# Does every ring have a subring that is a field?

Well, that question arose at the very moment I saw two examples. The first one was about an integrity domain which has a subring that is a field (I don't remember the specific example) and the second one is:

Let $$M = M_2(\mathbb{R})$$ be the set of all $$2 \times 2$$ matrixes with entries in $$\mathbb{R}$$ and $$1_M = \left[\begin{array}[cc] &1 & 0\\ 0 & 1\end{array}\right]$$. The set $$A = \left\{a \in \mathbb{R};\left[\begin{array}[cc] &a & 0\\ 0 & 0\end{array}\right]\right\} \subset M$$ is a subring of $$M$$ with unity $$1_A = \left[\begin{array}[cc] &1 & 0\\ 0 & 0\end{array}\right] \neq \left[\begin{array}[cc] &1 & 0\\ 0 & 1\end{array}\right] = 1_M$$. Furthermore, since $$A$$ is isomorphic to $$\mathbb{R}$$, $$A$$ is a field.

So, there can be subrings with different unities than the ring they are subsets to. The question is, if a ring does not have unity and none of the other properties, there can be a subring that has unity?

Or if it is possible, can someone give me an example of a ring without unity that has a subring with unity?

And an example of a ring with no additional properties that has a subring that is a field?

• What about the ring of integers $\mathbb{Z}$? Jan 18, 2021 at 15:56
• Instead use $$A = \{ \begin{pmatrix} a \\ & a \end{pmatrix} a \in \mathbb R\}$$ which is a subring which has the same multiplicative identity as $M_2(\mathbb R)$.
– D_S
Jan 18, 2021 at 16:00
• The ring of integers $\mathbb{Z}$ has unity $1$, I want an example of a ring without unity that has a subring with unity. Jan 18, 2021 at 16:02
• Dietrich, the question is similar, but still the $\mathbb{Z}_n$ have unit, the element $\overline{1}$. I don't know the translation from portuguese to english of "corpos" (literal translation is "bodies"), which are commutative rings with unit, without dividers of 0 and multiplicative inverses. Jan 18, 2021 at 16:08

Let $$R = \mathbb R \times 2\mathbb Z$$. Since $$2\mathbb Z = \{ 2n : n \in \mathbb Z \}$$ does not have a multiplicative identity, neither does $$R$$. However, $$R$$ contains $$\{ (x,0) : x \in \mathbb R\}$$ as a subring which is a field, isomorphic to $$\mathbb R$$.
• Great answer man. Which operations are defined on $R$? Jan 18, 2021 at 16:49
• Pointwise addition and multiplication: $(a,b) \cdot (c,d) = (ac,bd)$ and the same for addition.
• Cool. So it is a commutative ring with no dividers of $0$? Do you know if a ring without additional properties can have a subring that is a field? Also, thank you! You've already helped a lot :) Jan 18, 2021 at 17:21