# Rings and Ideals

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• Jan 15th 2009, 08:02 AM
Inti
Rings and Ideals
Need some help...

Let $\mathbb{Z}[\sqrt{-5}]$ the ring of gaussian integers.

1)Show that $<3>=<3;1+\sqrt{-5}><3;1-\sqrt{-5}>$
( $$ means the ideal generated by $a$)

2)Prove that the ideal on the right is prime.

Thank you folks!
• Jan 15th 2009, 04:42 PM
NonCommAlg
Quote:

Originally Posted by Inti

Let $\mathbb{Z}[\sqrt{-5}]$ the ring of gaussian integers.

1) Show that $<3>=<3,1+\sqrt{-5}><3,1-\sqrt{-5}>$

Hint: we have $3=3^2 - (1+\sqrt{-5})(1-\sqrt{-5}).$ thus: $<3> \subseteq <3,1+\sqrt{-5}><3,1-\sqrt{-5}>.$ for the other side of the inclusion, note that $(1+\sqrt{-5})(1-\sqrt{-5})=6 \in <3>.$

Quote:

2) Prove that the ideals on the right is prime.

Hint: define the map $f: \mathbb{Z}[\sqrt{-5}] \longrightarrow \mathbb{Z}_3$ by $f(a+b\sqrt{-5})=a-b.$ show that $f$ is a well-defined surjective ring homomorphism and: $\ker f = <3, 1+\sqrt{-5}>.$
• Jan 16th 2009, 04:35 AM
Inti
thereīs one thing thatīs not clear to me:
Iīve just proved that $\ker f = <3,1+\sqrt{-5}>$, which means that $\mathbb{Z}[\sqrt{-5}]/<3,1+\sqrt{-5}>$ is isomorphic to $\mathbb{Z}3$, which is a field.
That allows me to say that the ideal is prime?
As far as Iīm concerned, the quotient of a ring by a prime ideal is a domain.
Iīm kind of confused with that...
But the help was very helpful (Nod)

And I have one more question: is the product of two ideals the set of all the products of the elements of each ideal??

Thanks!
• Jan 16th 2009, 04:55 PM
NonCommAlg
Quote:

Originally Posted by Inti

thereīs one thing thatīs not clear to me:
Iīve just proved that $\ker f = <3,1+\sqrt{-5}>$, which means that $\mathbb{Z}[\sqrt{-5}]/<3,1+\sqrt{-5}>$ is isomorphic to $\mathbb{Z}3$, which is a field.
That allows me to say that the ideal is prime?

that means the ideal is maximal and we know that every maximal ideal is prime.

Quote:

And I have one more question: is the product of two ideals the set of all the products of the elements of each ideal??

no. if $I$ and $J$ are two ideals, then $IJ=\{\sum_{k=1}^na_kb_k: \ n \in \mathbb{N}, \ a_k \in I, b_k \in J \}.$ however, for example, if you want to prove that $IJ \subseteq K,$ where $K$ is an ideal, then you only need to prove that

$ab \in K$ for all $a \in I, b \in J.$