2
$\begingroup$

EDIT: The first paragraph has been indicated as inaccurate, please see this answer.

When we have something like $\mathbb{Z}_2[x]/(x^2 + x + 1)$, we understand that this means the set of polynomials over indeterminate $x$ where the coefficients are drawn from $\mathbb{Z}_2 = \{0,1\}$ with degree less than $\text{deg} (x^2 + x + 1)$.

So, in this set we can include polynomials like $x$, $x+1$ and even $1$, but not $x^2+1$ or $2x+1$. This is okay so far with me.

But in some lecture note (Rings and fields, Sergei Silvestrov, Spring term 2011, Lecture 5; page 3, Example 1 and subsequent ones) I find representations like, $\mathbb{Z}_2[x]/(x^2 + x + 1) = \{[0], [1], [x], [x + 1]\}$.

I fail to understand why the square brackets have been put around the polynomials.

Answer to this question says that this depends on the context and may be same as parentheses or may mean braces to represent sets.

Your help will be appreciated.

$\endgroup$
2
  • 2
    $\begingroup$ It's an equivalence class. All polynomials in $[0]$ are equivalent to the polynomial $0$ up to polynomial multiples of $x^2+x+1$. The nice thing is that the set of resulting equivalence classes inherits the structure of the original space. Hence the name quotient space. $\endgroup$ May 30 '13 at 10:07
  • $\begingroup$ @Raskolnikov Thanks for pointing to the right direction. Discussions in the book by Hungerford cleared things substantially. $\endgroup$
    – Masroor
    May 30 '13 at 19:49
2
$\begingroup$

Your first three lines aren't accurate: that symbol is for a field , represented by that quotient ring, the elements of which can be expressed as polynomials of degree zero or degree one in one of the roots of the polynomial $\;x^2+x+1\;$ over $\,\Bbb F_2\,$.

Thus, the elements in that "set" are not polynomials but expressions of the form $\,w, w+1,$ , etc., where $\, w^2+w+1=0\pmod 2\,$ , and precisely in order to avoid this confusion is that Silvestrov uses the symbols he does...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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