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Proving an element of Z[i] is irreducible?



As above (I didn't want to type it out again so hopefully this is clear from the picture.)

Any help with 2?
Original post by cliveb2016

Any help with 2?


If x is irreducible in the Gaussian integers, then N(x) is prime in the ordinary integers. Can you take it from there, using the first part?
Reply 2
Original post by Gregorius
If x is irreducible in the Gaussian integers, then N(x) is prime in the ordinary integers. Can you take it from there, using the first part?


Suppose a is irreducible in Z, then N(a) is prime in Z but N(a)=pq which is not prime in Z (contradiction) so a is not irreducible in Z and so a cannot be written as a product of 1 irreducible so it must be written as a product of more than one but strictly less than 2 irreducibles hence it must be the product of exactly 2 irreducibles.

Is that right?

I got stuck since I didn't know that
If x is irreducible in the Gaussian integers, then N(x) is prime in the ordinary integers.

I thought that implication is only true the reverse way i.e

N(x) is prime in the ordinary integers then x is irreducible in the Gaussian integers.

Could you explain why your statement holds. (I tried to prove with no luck)
Original post by cliveb2016

Could you explain why your statement holds. (I tried to prove with no luck)


It doesn't - I got it wrong!

So, let's try again: if x = a+ ib is a Gaussian integer, with a and b non-zero, then x irreducible implies N(x) prime. So you can apply the reasoning you have already worked out. Then you have to deal with the either a or b equals to zero cases.

Sorry about that.
Reply 4
Original post by Gregorius
It doesn't - I got it wrong!

So, let's try again: if x = a+ ib is a Gaussian integer, with a and b non-zero, then x irreducible implies N(x) prime. So you can apply the reasoning you have already worked out. Then you have to deal with the either a or b equals to zero cases.

Sorry about that.


No worries thanks for the help.

I think I can finish the proof with the claim "if x = a+ ib is a Gaussian integer, with a and b non-zero, then x irreducible implies N(x) prime."

Would you be able to elaborate why this is true?

thanks
Original post by cliveb2016

Would you be able to elaborate why this is true?


Getting into the realms of bookwork - page 69 of Baker's "Concise Introduction", for example. Did this question arise from an algebraic number theory course or has it been sprung on you in a lower level algebra course?

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