Determine prime in non-UFD

  • Thread starter bessletama
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I am trying to prove that [itex]11[/itex] is a prime in [itex]\mathbb{Z}[\sqrt{-5}][/itex].

I noticed that [itex]\mathbb{Z}[\sqrt{-5}][/itex] is not a UFD so I cannot show that it is irreducible then conclude it is prime.

I know that that an ideal is prime if and only if the quotient ring is a domain.
I was wondering if it is correct for me to show that
[itex]\mathbb{Z}[\sqrt{-5}]/(11)\cong\mathbb{Z}_{11}[x]/(x^2+1)[/itex]
If this is true then I can conclude that [itex]\mathbb{Z}[\sqrt{-5}][/itex] is a domain because
[itex]\mathbb{Z}_{11}[x]/(x^2+1)[/itex] is a finite field.
Thank you

EDIT: OMG, Made a huge typo originally. The ring is [itex]\mathbb{Z}[\sqrt{-5}][/itex] not [itex]\mathbb{Z}[\sqrt{5}][/itex]
 
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Answers and Replies

  • #2
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I am trying to prove that [itex]11[/itex] is a prime in [itex]\mathbb{Z}[\sqrt{5}][/itex].

I noticed that [itex]\mathbb{Z}[\sqrt{5}][/itex] is not a UFD so I cannot show that it is irreducible then conclude it is prime.

I know that that an ideal is prime if and only if the quotient ring is a domain.
I was wondering if it is correct for me to show that
[itex]\mathbb{Z}[\sqrt{5}]/(11)\cong\mathbb{Z}_{11}[x]/(x^2+1)[/itex]
If this is true then I can conclude that [itex]\mathbb{Z}[\sqrt{5}][/itex] is a domain because
[itex]\mathbb{Z}_{11}[x]/(x^2+1)[/itex] is a finite field.
Thank you

You could conclude that if you can show the isomorphism [itex]\,\mathbb{Z}[\sqrt{5}]/(11)\cong\mathbb{Z}_{11}[x]/(x^2+1)\,[/itex] .

In fact, you'd conclude something stronger: the ideal [itex]\,(11)\subset \Bbb Z[\sqrt 5]\,[/itex] is then maximal and thus prime.

DonAntonio
 
  • #3
mathwonk
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i guess the first step for me would be to try to find a square root of -5 in the ring Z11.
 

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