17 questions linked to/from The subring test
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### If $a \mid m$ and $(a + 1) \mid m$, prove $a(a + 1) | m$.

Can anyone help me out here? Can't seem to find the right rules of divisibility to show this: If $a \mid m$ and $(a + 1) \mid m$, then $a(a + 1) \mid m$.
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### Prove that $(n\mathbb Z, +, \times )$ are the only subrings of $(\mathbb Z, +, \times)$

I had to find all the subrings of the integers and then prove that there aren't any more. It's clear to me the $(n\mathbb Z, +, \times )$ is a subring of the integers for all $n$ element of the ...
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### Show that $A=\{x+y\sqrt{2}:x,y \in \mathbb{Z}\}$ is a commutative ring with unity, find the zero element, the unity and the negative of $a$

Show that $A=\{x+y\sqrt{2}:x,y \in \mathbb{Z}\}$ is a commutative ring with unity, find the zero element, the unity and the negative of an arbitrary $a$. First thing first, I need to show it is a ...
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### Proof Verification: Let S be a non-empty subset of a ring R. Then S is a subring of R if and only if S is closed under $-$ and $\times$.

Let S be a non-empty subset of a ring R. Then S is a subring of R if and only if S is closed under $-$ and $\times$. Proof: First, prove that S is a subgroup of R. Pick an arbitrary element $x$ from ...
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Let $R = \{a + b\alpha |\ a,b \in \mathbb{Z}\}\subseteq \mathbb{C}$ where $\alpha = \frac{1}{2}(1+\sqrt{-19})$ Is $R$ an integral domain? To show whether or not $R$ is an integral domain, letting $x ... 1answer 291 views ### Subring Test for non-empty subset$S ⊂ R$[closed] Show that a non-empty subset$S ⊂ R$is a subring of$R$if for all$r, s ∈ S$we have$r − s ∈ S$and$rs ∈ S$. (This makes it easier to verify a set is a ring, if you know the set lives in a larger ... 3answers 147 views ### Prove that$\{x + y \cdot \sqrt{3} | x, y \in \mathbb{Z} \}$is a ring (or not) How to prove that$(R, +, \cdot)$is a ring (or not), where$R = \{x + y \cdot \sqrt{3} | x, y \in \mathbb{Z} \}$? Update. Is this proof correct?$(R, +)$is an abelian group: Closure:$a, b \in R \...
Let's say I have the ring: $\mathbb{Z}[\sqrt{2}]=\{a+b\sqrt{2}\mid a,b\in\mathbb{Z}\}$. Now the question asked is to prove whether or not this ring is an integral domain. By definition: "An ...