Showing a Ring and an Ideal are equal

Question

Q: Let R be a commutative ring with unity. Prove that if A is an ideal of R and A contains a unit, then A=R.

This is my attempt at an answer: It suffices to show that all the elements in R are in A.

Let a be and element of A, and r be an element of R. Since A is an ideal of R then all elements ar and ra must be in A.

We know that A has a unit say 1, then we know 1r and r1 for all r in R, are in A.

Thus all elements in R are in A, therefore A=R.

Answer 1

The idea is correct. But, the details are lacking.

You know that $A$ contains a unit (not the unit element $1_R$).

Recall that a unit is an element $u$ that has a multiplicative inverse, that is there exists a $v$ such that $uv =1_R$.

So, you have such an element $u$. It might be $1_R$ itself, but it does not have to be. And if it is not there is a small step missing.

Answer 2

It's: $I=R \Leftrightarrow \exists x\in I: \; x\in U(R)$
The way "=>" is common The other way "<=" is alike yours: if $x\in I$ and $x\in U(R),\;$ because $I$=ideal, then $1_{R}=x\cdot x^{-1}\in I$. Then because $I \trianglelefteq R$=commutative, $\forall r\in R: \; \; r = 1_{R}\cdot r = r \cdot 1_{R} \in I$ => $I=R$. (but eventhough its alike your proof).