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Number Theory maximal ideal

Loiz

New member
Sep 16, 2012
3
hello all of you

Why is (2, 1+sqrt(-19)) a maximal ideal of Z[sqrt(-19)]? How to show such thing?

I thank you for any hints
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,705
hello all of you

Why is (2, 1+sqrt(-19)) a maximal ideal of Z[sqrt(-19)]? How to show such thing?

I thank you for any hints
There may be better ways to do this, using results from Galois theory. But if you want a bare-hands method, try this. Define a map $f:\mathbb{Z}[\sqrt{-19}] \to \mathbb{Z}/2\mathbb{Z}$ by $f(a+b\sqrt{-19}) = a-b \pmod2.$ Show that $f$ is a ring homomorphism with kernel $(2, 1+\sqrt{-19}).$ Since the range of $f$ is a simple ring, it follows that the kernel must be a maximal ideal.
 

Loiz

New member
Sep 16, 2012
3
There may be better ways to do this, using results from Galois theory. But if you want a bare-hands method, try this. Define a map $f:\mathbb{Z}[\sqrt{-19}] \to \mathbb{Z}/2\mathbb{Z}$ by $f(a+b\sqrt{-19}) = a-b \pmod2.$ Show that $f$ is a ring homomorphism with kernel $(2, 1+\sqrt{-19}).$ Since the range of $f$ is a simple ring, it follows that the kernel must be a maximal ideal.
Thank you very much for your help.

but i get stuck with this
kernel($f$)= {$a + b\sqrt{-19} : f(a+b\sqrt{-19})=0$} = {$a+b\sqrt{-19} : a -b \pmod2 $}={$a+b\sqrt{-19} : a=b \pmod2 $}

I am missing something crucial here cause i just dont understand why this is equivalent to the ideal $(2, 1+\sqrt{-19})$. What does an element in this ideal look like?

thanks again!
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,705
Thank you very much for your help.

but i get stuck with this
kernel($f$)= {$a + b\sqrt{-19} : f(a+b\sqrt{-19})=0$} = {$a+b\sqrt{-19} : a -b \pmod2 $}={$a+b\sqrt{-19} : a=b \pmod2 $}

I am missing something crucial here cause i just dont understand why this is equivalent to the ideal $(2, 1+\sqrt{-19})$. What does an element in this ideal look like?
The way I approached this problem was to start by writing an element $a+b\sqrt{-19}\in \mathbb{Z}[\sqrt{-19}]$ as $a+b\sqrt{-19} = (a-b) + b(1+\sqrt{-19}).$ If $a-b$ is even, say $a-b=2k$, then $a+b\sqrt{-19} = 2k + b(1+\sqrt{-19}).$ That is a linear combination of $2$ and $1+\sqrt{-19}$, and is therefore in the ideal.

Conversely, the set $J = \{a+b\sqrt{-19}\in \mathbb{Z}[\sqrt{-19}] : a-b \text{ is even}\}$ is an ideal in $\mathbb{Z}[\sqrt{-19}]$, which contains $2$ and $1+\sqrt{-19}$. To see that it is an ideal, you need to show that if $a+b\sqrt{-19}\in J$ and $x+y\sqrt{-19} \in \mathbb{Z}[\sqrt{-19}]$, then their product is in $J$. The calculation for that goes like this: $(a+b\sqrt{-19})(x+y\sqrt{-19}) = (ax-19by) + (ay+bx)\sqrt{-19})$, and $(ax-19by) - (ay+bx) = (a-b)x - (a+19b)y$. Since $a+19b = (a-b) + 20b$, which is an even number, it follows that $(ax-19by) - (ay+bx)$ is even, as required.

That shows that $J = (2, 1+\sqrt{-19})$. So to see whether an element $a+b\sqrt{-19}$ is in $(2, 1+\sqrt{-19})$, you just need to check whether $a-b$ is even or odd. That is what suggested to me the idea of defining the homomorphism $f:\mathbb{Z}[\sqrt{-19}] \to \mathbb{Z}/2\mathbb{Z}.$
 

Loiz

New member
Sep 16, 2012
3
The way I approached this problem was to start by writing an element $a+b\sqrt{-19}\in \mathbb{Z}[\sqrt{-19}]$ as $a+b\sqrt{-19} = (a-b) + b(1+\sqrt{-19}).$ If $a-b$ is even, say $a-b=2k$, then $a+b\sqrt{-19} = 2k + b(1+\sqrt{-19}).$ That is a linear combination of $2$ and $1+\sqrt{-19}$, and is therefore in the ideal.

Conversely, the set $J = \{a+b\sqrt{-19}\in \mathbb{Z}[\sqrt{-19}] : a-b \text{ is even}\}$ is an ideal in $\mathbb{Z}[\sqrt{-19}]$, which contains $2$ and $1+\sqrt{-19}$. To see that it is an ideal, you need to show that if $a+b\sqrt{-19}\in J$ and $x+y\sqrt{-19} \in \mathbb{Z}[\sqrt{-19}]$, then their product is in $J$. The calculation for that goes like this: $(a+b\sqrt{-19})(x+y\sqrt{-19}) = (ax-19by) + (ay+bx)\sqrt{-19})$, and $(ax-19by) - (ay+bx) = (a-b)x - (a+19b)y$. Since $a+19b = (a-b) + 20b$, which is an even number, it follows that $(ax-19by) - (ay+bx)$ is even, as required.

That shows that $J = (2, 1+\sqrt{-19})$. So to see whether an element $a+b\sqrt{-19}$ is in $(2, 1+\sqrt{-19})$, you just need to check whether $a-b$ is even or odd. That is what suggested to me the idea of defining the homomorphism $f:\mathbb{Z}[\sqrt{-19}] \to \mathbb{Z}/2\mathbb{Z}.$
Thank you so much! :)

Do you possibly know why it now follows that the multiplier ring of the ideal is not equal to $\mathbb{Z}[\sqrt{-19}]$? And that the ideal is not invertible?