Is $\mathbb{Z}[x_1,x_2,\ldots]$ the universal non-Noetherian ring?

Question

I recently learnt about the fact that $\mathbb{Z}[x_1, x_2,\ldots]$ is an example of a non-Noetherian ring. This example seems like it should in some sense be the "universal example" of a non-Noetherian ring. So, my question is if there is a result stating the existence of a morphism between $\mathbb{Z}[x_1,x_2,\ldots]$ and any given non-Noetherian ring, or something along those lines.

Answer 1

The answer is No. Any ring $R$ admits a homomorphism $\mathbb{Z}[x_1,x_2,\dotsc] \to R$ (just send all $x_i$ to $0$, for example, and use the unique homomorphism $\mathbb{Z} \to R$). It is also not possible to add reasonable assumptions to this homomorphism which enforce $R$ to be Non-Noetherian or that such a homomorphism needs to exist when $R$ is Non-Noetherian. For example, there is a monomorphism of rings $\mathbb{Z}[x_1,x_2,\dotsc] \to \mathbb{C}$, but $\mathbb{C}$ is a field and hence Noetherian. If $I$ is some infinite set, then $\mathbb{Z}[(x_i)_{i \in I}]$ is a Non-Noetherian ring, which does not embed into or admit an epimorphism from $ \mathbb{Z}[(x_i)_{i \in \mathbb{N}}]$ unless $I$ is countable. The ring $\mathbb{Z}[x_1,x_2,\dotsc]$ is Non-Noetherian, the quotient $\mathbb{Z}[x_1,x_2,\dotsc]/\langle x_1 \rangle$ is still Non-Noetherian, but the quotient $\mathbb{Z}[x_1,x_2,\dotsc]/\langle x_1,x_2,\dotsc \rangle = \mathbb{Z}$ is Noetherian. The category of Non-Noetherian rings is also badly behaved, there are no initial or terminal objects. This false impression that $\mathbb{Z}[x_1,x_2,\dotsc]$ is somehow universal among Non-Noetherian rings probably can be avoided by looking at more examples of Non-Noetherian rings, such as $\prod_{p \text{ prime}} \mathbb{F}_p$ or $C^{\infty}(\mathbb{R})$.