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We are having ring $\mathbb{Z}[\sqrt{-6}]$. Which of the sets is subrings of $\mathbb{Z}[\sqrt{-6}]$ and which are ideals?

  1. $\mathbb{Z}+5\mathbb{Z}[\sqrt{-6}]$
  2. $5\mathbb{Z}+\mathbb{Z}[\sqrt{-6}]$
  3. $2\mathbb{Z}+3\mathbb{Z}[\sqrt{-6}]$

I know that if the ring is ideal of some set, then it is automatically a subring of this set, but that works only one way. If we have a subring, we can't say that it is an ideal.

To prove that a subring is an ideal we must present for example element from first set as $a+5b\sqrt{-6}$, where $a,b \in \mathbb{Z}$ and multiply it on similar representation of $\mathbb{Z}[\sqrt{-6}]$: it will be $a+b\sqrt{-6}$. And then if result $\in \mathbb{Z}+5\mathbb{Z}[\sqrt{-6}]$ we can say that it is an ideal.

Am I doing right way? And can you help me with proving that set is subring? (What should we say, to prove that set is subring).

Thanks for any help!

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1 Answer 1

up vote 3 down vote accepted

To prove that a set $I\subset R$ is an ideal, you must take an element $x$ from $I$ and an element $r$ from $R$ and prove that $x\cdot r$ is still an element of $I$. For example, for your first example, take $a+5b\sqrt{-6}$ from the "ideal" and $c+d\sqrt{-6}$ from the ring. Multiply the two to get $(ac - 30bd) + (5bc + ad)\sqrt{-6}$. Is this an element of $\mathbb Z + 5\mathbb Z [\sqrt{-6}]$ or can you find a counterexample where this does not hold?

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Proving ideals quite clear for me. It will not be ideal, but it is subring. Which statements should we prove, to show ring is a subring? It seems the 2-nd ring not close by "*" operation. Is it enough to show that subring closed by '+' and '-' or we should prove some another statements? –  JesusChrist Jun 30 at 7:52
    
To prove that a set $K\subset R$ is a subring, you must prove that it is closed both under multiplication and adition and that it contains aditive inverses, so for $x,y\in K$, you must show that $-x, x+y$ and $xy$ are all elements of $K$. –  5xum Jun 30 at 8:00
    
Thanks a lot!!! –  JesusChrist Jun 30 at 8:02
    
@5xum Also closure under addition should be proved; in the given sets it is almost obvious, but nonetheless noting this is important. –  egreg Jun 30 at 14:39

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