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Let $R\neq S$ be rings with unity. Let $R$ be a subring (sharing the same unity) of $S.$ Let $\{0\}\neq K\subseteq R.$ Is it possible that $K$ is at the same time an ideal in $R$ and an ideal in $S?$

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Should your subrings have unity? Because some people don't care about my friend unity. – user21436 Jan 20 '12 at 13:11
0$\,\,\,\,\,\,$ – M P Jan 20 '12 at 13:28
@ymar $\{0\}$ is an ideal in every ring. – Dylan Moreland Jan 20 '12 at 14:05
You might search around for examples of "conductors" of ring extensions. – Dylan Moreland Jan 20 '12 at 14:15
An other trivial example is when $R=S$. It's a very good question! – emiliocba Jan 20 '12 at 14:29
up vote 2 down vote accepted

Let $O$ be a valuation domain of Krull dimension >1 and let $p\neq 0$ be a non-maximal prime of $O$. Then every proper ideal of the localization $O_p$ is a proper ideal of $O$ itself.

To show this it suffices to prove the inclusion $pO_p\subset O$: It then follows that $pO_p=p$ and since $pO_p$ is maximal, every ideal of $O_p$ is contained in $p$ and thus in $O$.

So assume that some element $\frac{a}{b}$, $a\in p$, $b\in O\setminus p$ is not in $O$. Since $O$ is a valuation domain one concludes $\frac{b}{a}\in O$, hence $b\in p$, a contradiction.

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Let $S$ be the set of sequences in $\mathbb{Z}/\mathbb{2Z}$, which is a ring with term-wise addition and multiplication. The set of sequences that are eventually constant is a subring R (it contains $1=(1,1,1,...)$). The set of sequences that are eventually zero is an ideal $K \subset R$ such that $K \lhd S$.

A simpler example works with $S=\mathbb{Z}/\mathbb{2Z} \times\mathbb{Z}/\mathbb{2Z} \times \mathbb{Z}/\mathbb{2Z}$, $R$ being the subring with the last two entries the same, and $K$ being the ideal with the last two entries zero.

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These are great examples. Thanks. – user23211 Jan 20 '12 at 16:45

Consider $S = \mathbb Q[x]$, the rational polynomials in one unknown. Define $R = \mathbb Z + \mathbb x\mathbb Q[x]$. Now $K = x\mathbb Q[x]$ is an ideal in both rings.

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More generally, given any graded ring $R=\oplus R_i$, and a graded subring $S=\oplus S_i$, such that, for some $n$, $\forall i\geq n: S_i=R_i$. Then $\oplus_{i\geq n} R_i$ is an ideal in $R$ and $S$. – Thomas Andrews Jan 20 '12 at 15:56
Thank you very much. I will accept Hagen's answer though, because yours has many upvotes anyway, and I think both should be on top. – user23211 Jan 20 '12 at 16:43
Yes, I upvoted @CruiskeenLawn because that was interesting to me. – hardmath Jan 20 '12 at 22:20

If $K$ is an ideal in $S$, then it's an ideal in $R$ (by definition).

Example: $$ K=\left\{\left(\begin{array}{cc}0&b\\ 0&c\end{array}\right):b,c\in\mathbb R\right\} $$ is a left-ideal in $ R=\left\{\left(\begin{array}{cc}a&b\\ 0&c\end{array}\right):a,b\in\mathbb R\right\} $ and in $S=M_2(\mathbb R)$.

Conversely, it's not true that if $K$ is an ideal in $R$ then it is neccesarly an ideal in $S$.

Counterexample: $$ K=2\mathbb Z\subset R=\mathbb Z\subset S=\mathbb Z[\sqrt{2}]=\{a+b\sqrt{2}:a,b\in \mathbb Z\}. $$ Clearly, $2\mathbb Z$ is not an ideal in $\mathbb Z[\sqrt{2}]$.

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But this doesn't answer my question! Of course an ideal in a subring doesn't have to be an ideal in the superring. I'm asking if it can be. – user23211 Jan 20 '12 at 13:51
It's a nice example now, but still not to the question I asked. I was asking about ideals, not left ideals. But this is something worth noticing, still. Thanks. – user23211 Jan 20 '12 at 15:16

Another example: $\mathbb Z$ embeds into $\mathbb Z[x]/(2 x)$, then $(2)$ is an ideal in both rings.

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