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Here's my proof:

Let the ideal $I = \{i_1, i_2,...\}$ and $J=\{j_1, j_2,...\}$ Then $I \cup J$ is an ideal only if for all $\nu, \mu$, $i_{\nu}+{j_\mu} \in I \cup J$. Let us assume that there exists an element $i_{\nu}$ of $I$ that does not belong to $J$. Then, for any $j_{\mu}$, either $i_{\nu}+{j_\mu} \in J$, which is a contradiction since it would imply that $i_{\nu} \in J$, or $i_{\nu}+{j_\mu} \in I$, which implies that $j_{\mu} \in I$. Letting $\mu$ run through the index of $J$, we get that $J \subset I$. Thus, our proof is complete.

However, here are some things bothering me:

  1. There seems something fishy about this proof, but I can't point it out. Is my proof correct?

  2. Is this a constructive proof or a proof by contradiction?

  3. Does this proof use the Axiom of Choice?

Thanks in advance for any help!

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    For me, it' perfectly correct. It is not a proof by contradiction, but I don't see why it should be constructive: you don't have to obtain an output. Last, it doesn't use the axiom of choice: you only prove that, if $I$ is not contained in $J$, then it's $J$ which is contained in $I$ – 2017-02-26
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    It’s not your question, but you should **not assume** that the ideals are countable, unless you are assuming the ring is countable (eg, the integers). Don’t index elements by natural numbers. Instead use phrases like “for all $j$ in $J$...” – 2017-11-17

2 Answers 2

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  1. Your proof uses contradiction to show that $j_\mu\in J$.
  2. It does not.

A less fishy proof:

Suppose that there are $i\in I\setminus J$, $j\in J\setminus I$. If $I\cup J$ is an ideal then $i+j\in I\cup J$, but $i+j\in I$ implies $j\in I$ and $i+j\in J$ implies $i\in J$. Both conclusions are contradictions, so there is no such pair $(i,j)$.

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    2: Not really. The phrase ‘by contradiction’ is often used indiscriminately. It is an argument of type `it can't be such and such, otherwise…`, which is more like contrapositive. – 2017-02-26
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This theorem says that if A, B are ideals of ring R and A is subset of B, then the union of A and B is also ideal. Notice that if A is subset of B, then the union of A and B is B itself. So, the union of A and B which is equal to B is also ideal, since B is ideal.