Sub Rings of the Real Field $\mathbb{R}$

Question

Are there proper sub-rings of $\mathbb{R}$ which are not fields , which contain $\mathbb{Q}$? Towards answering this, I am aware of the result that proper sub rings of $\mathbb{Q}$, properly containing $\mathbb{Z}$ are principal ideal domains. My hunch is that proper sub-rings of $\mathbb{R}$ properly containing $\mathbb{Q}$ should be a field. Can anyone help me please with the proof or with a counter example?

Answer 1

For instance, given any transcendental real number $\xi$, the ring $\Bbb Q[\xi]$ (i.e. the image in $\Bbb R$ of the polynomial ring $\Bbb Q[X]$ through the map $p(X)\mapsto p(\xi)$) is not a field.

Answer 2

You already have a correct answer for your question. But let me add few more words to it. If you want to construct a sub-ring of $\mathbb{R}$ that properly contains $\mathbb{Q},$ the simplest thing we can do is adjoin an irrational number $\zeta$ to $\mathbb{Q}$ and close the resulting set with rational linear combinations and products. But it turns out that if $\zeta$ is an algebraic number then $\mathbb{Q}[\zeta]$ is a field. Why? Because any algebraic number has a unique monic minimal polynomial with rational coefficients, say, of degree $n$. Then $\{1,\zeta,\cdots, \zeta^{n-1}\}$ form a basis for $\mathbb{Q}[\zeta]$ as a $\mathbb{Q}$-vector space. Moreover for any $a_0, a_1,\cdots a_{n-1}\in\mathbb{Q}$ (not all zero), you can find the (unique) multiplicative inverse of $a_0+a_1\zeta+\cdots+a_{n-1}\zeta^{n-1}$ by solving $$(a_0+a_1\zeta+\cdots+a_{n-1}\zeta^{n-1})(b_0+b_1\zeta+\cdots+b_{n-1}\zeta^{n-1})=1$$ for $b_0, b_1,\cdots b_{n-1}\in\mathbb{Q}.$ The well know examples for this construction are quadratic number fields, cyclotomic fields and generally splitting fields of rational polynomials.

So, we cannot construct a pure ring by adjoining one (or more) algebraic numbers. On the other hand, if we adjoin a transcendental number $\xi,$ the resulting set $\mathbb{Q}[\xi]$ is a ring (in fact, it is a Euclidean domain), but not a field. In particular, there is no polynomial $P$ over $\mathbb{Q}$ such that $\xi P(\xi)=1.$ In the same way, for any sub-field $F$ (other than rationals) of $\mathbb{R}$ you can find a ring $R$ that properly contains $F.$ Also, You can modify this idea to obtain some other rings, for example the $\mathbb{Q}$-vector space spanned by $\{1, \pi^2, \pi^3, \pi^4,\cdots \}$ (this is not even a GCD domain).

However, there are other more obscure rings of $\mathbb{R}$ that contains rationals. For an example see here.
I think, we can (non-canonically) embedded any ring of characteristic $0$ with cardinality less than $\mathfrak{c}$ as a proper sub-ring of $\mathbb{R}.$