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Let $A$ be a ring and $I$ an ideal of $A$. Prove that $I=A$ if, and only if, $I$ has an invertible element of $A$.

It's pretty easy to do this if we consider $A=\mathbb{Z}$: If $x\in I$ and $1\in I$, then by the definition of ideal we have $x+1\in I$. If we split $\mathbb{Z}$ into $\mathbb{Z}=\mathbb{Z}_{> 0}\cup\mathbb{Z}_{\leq0}$, it's easy to see by induction that, for $\mathbb{Z}_{>0}$, $x\in I\land x+1\in I\;\forall x\in\mathbb{Z}_{>0}$ implies that $I$ covers all elements of $\mathbb{Z}_{>0}$ and since every positive integer has a additive inverse, it follows that $I$ also covers all elements from $\mathbb{Z}_{\leq0}$, so $I = \mathbb{Z}$. For the converse it's even more simple: Since $1\in I$, we have for any $x\in \mathbb{Z}$ that $1\cdot x\in I$, so $I=\mathbb{Z}$.

But how can I reformulate this proof for the more general case?

A. Riba
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  • If $i\in I$ then for any $a\in A$, $ia\in I$. If $I$ has an invertible element then what else does $I$ contain? What can you establish with that element? – John Douma Jul 03 '23 at 15:41
  • @JohnDouma If $I$ has an invertible element $v$, then $x+vy\in I$ for $x,y\in I$. Depending on the choice of $y$ we would have $x+1\in I$, so $x\in I \land x+1\in I$, and then I could use the argument that I presented on my proof for the special case. Is that correct? – A. Riba Jul 03 '23 at 16:04
  • If $1 \in I$ then $I = A$ (why?). Prove that $1 \in I$ if an invertible element is in $A$. – CyclotomicField Jul 03 '23 at 16:19
  • This is not a quotient ring problem. Cosets of ideals won't help. You must concentrate on the ideal's multiplicative properties. – John Douma Jul 03 '23 at 18:41

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