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Let $R$ be a commutative ring with unity and $K$ a subring of $R$. I want to prove that If $K$ is a field and $R$ has no zero-divisors, then $1_R=1_K$ (in particular $1_R \neq 0$, so $R$ is an integral domain an a K-vector space).

Also I want a counterexample for when $R$ has zero-divisors.

I don't know how to show the claims that are mentioned about. Any idea would be apprieciated.

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It's sufficient to assume only that $R$ has no zero divisors. This is because if $0\neq 1_K\neq 1_R$, then $1_K$ is an idempotent of $R$, hence $$(1_K-1_R)1_K=1_K1_K-1_R1_K=1_K-1_K=0$$ This contradicts the assumption that $R$ has no zero divisors.

This also allows for relatively easy construction of your counterexample. If you want something explicit, we can take $R=\mathbb{Z}_2\times \mathbb{Z}_2$ and $K=\mathbb{Z}_2\times\{0\}$.

Edit: There's a bit of a subtlety both of us missed: If you don't require that $R$ have unity distinct from the additive identity, our proofs won't work. In that case the assumption that $K$ is a field comes in, because in a field $1\neq 0$.

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  • $\begingroup$ Why is that $\;1_R1_K=1_K\;$ ? I'd say that $\;1_R1_K=1_R\;$ ... $\endgroup$ – DonAntonio Oct 14 '16 at 11:58
  • $\begingroup$ @DonAntonio because $1_Rr=r$ for all $r\in R$, including $1_K$. $\endgroup$ – Matt Samuel Oct 14 '16 at 11:59
  • $\begingroup$ @Ma Of course. Thank you. +1 $\endgroup$ – DonAntonio Oct 14 '16 at 12:00
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The question statement seems a little overcomplicated, so I'd like to rephrase and answer this way:

Lemma If $R$ is a commutative domain (i.e. not necessarily with identity, and if $a,b$ are any elements such that $ab=0$ then one of $a,b$ is zero) then there are at most two idempotents: $0$ is always one of them, and if there is a second one, it is the identity of the ring and we'll call it $1$.

Proof: If $e$ is a nonzero idempotent, then $e(ex-x)=0$ implies $ex-x=0$ for all $x$ in the ring, so that $e$ is the identity of the ring.

In your original question, stipulating that $K$ is a field is just a kludgy way to get it to have identity. Since $R$ and $K$ are both domains (in the sense I mentioned above) then both of them only have two idempotents: they must share two elements $0$ and $1$ which act as the additive and multiplicative identities in both rings, thanks to the lemma above.

Corollary (enhanced rewritten version of question) If $R$ is a commutative domain and $K$ is a subring with nonzero identity, then $R$ has identity, and it is the same as $K$'s.

For a counterexample, you could consider a field $F$ and $R=F\times F$ where you have the subring $\{0\}\times F$ (isomorphic to $F$) with identity $1_K=(0,1)\neq (1,1)=1_R$.

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Actually it is enough to know, that $R$ has no zero-divisors or even that $K$ contains no zero-divisors of $R$: Assume you have an element $1_K \in K$ with $1_K k=k$ for all $k \in K$. Hence you have $(1_R -1_K)k=k-k=0$. Taking any $k \neq 0 \in K$ and using $k$ is no zero-divisor you are done.

Your counterexample:

As a non-commutative example take $R= \mathbb R^{2\times 2}$ (for commutative take $R=\{\begin{pmatrix} x&0\\0&y \end{pmatrix} | x,y \in \mathbb R\}$, which is practically the same example as the others stated) and $K=\{\begin{pmatrix} x&0\\0&0 \end{pmatrix} | x \in \mathbb R\}$. You can easily see, that $1_K=\begin{pmatrix} 1&0\\0&0 \end{pmatrix}\neq E_2 = 1_R$

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