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I am reading a paper about rings (http://malaschonok.narod.ru/publ/ma01.ps, 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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    You can see [this](http://math.stackexchange.com/questions/69050/question-on-quotient-group) one too for more concrete interpretation.2012-07-16

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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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    @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."2012-07-16