# Invariant Properties of Isomorphic Rings

I'm a second year maths student, looking at Rings and Modules questions for my exam.

A property $$P$$ of rings is invariant under isomorphism if whenever $$R$$ is a ring with property $$P$$, and $$S$$ is a ring isomorphic to $$R$$, then $$S$$ also has property $$P$$.

Is there a property $$P$$ of rings invariant under isomorphism such that $$x \in R$$ is irreducible if and only if $$R/\langle x \rangle$$ has property $$P$$?

I know that for PIDs, $$x \in R$$ is irreducible if and only if $$R/\langle x \rangle$$ is a field. So, for a counterexample, I've been trying to look at rings which are not PIDs, but it's been very difficult to find a good isomorphic pair $$R,S$$ for which the statement doesn't hold.

I note that $$x \in R$$ is prime if and only if $$R/\langle x \rangle$$ is an integral domain seems to be a property of rings invariant under isomorphism. This is because isomorphisms preserve integral domains.

Any help would be greatly appreciated!

(All rings are commutative throughout.)

The answer is no. Let's be clear about what it would take to find a counterexample; as written, it's not immediately clear how to negate this statement, since it is phrased in terms of the existence of a certain property, and it's not clear what it would mean for this property to not exist. I suspect you might be confused about this because you write:

it's been very difficult to find a good isomorphic pair $$R,S$$ for which the statement doesn't hold.

But to find a counterexample it doesn't matter whether $$R$$ and $$S$$ are isomorphic or not.

Anyway, here is a rephrasing of the desired statement which makes no explicit mention of properties:

If $$R, S$$ are two rings, $$x \in R, y \in S$$ are two elements, $$x$$ is irreducible, and $$R/x \cong S/y$$, then $$y$$ is also irreducible.

Take a moment to convince yourself this is equivalent to the statement in terms of properties. The point of rewriting the statement this way is that it's clearer how to negate it:

There exist two rings $$R, S$$ and elements $$x \in R, y \in S$$ such that $$x$$ is irreducible and $$R/x \cong S/y$$ but $$y$$ is not irreducible.

So, let's look for such rings. Since primes and irreducibles are the same in UFDs we need rings which are not UFDs. We can take, for example, $$R = \mathbb{Z}[\sqrt{-5}]$$ and $$x = 3$$ (this is the example of an irreducible element which is not prime given by Wikipedia), which is irreducible because its norm is $$9$$ and there are no elements with norm $$3$$. However, it is not prime because

$$R/x \cong \mathbb{F}_3[t]/(t^2 + 5) \cong \mathbb{F}_3[t]/(t^2 - 1) \cong \mathbb{F}_3^2$$

is not an integral domain. Now we can find an element $$y$$ in another ring $$S$$ which is not prime such that $$S/y \cong \mathbb{F}_3^2$$, and if $$S$$ is a UFD then $$y$$ is also not irreducible. Here we can just take, for example, $$S = \mathbb{F}_3[t], y = t^2 - 1$$.