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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 $I$.

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    Hi Emad. I think the problem asks to prove that the ideal must be equal to $R$, not that it is included in $R$ – 2017-01-31
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    Hey Daniel thank you for much, it turned out to be a type. I dont know how but I added the word "in $R$" rather it should be just $R$ which makes more sance. Thank you very much – 2017-01-31
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    You're welcome! you should also thank Falcon for his answer, since he also clarify that no every ideal is a subring – 2017-01-31

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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.

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    Yes your right, it was a typoe. The question asks Prove that any ideal of R containing 1 must be R not in R. Thank you very much man – 2017-01-31
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    So its more proper to say that an ideal is a subset of $R$ rather than a subring. Thank you for clarifying that too – 2017-01-31
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    @EmadZamout Well.. being a sub-module is much stronger than being a subset! – 2017-01-31
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    @EmadZamout Exactly, if you don't know modules keep it simple saying that an ideal is a subset of $R$ satisfying some properties. I just wanted to specify that an ideal is indeed a sub-structure. It is my pleasure to have helped. – 2017-01-31