# Math Help - UFDs, primes and irreducibles - Dummit and Foote - Chapter 8

1. ## UFDs, primes and irreducibles - Dummit and Foote - Chapter 8

I am reading Dummit and Foote Chapter 8, Section 8.3 UFDs.

On page 284 (see attached) Dummit and Foote prove Proposition 10 which shows the following:

"In an integral domain a prime element is always irreducible."

D&F then state:

"It is not true in general that an irreducible element is necessarily prime."

They give as an example the element 3 in the quadratic integer ring $R =\mathbb{Z} [\surd -5 ]$

They assert that the element 3 is irreducible but not prime.

I am struggling to show rigorously that 3 is irreducible in R, despite D&F's reference to the calculations on page 273 (see attached)

That is to show that whenever $3 = ( a_1 + b_1 \surd -5 ) (a_2 + b_2 \surd -5 )$

then one of $( a_1 + b_1 \surd -5 ) , (a_2 + b_2 \surd -5 )$ must be a unit in $\mathbb{Z} [\surd -5 ]$

Peter

2. ## Re: UFDs, primes and irreducibles - Dummit and Foote - Chapter 8

Originally Posted by Bernhard
"It is not true in general that an irreducible element is necessarily prime."

They give as an example the element 3 in the quadratic integer ring $R =\mathbb{Z} [\surd -5 ]$

They assert that the element 3 is irreducible but not prime.

I am struggling to show rigorously that 3 is irreducible in R, despite D&F's reference to the calculations on page 273 (see attached)

You've had this shown to you, that 3 is irreducible in this ring, at least twice in the last month. Look them over again:

Existence of Greatest Common Divisors - Dummit and Foote Ch8

Euclidean Domains - Dummit and Foote - Chapter 8 - Section 8.1 - Example on Quadratic

3. ## Re: UFDs, primes and irreducibles - Dummit and Foote - Chapter 8

Thanks for pointing that out ... And apologies for forgetting this ... My day job intrudes on my Maths to the point I sometimes have to revisit problems.

Will now go to the posts you mention

Thanks again,

Peter

4. ## Re: UFDs, primes and irreducibles - Dummit and Foote - Chapter 8

if 3 were reducible in Z[√(-5)], say 3 = ab where a,b are not units, we would have N(3) = 9 = N(ab) = N(a)N(b) (here N is the field norm, or the square of the complex modulus. this is always a positive integer).

if N(a) OR N(b) = 1, then it must be ±1, and both of these are units.

hence N(a) = N(b) = 3.

suppose a = c + d√(-5) where c,d are integers.

N(a) = 3 implies c2+5d2 = 3.

thus d = 0, (or else N(a) > 3) and c2 = 3 has no integer solution.