If A is an Ideal of a ring R and the unity 1 belongs to A, prove that A=R.
It is a sufficient condition to show that $A\subseteq R$ and $R\subseteq A$.
Indeed, it is trivial to see that $A\subseteq R$ by the Theorem of Ideal test. $A$ must be a non-empty subset of a ring R. Since $A$ is an Ideal of a ring R, it is true that $A$ is a non-empty subset of a ring R.
Now, $A\subseteq R$.
So every element a in A is also an element in R. Let $a = r \in R$
Then, $\left [ a\cdot r \right ]\equiv \bar{r} \in R$. Recall that for a ring R it is true that $\left ( R,+ \right )$ is Abelian so closure under addition follows. Indeed, $\bar{r} \in R.$
Thus, $R\subseteq A.$
And so, $A=R$.
Note that I have not utilised the unity. Potential error exists. Any help in the form of Hint is appreciated.
Thanks in advance.
For any $r\in R,$ and any $a\in I,$ we have $r.a\in I.$ Take $a=1.$ Then $r.1=r \in I;$ For any $r\in R.$