# Maximal subring

I have a quick question on a problem out of Dummit and Foote (problem 7.5.6 for those who might have the text):

Prove that the real numbers, $\mathbb{R}$, contain a subring $A$ with $1 \in A$ and $A$ maximal (under inclusion) with respect to the property that $1/2 \notin A$ [Hint: Use Zorn's Lemma].

My question is in how I should interpret this. Are we looking for a subring with the properties (1) it contains $1$ and (2) it is maximal with respect to 'not containing 1/2?'

Or should we be looking for a subring $A$ with the single property (1') it is maximal with respect to 'containing $1$' and 'not containing $1/2$?'

In the first interpretation, showing (2) is quite straitforward: as the hint suggests, apply Zorn's Lemma to the collection of all subrings that don't contain 1/2. The problem comes in trying to prove (1), that the maximal element contains the identity. If this is how I should interpret the problem, then this is where I'm hung up. The subfield generated by $A$ certainly contains $1$, and if I could show $1/2$ is not in that subfield, then I'd be done since $A$ is maximal. But is this even true? Is there another way I could show this?

If I should interpret the problem in the second way (i.e., prove property (1') only), then the problem is, similar to above, a straitforward application of Zorn's Lemma to the collection of all subrings that contain $1$ but not $1/2$.

So, to summarize my question:

How should I interpret this problem? If I should interpret it the first way I mentioned, then I need help showing the maximal element contains the identity. If I should interpret the problem the second way, then I don't have any problem.

Any advice is greatly appreciated.

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"the set of rings containing $1$ and not $1/2$ has maximal elements". – Mariano Suárez-Alvarez Jan 18 '12 at 6:56
It doesn't matter. Remember the key word maximal. You can heap on any further conditions you like, it will not affect maximality as long as you can keep $1/2$ out. – André Nicolas Jan 18 '12 at 6:57
Usually, the construction «maximal with respect to the property P» is used confusingly, sadly. – Mariano Suárez-Alvarez Jan 18 '12 at 7:02

## 1 Answer

From the statement of Zorn's Lemma given in Dummit & Foote, I would use the second interpretation, for exactly the reasons you give. Zorn's Lemma will give you a subring of $\mathbb{R}$ which is maximal with respect to not containing $\frac{1}{2}$, but you can't really control whether this subring contains $1$.

You should note that there is an a priori slightly more general (but actually equivalent) form of Zorn's Lemma that would enable you to use the first interpretation:

Zorn's Lemma (second form). Suppose that $( A , \leq )$ is a partially ordered set in which every chain has an upper bound. Then given any element $a \in A$ there is a maximal element $b$ of $A$ with $a \leq b$.

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Thanks so much everybody, it makes much more sense now. I did not realize that the maximal subring I obtain in the proof of the second interpretation satisfies the first interpretation. – John Myers Jan 18 '12 at 7:22