A commutative ring $R$ is called a principal ideal domain (PID) if every ideal of $R$ can be generated by a single element. If $R$ is a principal ideal domain, is every subring of $R$ a principal ideal domain? No, definitely not. That is because you can take any integral domain that is not a principal ideal domain, like $\Z[x]$, and take its fraction field. Its fraction field is a PID and the original ring sits inside it as a subring.
Another more interesting example is the ring $\Q[x]$ of polynomials with rational coefficients. It is a PID, yet the subring $\Q[x^2,x^3,x^4,\dots]$ is not. The ideal $(x^2,x^3)$ in this ring is not a principal ideal. By the way, is the ring $\Q[x^2,x^3,\dots]$ Noetherian? Does there exist an ideal in it that needs at least three generators?
What about products? Obviously, the product of two nonzero integral domains is no longer an integral domain. However, is the product of a collection of principal ideal rings still a principal ideal ring? (Note: a principal ideal ring (PIR) is a commutative ring in which every ideal is principal.) For finite products, this is true: a finite product of PIRs is a PIR. I’ll leave that one to the reader.
But what about infinite products? Then things are different. An infinite product of principal ideal rings need not be a principal ideal ring. A simple example is $\Z^\omega$, and the proper ideal $I$ generated by the elements $(2,0,0,\dots), (0,2,0,\dots),(0,0,2,\dots),\dots$. It is certainly a proper ideal, since every element in this ideal has every coordinate even. However, $I$ cannot be principal, for every element of $I$ has only finitely many coordinates nonzero, whereas a generator would necessarily have all coordinates nonzero.
Hey, are principal ideal domains closed under ultraproducts? No, they’re not. For example, consider Peano rings: these are rings that are elementarily equivalent to the integers $\Z$. What’s elementary equivalence? Two structures are elementarily equivalent if they satisfy the same first order sentences in the language of the structures. So that means a Peano ring is one that satisfies exactly the same first-order ring sentences as the integers. One of the basic properties of Peano rings is that every Peano ring not isomorphic to the integers is not Noetherian. Therefore, an ultrapower of $\Z$ with respect to non-principal ultrafilters is not a PIR. In fact, you can somewhat explicitly write down an ideal that is not principal in such a ring!