Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm having a hard time wrapping my head around quotient rings that isn't something of the form $\mathbb{Z}/n\mathbb{Z}$.

Is it correct to think of $\mathbb{Z}[x]/(x-1)$ as the ring of functions with integer coefficient with zeros at one? So this would be the ring of functions with solutions $x=1$

share|cite|improve this question
Try… it has some nice examples as well as some more general results. – Robert Cardona Feb 12 '13 at 5:15
up vote 4 down vote accepted

First of all, the ring $\,R:=\Bbb Z[x]/(x-1)\,$ is a quotient ring of polynomials , not functions.

You can look at $\,R\,$ as the ring of residues of polynomials modulo $\,x-1\,$ , in a very similar way as $\,\Bbb Z/n\Bbb Z\,$ is the ring of residues modulo an integer $\,n\,$.

How do the element in $\,R\,$ look? Well, using Euclides algorithm, divide with residue any integer polynomial by $\,x-1\,$ :

$$p(x)\in\Bbb Z[x]\,\,,\,\,p(x)=g(x)(x-1)+r(x)\,\,,\,\,\deg r<1\,\,\,or\,\,\,r(x)=0$$

Thus, in the quotient ring, we can write


which means $\,p(x)=r(x)\pmod{(x-1)}\,$

But, of course, $\,\deg r<1\Longleftrightarrow r(x)=$constant or $\,r(x)=0\,$, so now it isn't hard to prove that $\,R\cong\Bbb Z\,$ ...

share|cite|improve this answer

You can think of $R/I$, for an ideal $I$ in $R$, as if it is $R$ where certain things that were not zero before might now be zero. The 'new' things that are zero are the things in the ideal. The operation is just as before, only done on representatives. By this you quotient $I$ away.

In more detail, if you look at the construction of the quotient ring then you'll see that the definition is very simple, and it's just the details in proving that it really is a ring that are a bit cumbersome. The elements in the quotient ring $R/I$ are all of the form $r+I$ with $r\in R$. Addition and multiplication are performed on representatives: $(r+I)+(s+I)=(r+s)+I$ and $(r+I)(s+I)=(rs)+I$. The crucial thing to remember is that now equality of elements in the quotient is not so obvious. Namely, $r+I=s+I$ precisely if $r-s\in I$.

So, the way to think of the abstract quotient construction of $R/I$ is that "it's just $R$ where everything in $I$ is forced to be $0$". Of course the isomorphism theorem related this construction to kernels and images, giving you another way to think of quotients.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.