I'm trying to solve this group theory problem, and I'm really not sure how to approach this. The question is:
Show that $\mathbb{F}_7[\sqrt{-1}] := \{a+b\sqrt{-1}$ | $a,b \in \mathbb{F}_7\}$ is a ring.
I have been stuck on this problem for hours and I really cannot figure it out.
This is my progress so far:
$(a+b\sqrt{-1}) + (c+d\sqrt{-1}) = (a+c) + (b+d)\sqrt{-1}$
I'm not sure what I'm doing, I'll appreciate it if anyone can help me out. Thanks in advance!
It would probably help to know that $\mathbb{F}_{7}$ denotes the field of $7$ elements. Moreover, since $7$ is a prime number $\mathbb{F}_{7} \cong \mathbb{Z}_{7}$, the integers modulo $7$. To show that a set is a ring, we have to show that it is 1.) closed under addition, 2.) multiplication, 3.) left/right distributive, and 4.) each element has an additive inverse. Let $R = \mathbb{F}_{7}[i]$, where $i = \sqrt{-1}$.
1.) Let $a + bi , c + di \in R$. Then $(a+bi) + (c+di) = (a+c) + (b+d)i \in R$, since $\mathbb{F}_{7}$ is a field and hence closed under addition.
2.) Let $a+bi , c+di \in R$. $(a+bi)(c+di) = (ac + bdi^{2}) + (ad+bc)i = (ac-bd) + (ad+bc)i \in R$, once again by the fact that $\mathbb{F}_{7}$ is a field, and is closed under addition and multiplication.
3.) I will show left distributivity, right distributivity is similar. Let $a+bi, c+di, e+fi \in R$. Then $(a+bi)(c+di + e+fi) = (a+bi)((c+e)+(d+f)i) = (a(c+e) - b(d+f)) + (b(c+e) + a(d+f))i) = (ac+ae-bd-bf) + (bc+be+ad+af)i = (a+bi)(c+di) + (a+bi)(e+fi)$.
4.) Let $a+bi \in R$ then its additive inverse is $-a-bi \in R$, because $\mathbb{F}_{7}$ is a field and so the additive inverse exists and it is unique.