# Can an integral domain be embedded in a proper quotient of itself?

Does there exist an integral domain $R$ which has a proper ideal $J$ so that there exists an injective ring homomorphism $\phi \colon R \to R/J$?

If yes, what are suitable assumptions on $R$ to exclude such a behaviour? Especially, if $R$ is a finite dimensional complete Noetherian integral domain, can such behaviour occur?

Take the integral domain $R = k[x_1,x_2,\ldots]$ and let $J = (x_1)$. Then $R/J \cong k[x_2,x_3,\ldots]$ which is isomorphic to $R$ because you have infinitely many indeterminates.

Here are some thoughts for your second question: I think if you throw in the assumption that $R$ is a finitely generated $k$-algebra which is also an integral domain, then there is no non-zero prime ideal $\mathfrak{p}$ such that we have an injection $R \to R/\mathfrak{p}$. Indeed, if $\mathfrak{p}$ is non-zero its height is at least one, so that the formula $$\operatorname{ht}(\mathfrak{p}) + \dim R/\mathfrak{p} = \dim R$$ says $\dim R/\mathfrak{p} < \dim R$ and so you can't have an injection $R \to R/\mathfrak{p}$. For finitely generated $k$-algebras $R,S$ that are domains with $R \subseteq S$, we have $\dim R \leq \dim S$ as pointed out by YACP.

• Thank you, that answers the first part. – Sebastian Sep 20 '13 at 15:34
• I'd be very thankful if you could give me some more details. Why exactly is it not possible to have injection $R \to R/ \mathfrak{p}$? What do you need to argue that the ring cannot inject into a lower dimensional one and where do you use the assumption that $\mathfrak{p}$ is prime? Thanks a lot in advance. – Sebastian Sep 21 '13 at 13:08
• Yes, that seems to be the issue of the whole question.I would not have thought that the notion dimension behaves that "anti-intuitive". – Sebastian Sep 21 '13 at 14:19
• @user38268 Your answer (now deleted) seems to be okay. For finitely generated $k$-algebras $R\subset S$ which are integral domains we have $\dim R\le\dim S$. (Actually I've thought that you counted on this!) – user26857 Sep 21 '13 at 19:44
• @user38268 Simply use the fact that their dimension coincides with the transcendence degree of their fraction fields. – user26857 Sep 22 '13 at 8:53

EDIT: The following arguments are incomplete (and thus my response might false) - see the comments.

It is enough to assume that $R$ is a finite dimensional Noetherian integral domain in order to exclude this behaviour. In Eisenbud's book (p.219) one can find the following theorem:

If $R\subset S$ are Noetherian rings such that $S$ is a finitely generated $R$-module, then $\dim R = \dim S$.

So if $R$ is Noetherian, and $\phi$ is as above, then $R/J$ is a finitely generated $\phi(R)$-module, which thus has the same dimension as $R$.

On the other hand, for any ideal $J$ of $R$ one has

$$\dim\ R/J + \operatorname{ht}(J) \le \dim R.$$

Since any proper ideal has height greater/equal one (here we use the domain assumption), we obtain the contradiction $$\dim R = \dim R/J \le \dim R -1.$$

• Is it obvious that "$R/J$ is a finitely generated $ϕ(R)$-module"? – user26857 Sep 20 '13 at 18:18
• $R/J$ is f.g. over $R$ because $R$ is Noetherian, and since $R \cong \phi(R)$ as rings, it is also f.g. over $\phi(R)$. – Sebastian Sep 20 '13 at 18:27
• Thank you - the domain assumption will be necessary for this - I'll edit it. – Sebastian Sep 20 '13 at 18:29
• I'm not convinced at all by your arguments about the finiteness of $R/J$ over $\phi(R)$. This has nothing to do with the fact that $R$ is noetherian: $R/J$ is always finitely generated (cyclic) over $R$. It seems to me that $R/J$ doesn't keep the track of its $R$-module structure when change to $\phi(R)$. – user26857 Sep 20 '13 at 18:41
• Sorry, you are absolutely right, it seems that this does not hold. I have to think about it, but won't have time in the next two days. – Sebastian Sep 21 '13 at 12:56