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I am reading a paper about rings (, page 3). In this paper the term "quotient-ring" appeared.

What is a quotient-ring?

(Note: The text in the original paper has "quatient-ring" twice at the top of page 3, rather than "quotient ring")

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I believe you're looking for quotient rings. Is there a specific point in the wikipedia article that needs clarification, or is it just generally too abstract (totally understandable)? – Robert Mastragostino Jul 16 '12 at 2:57
I have that background. The link of the paper is, page 3. – juaninf Jul 16 '12 at 3:49
@Juan: Why did you edit to re-introduce the spelling error "quatient" instead of "quotient"? The manucript in question clearly has a typo, to go with the rest of the bad grammar. The very next sentence says "quotient-ring" again. – Arturo Magidin Jul 16 '12 at 13:00
As for other algebraic structures, a quotient ring is the set of equivalence classes of a ring under an equivalence relation that is compatible with the ring operations; a ring structure can then be defined in the quotient. This happens exactly when the equivalence class of $0$ is an ideal of the ring. – lhf Jul 16 '12 at 16:15
You can see this one too for more concrete interpretation. – Iyengar Jul 16 '12 at 18:24
up vote 6 down vote accepted

The statements at the bottom of page 2 and the top of page 3 are:

Proposition 1. QuasiEuclidividy is inherited by its quotient-ring

The quotient-ring of the ring of principal ideal is a ring of principal ideal.

I will note that the English is rather poor (the author is from Russia, so that may explain it). The first sentence should be "to quotient rings" or "to quotients", not "by its quotient-ring" (since there is nothing the 'its' can refer to). The second sentence should read: A quotient of a principal ideal ring is a principal ideal ring.

(The paper does say "quatient-ring", but this is pretty clearly a typo, to go with the rest of the bad grammar in the manuscript).

(I wanted to make sure they did not mean "ring of quotients" before answering)

A "quotient-ring" is just the quotient of a ring by an ideal. Given a ring $R$ and an ideal $I$ of $R$, the set of equivalence classes $R/I$ modulo the equivalence relation $$a \equiv b \pmod{I}\iff a-b\in I$$ is a ring under class operations: if $[x]$ is the equivalence class of $x$, the operations on $R/I$ defined by $$\begin{align*} {}[a] + [b] &= [a+b]\\ {}[a]\cdot[b] &= [a\cdot b] \end{align*}$$ are well defined and make $R/I$ into a ring, and the map $R\to R/I$ given by $a\mapsto [a]$ is an onto ring homomorphism, and the usual Isomorphism Theorems apply.

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thanks Arturo by your response, taking advantage of your explain I don't understand why the author write $E_{\alpha'}=\phi (E_\alpha)$ (pag. 2). This because, he defined $\phi$ than $\phi: R \longrightarrow R'$, and $E_\alpha$ is a matrix and not element in $R$. – juaninf Jul 16 '12 at 13:10
@Juan: Sorry, but I'm not about to wade through a badly-written paper in an area that does not interest me. Moreover, this is a separate question from what you asked in this post, and currently, I can only read that paper currently with an on-line Postscript viewer, which leaves much to be desired. If this had been your original question, rest assured I would not have bothered trying to answer you. – Arturo Magidin Jul 16 '12 at 13:14
understand Arturo thanks by your reply equal – juaninf Jul 16 '12 at 13:47
@Juan: Si queres decir "Igual, gracias por tu respuesta", la traduccion correcta no usa "equal", sino mas bien la expresion seria "Thank you for your reply all the same." – Arturo Magidin Jul 16 '12 at 18:04

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