Let $R$ be a unital ring and $S\le R$ be a unital subring with $1_S = 1_R$.

Then it can be shown that $S^\times\subset R^\times\cap S$, where $R^\times$ and $S^\times$ are the sets of units of respective rings. (Statement *)

There can be found a ring $R$ and its subring $S$ such that $R^\times\cap S\ne S^\times$. That is, $\exists a\in R^\times\cap S$ such that $a\not\in S^\times$. (Statement **)

I have a couple of questions, as I'm feeling quite confused:

(1) Is it possible that $S\le R$ and yet $1_S\ne 1_R$?

(2) I can't think of an example of a ring and a subring that satisfy the Statement **. Can someone please give an example?

  • 1
    $\begingroup$ $\mathbb{Z}_5$ is not a subring of $\mathbb{Z}$ $\endgroup$ – Hayden Feb 20 '17 at 1:35
  • $\begingroup$ 1. $\mathbb{Z}_5\subset\mathbb{Z}$. 2. Take elements $a,b\in\mathbb{Z}_5$. Then $ab\in\mathbb{Z}_5$. 3. $a+(-b) = (a-b\mod 5)\in\mathbb{Z}_5$. Why not a subring? proofwiki.org/wiki/Subring_Test $\endgroup$ – sequence Feb 20 '17 at 1:42
  • 2
    $\begingroup$ the first point is debatable, but if you're saying $\mathbb{Z}_5 = \{0,1,2,3,4\}$, then sure. For the second, $4\cdot 4 = 16\notin \mathbb{Z}_5$ and $4+4=8 \notin \mathbb{Z}_5$. You can't just switch what the operations are half-way through the calculation. Saying that $(S,+_S,\cdot_S,0_S,1_S)$ is a subring of $(R,+_R,\cdot_R,0_R,1_R)$ means that $S\subset R$, $0_S=0_R$, $1_S=1_R$, $+_S = +|_{S\times S}$, and $\cdot_S = \cdot_R|_{S\times S}$. Alternatively, given a subset $S\subset R$, then it induces a subring as above exactly when $0,1\in S$ and $S$ is closed under the operations. $\endgroup$ – Hayden Feb 20 '17 at 1:46
  • $\begingroup$ @Hayden $ab \mod 5\in \mathbb{Z}_5$, but I now see that $\cdot_{\mathbb{Z}_5}$ is different from $\cdot_\mathbb{Z}$ due to the $\mod 5$ operator. So this is enough to see that it's not a subring. I'll now edit my post. $\endgroup$ – sequence Feb 20 '17 at 1:50

For (1), consider the (unital) ring $\mathbb{R}^2$ where addition and multiplication are calculated coordinate-wise, i.e. $$(a,b)+(c,d) = (a+c,b+d)$$ $$(a,b)\cdot (c,d) = (a\cdot c,b\cdot d)$$ Then the subset $\mathbb{R}\times \{0\} = \{(a,0) \mid a\in\mathbb{R}\}$. In this case, $\mathbb{R}\times \{0\}$ is a subring of $\mathbb{R}^2$ since it is closed under addition and multiplication and share the same additive identity. However, it is also unital with unit $(1,0)\neq (1,1)$, so it isn't a unital subring of $\mathbb{R}^2$.

For (2), consider the ring $\mathbb{R}$ with the usual operations. It has unital subring $\mathbb{Z}$. Then $\mathbb{R}^\times = \mathbb{R}\setminus\{0\}$, and clearly $\mathbb{R}^\times \cap \mathbb{Z} = \mathbb{Z}\setminus \{0\}$, though $\mathbb{Z}^\times = \{\pm 1\}$.


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