Answer to Question #154694 in Combinatorics | Number Theory for Sohel

Question

If a is prime, then a is irreducible, but not conversely test the validity of the statement.

Answer 1

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[$\sqrt{-3}$ ] =

{a + ib$\sqrt3$ : 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$\sqrt{3}$ )(c + id$\sqrt{3}$ );

 take complex conjugates to get

2=(a − ib$\sqrt{3}$ )(c − id$\sqrt{3}$ ).

Now multiply these two equations to obtain

4 = ($a^2$ + 3$b^2$ )($c^2$ + 3$d^2$ ).

Each factor on the right must be a divisor of 4, and there is no way that $a^2$ + 3$b^2$ can be 2.

Thus one of the factors must be 4 and the other must be 1. If, say, $a^2$ + 3$b^2$ = 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$\sqrt{3}$ )(1 − i$\sqrt{3}$ ) = 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.