Question

Consider the subring of R
\(\mathbb Z[\sqrt2]=\{a+b\sqrt2|a, b\in \mathbb Z\}\)
Where addition and multiplication are as defined as usual in \(\mathbb R\). (You don't have to prove that it is a subring of R)
Show that \(\mathbb Z[\sqrt2]\) is NOT a field
Please, help

Answer 1

i think i remember this maybe
gimmick is that \(\frac{1}{2}\) is not in the domain, so \(2\) does not have a multiplicative inverse, so not a field

Answer 2

I am cool with that too:)

Answer 3

i think also \(\sqrt{2}\) does not have an inverse
try solving
\[\sqrt{2}(a+b\sqrt2)=1\]

Answer 4

This is what I try
Let a+bi in the set
suppose c +di is the multiplication inverse of a +bi, that is (a+bi)(c+di) =1
and I end up at c is not in Z, hence element in the set has no multiplication inverse, then, it is not a field.
But I am not sure

Answer 5

make sure you rule out the case of the multiplicative identity, because 1 is its own inverse....

Answer 6

you only need to show one example of why its not a field, and I think the first example given was the easiest.

Answer 7

1 is multiplicative identity,

Answer 8

every element in the set *1 = itself.

Answer 9

I know, but you just said that you took two arbitrary elements and assumed there was a inverse, and then showed c was not in Z, but this is not the case for 1

Answer 10

:)
that is my final exam this morning. This is only one problem I was not sure. It's too late to correct.

Answer 11

ahh well, it shows that you know what you are trying to do....I doupt much will be taken