I'm just starting to learn about quotient rings. I was able to think about what type of of elements are generated by some $a$ when $\langle a \rangle$ is simply an integer, and also with simple quotient rings, but I just don't get this example.

My question about the quotient ring given, $(\mathbb{Z}/2\mathbb{Z})[x]/\langle x^2+x+1\rangle$ are:

  1. What kind of elements are in $\langle x^2+x+1 \rangle$? Applying the definition, I get something like $\{(x^2+x+1)q(x) \colon q(x) \in (\mathbb{Z}/2\mathbb{Z})[x]$}. I cannot readily list out the elements though. What is an intuitive way of thinking about the principal ideal generated by a polynomial?
  2. What type of elements are in $(\mathbb{Z}/2\mathbb{Z})[x]/\langle x^2+x+1\rangle$?
  3. The author claims $x^2 = x+1$. I don't see why.

I am self learning this, so this should help illuminate how I may think about quotient rings as I proceed. Thanks in advance.

  • 1
    $\begingroup$ your quotient ring is $\mathbb{F}_4$ I think since $x^2+x+1$ has no roots over $\mathbb{Z}_2$. As for why $x^2=x+1$ this is because a quotient ring is basically a ring formed by taking your original ring and breaking it up into equivalence classes with the ideal you're quotienting out by being the $0$ class. Thus $x^2+x+1=0$. As for intuition you're unlikely to find much in isolation, but as you study more algebra and see how this piece fits into larger and more general frameworks your intuition about it will increase. $\endgroup$ – Thoth Oct 22 '15 at 8:44
  • $\begingroup$ The principal ideal generated by an element $a$ in a commutative ring $R$ is by definition $aR=\{ar\mid r\in R\}$. This applies accordingly to $R=(\mathbb Z/2\mathbb Z)[x]$, so your description is correct and there is not much you can do about this. $\endgroup$ – user26857 Oct 22 '15 at 8:58

The golden rule:

In a quotient ring $R/I$ a residue class $\hat a$ is zero iff $a\in I$.

So (the residue class of) $x^2+x+1$ is $0$ in $(\mathbb Z/2\mathbb Z)[x]/(x^2+x+1)$. By an abuse of notation we can write $x^2+x+1=0$ (instead of $\widehat{x^2+x+1}=\widehat{0}$), and since in $\mathbb Z/2\mathbb Z$ we have $1=-1$ then $x^2=x+1$.

In order to find the form of the elements of $(\mathbb Z/2\mathbb Z)[x]/(x^2+x+1)$ start with a polynomial $f\in(\mathbb Z/2\mathbb Z)[x]$ and write $f(x)=(x^2+x+1)g(x)+r(x)$ with $\deg r\le1$. Now, by taking the residue classes modulo the ideal $(x^2+x+1)$ we get $f(x)=r(x)$, so the elements of $(\mathbb Z/2\mathbb Z)[x]/(x^2+x+1)$ are of the form $ax+b$ with $a,b\in\mathbb Z/2\mathbb Z$. (These are in fact only four: $0$, $1$, $x$, $x+1$.)

  1. The set you're written is correct. In any commutative ring ideal generated by one element $a$ is just $<a>=\{ra|r\in R\}$

  2. Ideal and quotient ring is similar to normal subgroup and quotient group , the element in $R/I$ (here your $R$ is $Z/2Z[x]$ and $I$ is $<x^2+x+1>$) is of the form $I+r$ with $r\in R$ and $I+r_1=I+r_2 \iff r_1-r_2\in I$.

  3. It's a short hand notation of $<x^2+x+1>+x^2=<x^2+x+1>+x+1$, which is equivalent to $x^2-x-1\in <x^2+x+1>$. In fact $x^2-x-1=x^2+x+1$ because the coefficient is in $Z/2Z$


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