a + ib : a, b ∈ Z}; in R,2 is irreducible but not prime.
To see this, first suppose that we
have a factorization of the form
2=(a + ib )(c + id );
take complex conjugates to get
2=(a − ib )(c − id ).
Now multiply these two equations to obtain
4 = ( + 3 )( + 3 ).
Each factor on the right must be a divisor of 4, and there is no way that + 3 can be 2.
Thus one of the factors must be 4 and the other must be 1. If, say, + 3 = 1, then
a = ±1 and b = 0. Thus in the original factorization of 2, one of the factors must be a
unit, so 2 is irreducible.
Finally, 2 divides the product (1 + i )(1 − i ) = 4, so if 2
were prime, it would divide one of the factors, which means that 2 divides 1, a contradiction
since 1/2 is not an integer.
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