If I take a Ring modulo a prime-like element, does it become a field? Let's say we have a ring, $R$. We call an element $p$ of $R$ "prime-like" if for all $a$ and $b$ such that $p=ab$, exactly one of $a$ or $b$ is multiplicatively invertible. For example, in the integers, $-7$ is prime-like, because in $-7=-1*7$ and $-7=1*7$, exactly one of the terms is invertible. For real polynomials, $x^2+1$ is prime-like, because it can only be expressed as a product by $(\frac1ax^2+\frac1a)a$ for a nonzero constant $a$, and $a$ is invertible and $\frac1ax^2+\frac1a$ is not. Not that $0$ is never prime-like since $0=0*0$ (both are noninvertible), and an invertible element, $w$ is never prime-like since $w = w*1$ (both $w$ and $1$ are invertible).
We take a ring modulo an element $r$ by taking an equivalence relation where $r_1=r_2$ in the new ring if there is an $r_3$ such that $r_1-r_2=r_3*r$ in the old ring. This new ring can be shown to in fact be a ring, and there is a ring homomorphism from the old ring to the new ring ($r$ gets mapped to $0$).
If $r$ is prime-like, is the new ring a field? For example, if you take the integers modulo or a prime $p$ (which is the same as modulo $-p$), it becomes a prime finite field. If you take the real polynomials modulo $x^2+1$, you get the complex number field.
Is it true in general that rings modulo prime-like elements are fields?
 A: As the other answer remarked, modding out by an irreducible (aka "prime-like") element doesn't give a field in general. The obvious counterexamples are polynomial rings; either a polynomial ring over another non-field in a single indeterminate, like $\mathbb{Z}[x]$; or a polynomial ring in two variables over a field like $K[x,y]$ when $K$ is a field. Here, if you mod out by one of the indeterminates - say $y$, which is an irreducible element - then you still have a polynomial ring $K[x]$ which is definitely not a field! There are also other examples of weird rings with this property too that are not polynomial rings.
However, there is a subclass of rings which does have the property that taking the quotient by an irreducible element gives a field. These are principal ideal domains, which are rings where every ideal is generated by a single element. If an ideal is generated by an irreducible element then this ideal is maximal and the corresponding quotient is a field. The examples above for polynomials fail because neither $\mathbb{Z}[x]$ nor $K[x,y]$ is a principal ideal domain; for example, in the first case given a prime number $p$ and an irreducible polynomial $f\in \mathbb{Z}[x]$ that remains irreducible modulo $p$ one can show that the ideal $(p,f)$ is not principal - it's not generated by a single element of $\mathbb{Z}[x]$. So modding out by one of the elements generating this ideal still leaves some nontrivial ideal structure - and fields have no nontrivial ideals. In the case where your ring is $K[x,y]$, the ideal $(x,y)$ is also not principal, so modding out by one of these indeterminates doesn't "kill off" the other one, even though both are irreducible.
However, if you are in a principal ideal domain - for example, $K[x]$ for a field $K$, and you mod out by the principal ideal generated by an irreducible polynomial $f\in K[x]$ - which satisfies your definition of "prime-like" - then you get a field, which is a field extension of $K$ obtained by adjoining to $K$ a root of this polynomial.
A: What you call "prime-like" is usually called irreducible. In general, irreducible elements and prime elements need not coincide, and whenever they do not, you will get an example where the quotient in question is not a field.
If you "combine" your examples and consider integer polynomials, then any prime number $p$ is still irreducible, but if you take the quotient by the ideal generated by it (which is what you are doing), then you get $(\mathbb{Z}/p\mathbb{Z})[x]$ which is not a field.
