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This is a problem I've made up, which I cannot unfortunately solve. Any help will be appreciated.

Let $R$ be a commutative ring with unity and $\operatorname{char} R=0$. I want to find the ring $\hat{R}$ satisfying:

  • $R \subseteq \hat{R}$ or there is an injection $i: R \hookrightarrow \hat{R}$
  • For every non-zero integer $n$ and $r \in R$, the equation $nx = r$ can be solved in $\hat{R}$
  • If $S$ satisfies the above two conditions then $\hat{R} \subseteq S$ (or there is an injection etc.)

Let's call these minimal extension rings. Now, if $R$ is an integral domain, then its quotient field is an extension ring (need not be minimal), so the class of extension rings for domains is non-empty. $\mathbb Z$ has $\mathbb Q$ as its minimal extension ring.

Are all minimal extension rings of integral domains quotient fields?

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  • $\begingroup$ The quotient field of an integral domain won't necessarily satisfy your second requirement for (minimal) extension rings, unless the characteristic is 0. If the characteristic is $p\neq0$ then $px=1$ has no solution. $\endgroup$ – Andreas Blass Oct 25 '14 at 21:37
  • $\begingroup$ @AndreasBlass Thank you for the correction $\endgroup$ – erkenntnis Oct 25 '14 at 21:50
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We can define $\hat{R}$ as the localization $(\mathbb{Z} \setminus \{0\})^{-1} R$ at the multiplicative subset $\mathbb{Z} \setminus \{0\}$ of $R$; where we use the canonical embedding $\mathbb{Z} \to R$. Equivalently, we have $\hat{R} = \mathbb{Q} \otimes_{\mathbb{Z}} R$. The universal property is: If $f : R \to S$ is a homomorphism of commutative rings and $S$ has the property that every $n \in \mathbb{Z} \setminus \{0\}$ is a unit in $S$ (equivalently, equations $ns=s'$ can be solved), then $f$ has a unique extension to a ring homomorphism $\hat{R} \to S$.

If $R$ is an integral domain, then $\hat{R}$ is contained in $Q(R)$, but usually it is a proper subring. Consider $R=k[x]$ for example (where $k$ is a field), here $x^{-1} \notin \hat{R}$.

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  • $\begingroup$ @user26857 I don't understand your comment. $\endgroup$ – Martin Brandenburg Oct 26 '14 at 9:06

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