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

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)

Can someone please help me produce a formal and rigorous demonstration that 3 is irreducible.

That is to show that whenever

then one of must be a unit in


Peter

Quote:

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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

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)

Can someone please help me produce a formal and rigorous demonstration that 3 is irreducible.

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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:

http://mathhelpforum.com/advanced-al...foote-ch8.html

http://mathhelpforum.com/advanced-al...quadratic.html

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

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.

Thanks Deveno ... Most helpful

Peter