(January 2016) Define what it means for a subset \(I\subseteq R\) to be an ideal of \(R\). Note: If you use other technical terms like “closed,” “subring,” “subgroup,” etc., you must fully define those terms as well.
Answer: \(I\subseteq R\) is an ideal if:
\(I\ne\varnothing\).
For all \(x,y\in I\), we have \(x-y\in I\).
For all \(x\in I\) and \(r\in R\), we have \(rx\in I\) and \(xr\in I\).
DONE! Comment.
We often see answers to this question that look like the following: Answer: \(I\subseteq R\) is an ideal if:
\(I\) is a subgroup of \(R\) under \(+\).
For all \(x\in I\) and \(r\in R\), we have \(rx\in I\) and \(xr\in I\).
But you cannot say “DONE!” at this point. Although it is technically a correct definition, it does not follow the instructions, since it uses the technical term “subgroup”.
But if you add: 3. \(I\) is a subgroup of \(R\) under \(+\) if it forms a group under \(+\) itself.
Then you are not done, since “group” is also a technical term that you have to define. This is why you should strike this out.
But if you replace it with: 3. \(I\) is a subgroup of \(R\) under \(+\) if it is nonempty and closed under subtraction.
Then you are still not done since “closed” is also a technical term! So you would have to add: 4. \(I\) is closed under \(-\) if for all \(x,y\in I\), we have \(x-y\in I\). FINALLY DONE! Conclusion.
Hopefully the moral is clear here. You should definitely think of the concepts like “subgroup under \(+\)” and “closed under subtraction” in your head, but you’re going to have to write out their meanings anyhow. So by all means think “closed,” but write “for all \(x,y\in I\), we have \(x-y\in I\).”
Solution
A subset \(I\subseteq R\) is a two-sided ideal precisely when the following three explicit conditions hold:
\(I\ne\varnothing\).
For every \(x,y\in I\), one has \(x-y\in I\).
For every \(x\in I\) and \(r\in R\), one has both \(rx\in I\) and \(xr\in I\).
1. Conditions 1 and 2 give exactly the required additive structure.
Choose \(a\in I\), possible by condition 1. Then condition 2 gives \[ 0=a-a\in I. \] For \(x\in I\), \[ -x=0-x\in I. \] For \(x,y\in I\), since \(-y\in I\), \[ x+y=x-(-y)\in I. \] Thus \(I\) contains \(0\) and is closed under addition and additive inverses. Conversely, any subset with those additive properties is nonempty and is closed under subtraction, because \[ x-y=x+(-y). \]
2. Condition 3 is exactly two-sided absorption by ring elements.
The definition of a two-sided ideal requires that multiplying an element of \(I\) by an arbitrary element of \(R\) on either side remains in \(I\). Condition 3 states precisely \[ rI\subseteq I \qquad\text{and}\qquad Ir\subseteq I \] for every \(r\in R\).
Hence the three displayed conditions are a complete definition of a two-sided ideal and use no further technical term that needs expansion.