Question #154694

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


1
Expert's answer
2021-01-12T17:05:58-0500

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[3\sqrt{-3} ] =

{a + ib3\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 + ib3\sqrt{3} )(c + id3\sqrt{3} );


 take complex conjugates to get


2=(a − ib3\sqrt{3} )(c − id3\sqrt{3} ).


Now multiply these two equations to obtain


4 = (a2a^2 + 3b2b^2 )(c2c^2 + 3d2d^2 ).


Each factor on the right must be a divisor of 4, and there is no way that a2a^2 + 3b2b^2 can be 2.


Thus one of the factors must be 4 and the other must be 1. If, say, a2a^2 + 3b2b^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 + i3\sqrt{3} )(1 − i3\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.


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