Ring of even integers

Question

So I am given a ring of even integers. It is an isomorphism of Z to R that is defined by f(x)=2x+4. I am trying to find the O(R) and the 1(R). I just set the function equal to 0 or 1 but apparently that is not the right thing to do in this case. Any rational as to what I am doing wrong or the correct way to think about it to set it up right? Should you plug in 0 for 2 instead?

Answer 1

Since $f$ is an isomorphism, it sends $0$ to $0(R)$ and $1$ to $1(R)$. So you're correct in setting the function equal to $0$ and $1$, respectively (that is, taking $f(0) = 4$ and $f(1) = 6$).

Let's find the units of $R$. The units of $\mathbb Z$ are $-1$ and $1$, and since this is an isomorphism, $f$ maps units to units, so the units of $R$ are $f(-1) = 2$ and $f(1) = 6$

Answer 2

The addition on $R$ has not been specified. If $f$ is to be an additive group isomorphism, then the addition $\oplus$ on $R$ has to be defined by $2s\oplus 2t=2s+2t-4$.

This is because $2s=f(s-2)$ and $2t=f(t-2)$. Since $f$ is an additive group isomorphism, we have $f((s-2)+(t-2)=2s\oplus 2t$. But $$f((s-2)+(t-2))=2s+2t-8+4=2s+2t-4.$$

Now the zero-element is not hard to identify. Call it $2z$. Then we want $2s\oplus 2z=2x$. That gives $2z=4$.

Do something similar to identify the multiplication $\otimes$ on $R$.