Every non-unit is in some maximal ideal

Question

I am trying to prove that every non-unit of a ring is contained in some maximal ideal. I have reasoned as follows: let $a$ be a non-unit and $M$ a maximal ideal. If $a$ is not contained in any maximal ideal, then the ideal $\langle M, a \rangle$ (that is, the ideal generated by $M$ and $a$) strictly contains $M$, a contradiction.

However, there is a detail I'm unsure about. It's easy to see that $\langle a \rangle$ is a proper ideal for any non-unit $a$, but how can I be sure that $\langle M,a \rangle$ is also a proper ideal? Couldn't there, for example, exist some $m \in M$ such that $1=m+a$, so that $\langle M,a \rangle$ is the whole ring?

Answer 1

Let $a$ non-unit. Consider the ideal $\langle a \rangle$. Since $a$ is not a unit this ideal does not contain $1$ so it is not the full ring $R$. Now, let's look at the ideals of $R$ that are not equal to $R$. In particular, $\langle a \rangle$ is one of them. The question is whether $\langle a \rangle $ can be enlarged to an ideal again not $R$ which itself cannot be enlarged any further. OK, if $\langle a \rangle$ itself cannot be enlarged any further, done, it is maximal. If not, then it can be enlarged to say some ideal $I_1$. If $I_1$ is maximal, done. If not, enlarge it to $I_2$. If $I_2$ is maximal, done. If not, enlarge it to $I_3$, and so on.

What can happen:

one obtains at some point an ideal $I_n$ that is maximal. Done

one never stops, and we get

$$\langle a \rangle \subset I_1 \subset I_2 \subset I_3 \subset \ldots $$ strict inclusions.

Now one can argue that in a lot of commonly met rings this can never happen. This is true, it can never happen in a quite large family of rings, called noetherian. But still, in principle one can go on for ever like this. What to do? Let's look at the union of all the $I_n$. It is an ideal itself, and moreover, it is not $R$. Why? I mean in principle it may just be possible that they exhaust $R$. But, attention. None of the $I_n$ contain the element $1$. This is the key. It follows that the union also does not contain $1$ and hence we got an even bigger ideal, still not $R$. If this is maximal, now we are done. What if not? Well, we should just keep on going and increasing at each step to a large one if we haven't yet reached a maximal. OK, one should be careful: how many steps can we perform? Already we have performed something like countably infinitely many. Well, is there a possibility to keep on going for ever? Certainly one cannot just go on forever in this extended sense, when the number of steps can be labelled by any $\it{well\ ordered \ set}$. Well, this is the content of a tricky method called the Zorn lemma. I will end now by saying that at some (perhaps very far away) point the procedure will stop and we will get our maximal ideal containing $\langle a \rangle$

Answer 2

Letting $M$ be just any maximal ideal of $R$ is not a good idea, since it will not have any special relation to $a$.

Since you appear to know that every nonzero commutative ring contains a maximal ideal (which is fact is just as hard to prove as what you want to prove), you could reason as follows, avoiding (another) use of Zorn's lemma. The quotient ring $R/\langle a\rangle$ is nonzero since $a$ is a non-unit. It therefore has a maximal ideal $I$, and you can take $M$ to be the inverse image of $I$ under the canonical projection $R\to R/\langle a\rangle$. You can see directly that it is maximal, or alternatively reason that $R/M\cong(R/\langle a\rangle)/I$ is a field.

Answer 3

Yes, it can happen that $\langle M,a\rangle$ is the whole ring, so your proof doesn't work. Try to reason as follows: let $\Sigma$ be the collection of ideals which contain $a$. Now use Zorn's lemma.