Elements of quotient ring

Question

Consider the rings $A_1 = \mathbb{Z}[\sqrt{3}]$ and $A_2 = \mathbb{Z}[\sqrt{2}]$. I am asked to find how many elements are there in $R_i$, being $R_i = A_i/2A_i$. For the first case, what I have tried by far is that given $x,y \in A_1, x R y \Leftrightarrow x - y \in 2A_1$, this is, if, given that $x = a + b\sqrt{3}, y = c + d\sqrt{3}$, $a$ and $c$ have the same parity, and $b$ and $d$ have the same parity. Hence, there are four equivalence classes, and thus four elements of that quotient ring: $R_1 = \{[0], [1], [\sqrt{3}], [1 + \sqrt{3}] \}$. This seems correct for me. However, when I try to do operations with those classes, for example, I find that $\sqrt{3} \cdot (1 + \sqrt{3}) = 3 + \sqrt{3}$, which is in the same equivalence class as $1 + \sqrt{3}$. Therefore, there would be two neutral elements for the product in this case. That is what really annoys me, and I find it weird. What have I wrong? By the way, I haven't mentioned $A_2$ because I figure out it is almost the same. I am also asked to find all the ideals of each ring. I would be grateful I someone could give me a hand.

Answer 1

I don't think there is anything wrong, there aren't two multiplicative identity elements. It is just that these two elements when multiplied together don't change $1+\sqrt{3}$. As an example why this shouldn't surprise you too much, if you consider the ring of matrices, then just because $AB=B$ does not imply that $A$ is the identity matrix. For example, we can observe that $$ \begin{bmatrix} -1 & 1\\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2\\ 2 & 4 \end{bmatrix} = \begin{bmatrix} 1 & 2\\ 2 & 4 \end{bmatrix}$$

However, what one may prove is that in a ring if $xy = y$ and $x\neq 1$, then $y$ is a zero divisor. Indeed, what you should observe is that $xy-y=0$, so then $(x-1)y=0$, so as $x\neq 1$, we see that $y$ is a zero divisor.

Thus, for your case, I believe you are correct, and your observation shows that $[1+\sqrt{3}]$ is a zero divisor, so you can conclude that $A_1/2A_1$ is not a division ring, so the ideal $2A_1$ is not a prime ideal.

Answer 2

You may work as follows, to also get the structure of the quotient rings, not only their number. Quotients of rings like $R/J$ (ring $R$ modulo and $R$-ideal $J$) are also displayed as fractions, $\frac RJ$, below. Principal ideals like $(2)$, are also written using only the generator. Isomorphisms are written as equalities. $$ \begin{aligned} \frac {A_1}2 &= \frac{\Bbb Z[x]/(x^2-3)}2 = \frac{\Bbb Z[x]}{(x^2-3,\ 2)} = \frac{\Bbb Z[x]/2}{x^2-3} = \frac{(\Bbb Z/2)[x]}{x^2-3} = \frac{\Bbb F_2[x]}{x^2-3} = \frac{\Bbb F_2[x]}{(x+1)^2} \ , \\ \frac {A_2}2 &= \frac{\Bbb Z[x]/(x^2-2)}2 = \frac{\Bbb Z[x]}{(x^2-2,\ 2)} = \frac{\Bbb Z[x]}{(x^2,\ 2)} = \frac{\Bbb Z[x]/2}{x^2} = \frac{(\Bbb Z/2)[x]}{x^2} = \frac{\Bbb F_2[x]}{x^2} \ , \end{aligned} $$ and in both cases the generating square root ($\sqrt 3$, respectively $\sqrt 2$) goes to $x$ (taken modulo the nilpotent ideal generated by $(x+1)^2$, respectively by $x^2$). The possible rests modulo the respective polynomial of degree two have degree one, so there are four elements in each quotient ring.