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A discrete valuation ring (DVR) is a principal ideal domain (PID) with exactly one non-zero maximal ideal.

Is $D=\left \{ x\in\mathbb{R}: \left | x \right |\leq 1 \right \}$ a discrete valuation ring?

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    $\begingroup$ No, it is not a ring. $\endgroup$
    – emiliocba
    Mar 4, 2016 at 12:42
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    $\begingroup$ It is not even a ring. $\endgroup$ Mar 4, 2016 at 12:42
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    $\begingroup$ Love the chance to write this: $1+1 = 2$. $\endgroup$ Mar 4, 2016 at 12:44

1 Answer 1

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Consider the ring axiom for closure under addition.

Indeed $1 \in D$ since $1 \in \mathbb{R}$ and $|1|=1 \leq 1$.

Now consider $1+1$ using $+$ defined on $\mathbb{R}$, $1+1=2\notin D$ since $|2|=2 \nleq 1$ hence $D$ does not satisfy closure under addition and so $D$ is not a ring.

So obviously $D$ cannot be a discrete valuation ring.

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