Define an ideal of a rng (ring without multiplicative identity) as an additive subgroup of the rng that is closed under multiplication on both sides by elements of the rng. Show that if $N$ is a rng, and $x \in N$ with $x \neq 0 $, then there is a largest ideal of $N$ that does not contain $x$. Explain why the case $x=0$ must be excluded.
Well, for explaining why $x=0$ must be excluded, I said that every additive subgroup must contain $0$, so every ideal of a rng contains $0$, so there is no largest ideal of $N$ that does not contain $0$.
For the other part, I'm not really sure. I tried to use Zorn's Lemma, but I'm not sure if this is right because I'm not sure if a "largest element" is the same as the "maximal element" given by Zorn's Lemma. Is there a better way?
Clearly there is some ideal of $N$ that does not contain $x$ because the trivial subgroup $\{0\}$ is an ideal that does not contain $x$, but I'm not sure how to find the largest one.
The poset you are considering is the set of all ideals not containing $x$ ordered by inclusion. It contains $\{0\}$ but not $N$.
If you have already checked that the union of a chain of elements of this poset lies in the poset, then Zorn's lemma kicks in and tells you there are maximal elements not containing $x$.
This does not a priori mean that such an ideal is a maximal ideal: it just means that any strictly larger ideals must pick up $x$ (but we don't know yet if they have to be $N$.)
There is no reason to expect such an ideal to be unique either. Let $F$ be a field and $N$ be the rng $\oplus _\mathbb N F$ of finitely nonzero tuples. Let $x$ be nonzero on $5$ places and zero elsewhere. Every ideal of the form $M_j=\{y\in N\mid y_j=0\}$ is a maximal ideal of $N$, and $x$ is not in five of these.
So it is not really correct to say "there is a largest" like that (meaning a maximum member), but it is right to say there are maximal elements like that.