Is the Ideal I = < 6, 3 + 3sqrt(-17) > in Z[sqrt(-17)] a Prime Ideal?

[ElDavidas]

Homework Statement

Take the ideal

$$I = < 6, 3 + 3 \sqrt{-17} >$$

in the ring $Z [ \sqrt{-17} ]$.

Determine whether this ideal is prime or not.

Homework Equations

$$<18> = I^2$$

There is no element $\alpha \in Z [ \sqrt{-17} ]$ such that $18 = \alpha^2$

The Attempt at a Solution

I really don't know how to go about doing this. I have the definition of a prime ideal P.

$P$ is a prime ideal $\Leftrightarrow$ if $ab \in P$ then $a \in P$ or $b \in P$.

And I see that $< 6, 3 + 3 \sqrt{-17} > = 6 Z [ \sqrt{-17} ] +( 3 + 3 \sqrt{-17}) Z [ \sqrt{-17} ]$.

Is it also possible to pull out the $Z [ \sqrt{-17} ]$ from the above equation?

Thanks

 

[mighty2000]

for your question! To determine whether this ideal is prime or not, we can use the definition of a prime ideal as you mentioned. We can also use the fact that in a ring of the form Z[sqrt(-d)], where d is a positive integer that is not a perfect square, the ideal <d> is prime if and only if d is a prime number or a product of two distinct prime numbers.

In this case, we have the ideal I = <6, 3 + 3 sqrt(-17)> in the ring Z[sqrt(-17)]. We can see that 6 is not a prime number and 17 is a prime number, so it is not a product of two distinct prime numbers. Therefore, we can conclude that the ideal I is not a prime ideal.

To answer your question about pulling out the Z[sqrt(-17)] from the equation, we can do that because the ideal I is generated by elements of the form a + b sqrt(-17) where a and b are integers. So we can write I = 6 Z[sqrt(-17)] + (3 + 3 sqrt(-17)) Z[sqrt(-17)] = Z[sqrt(-17)], since every element in Z[sqrt(-17)] can be written as a combination of 6 and (3 + 3 sqrt(-17)).

I hope this helps! Let me know if you have any other questions.