Let $R$ be a ring and $I$ a subgroup under addition, show $I$ is an ideal

Question

Let $R$ be a ring and $I$ be a subgroup under addition. Prove if for every $a,b \in R$

1) $(a + I) + (b + I) = (a + b ) + I$ and

2)$(a + I)(b + I) = ab + I$

then $I$ is an ideal.

attempt proof:

Suppose that that for every $a,b \in R$

1) $(a + I) + (b + I) = (a + b ) + I$ and

2)$(a + I)(b + I) = ab + I$

.

Then need to show $I$ is an ideal. That is if $a \in R$ and $i \in I$ , then $ai \in I$ and $ia \in I$.

Let $a \in R$ and $i \in I$ then $i + I = 0 + I$ and $a + I = a + I$ . THen $ ia + I = 0(a) + I = 0 + I$ for some $i$. THen $ia - 0 \in I$ so that $ia \in R$. Similarly for $ai$. So $I$ is an ideal

Can someone please verify this? Any help or better approach would be very appreciate it. I am not sure about the forward direction proof. Thank you!

Answer 1

Your proof looks right except for the typo $ia\in R$ I suppose you wanted to write $ia\in I$ and it is finished !