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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!

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1 Answer 1

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(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, $S = \mathbb{F}_3[t], y = t^2 - 1$.

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