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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?

Thanks in advance!

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    $\begingroup$ What about the ring of integers $\mathbb{Z}$? $\endgroup$
    – user299843
    Jan 18, 2021 at 15:56
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    $\begingroup$ Instead use $$A = \{ \begin{pmatrix} a \\ & a \end{pmatrix} a \in \mathbb R\}$$ which is a subring which has the same multiplicative identity as $M_2(\mathbb R)$. $\endgroup$
    – D_S
    Jan 18, 2021 at 16:00
  • $\begingroup$ The ring of integers $\mathbb{Z}$ has unity $1$, I want an example of a ring without unity that has a subring with unity. $\endgroup$
    – Fractal Admirer
    Jan 18, 2021 at 16:02
  • $\begingroup$ 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. $\endgroup$
    – Fractal Admirer
    Jan 18, 2021 at 16:08

1 Answer 1

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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 $\mathbb R$.

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  • $\begingroup$ Great answer man. Which operations are defined on $R$? $\endgroup$
    – Fractal Admirer
    Jan 18, 2021 at 16:49
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    $\begingroup$ Pointwise addition and multiplication: $(a,b) \cdot (c,d) = (ac,bd)$ and the same for addition. $\endgroup$
    – D_S
    Jan 18, 2021 at 17:00
  • $\begingroup$ 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 :) $\endgroup$
    – Fractal Admirer
    Jan 18, 2021 at 17:21
  • $\begingroup$ What "additional properties" do you mean? $\endgroup$
    – D_S
    Jan 18, 2021 at 17:40
  • $\begingroup$ The unity, commutativity of product, no dividers of 0 and multiplicative inverses. $\endgroup$
    – Fractal Admirer
    Jan 18, 2021 at 17:55

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