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I just don't get this. Looking through past papers, I came across this problem, Q. Let $F_2$ be a field with 2 elements. Let $P=x^3+x^2+1\in F_2[X]$. $I$ is the ideal of $F_2[X]$ generated by $P$. List all elements of the factor ring $F_2[X]/I$.

My answer; technically, I do get it correct. Having come across similar problems multiple times, I just guessed it'll be in the form $a+bx+cx^2$, where each coefficient is from the said field. Seems like the 2 elements are 0,1 so I can just list all combinations of it.

But, my question is, say what does $1+x^4+I$ become? I know I should treat $x^3+x^2+1=0$, but substitution doesn't help. And I cannot "decompose" or maybe factorize $1+x^4$ such that I can get a multiple of $x^3+x^2+1$ and let it be "absorbed' to $I$. It just confuses me, how I should specifically manipulate $1+x^4$, $1+x^5$, $x^5$ or anything like that.

Well, if I haven't explained my confusion clearly, simply put; Would someone please give me steps to obtaining what $1+x^4+I$, $1+x^5+x^7+I$ and so might be, and how I should think about doing it???

Maybe because it's extremely abstract but in the past 3 months, including my lecturer and textbook, no one is really able to explain it with extensive clarity really, or maybe math is just not my thing. This idea of Factor Rings, while I can give definitions if asked, just doesn't click.

It would be great if someone can help me out, thanks so much in advance

Pedro
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Melba1993
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  • Thanks everyone, that cleared it up for this question....still a little confusing when it comes to Factor rings, i still can't visualize how it's gonna look like for every single one but help appreciated :) – Melba1993 May 10 '15 at 19:16

3 Answers3

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In the quotient you have : $x^3=x^2+1$, hence $x^4=x^3+x=x^2+x+1$, &c.

Bernard
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Use the division algorithm: For $f, g \in F[x]$, there exist $q, r \in F[x]$ such that $f(x) = q(x)g(x)+r(x)$ with $\deg r < \deg g$.

For the example: we wish to interpret $f(x) = x^4+1$ modulo $g(x) = x^3+x^2+1$. Using the division algorithm, we have $$f(x) = (x+1)g(x)+(x^2+x),$$ so $(x^4+1)+I = (x^2+x+(x+1)g(x))+I = (x^2+x)+I$ because $g(x) \in I$.

Dustan Levenstein
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    Good answer. Small addendum: $F[x]$ is a Euclidean domain because $F$ is a field. If you want to use a division algorithm over more general polynomial rings, the leading coefficient of the polynomial you are dividing by must be a unit. – Alex Wertheim May 10 '15 at 01:18
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An important idea behind understanding factor rings $R/I$ is being able to simplify elements in the ring $R$ to certain forms after assuming (pretending, really) everything in $I$ equals $0$.

In $\Bbb F_2[x]/(x^3+x^2+1)$, we are taking $\Bbb F_2[x]$ and assuming that $x^3$ can be simplified to $x^2+1$, and as a consequence every power $x^n$ of $x$ higher than $2$ can be reduced to an expression involving strictly lesser powers (since $x^n=x^{n-3}x^3=x^{n-3}(x^2+1)=x^{n-1}+x^{n-3}$). Thus, whatever powers are present in a polynomial, we can keep simplifying the highest powers down to lower powers until eventually there are no more powers that are degree $3$ or higher, so everything looks like a quadratic in $x$, just as you guessed.

Moreover, no two quadratics can represent the same element, since their difference is quadratic which can't be in the ideal (as a quadratic can't be a multiple of a cubic) hence is nonzero.

anon
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