Prove that any ideal of $R$ containing $1$ must be $R$

Question

Let $R$ be any ring with unity. Prove that any ideal of $R$ containing $1$ must be $R$.

Hello guys. This is my assignment question so please don't give me a solution.

I am just not sure what this question is asking. If someone can clarify it it would be great! From what I understand that, the question is asking to prove that for any ideal of $R$ containing $1$ then that ideal must be in $R$. Isn't any ideal by definition a subring in $R$? If $1$ is an element of that ideal, then that ideal is basically $R$ since every element in $R$, is in $I$.

Answer 1

There must be a typo in the question you were asked, as you stated the correct statement is:

Let $I$ be an ideal of $R$, if $1_R\in I$, then $I=R$.

However, note that an ideal of $R$ is not a subring of $R$ in general, precisely because a subring is asked to contain $1_R$. Nevertheless, an ideal of $R$ is a sub-$R$-module.