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In Dummit & Foote:s "Abstract Algebra" (iirc 3rd ed.), p. 587, we are given that $x^3+x^2$ is a zero of $x^4+x+1$ as viewed inside $K := \mathbb{F}_2[x]/(x^4+x^3+1)$. I have a hard time finding that this holds:

One can note that \begin{align*} x^4+x+1-(x^4+x^3+1) &= x-x^3 =\\ &= x+x^3, \end{align*} so it is enough to check that $$(x^3+x^2)+(x^3+x^2)^3 \equiv 0 \ \mod (x^4+x^3+1) \qquad(1).$$

If we expand the left-hand side in $(1)$, we get \begin{align*} (x^3+x^2)+(x^3+x^2)^3 &= x^2+x^3+x^6+x^7+x^8+x^9 \qquad(2). \end{align*}

One can then use the relation $x^4 = x^3+1$ inside $K$ to simplify $(2)$; but I fail to find that this equals $0$. Dummit & Foote claims that a "simple computation" shows that this holds. What am I missing?

Ben123
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    Your deduction to $(1)$ is false. Indeed, $(1)$ is not true. – Dietrich Burde Oct 21 '24 at 18:28
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    Why would anyone use such confusing notation :-(. I would write $K=\Bbb{F}_2(\alpha)$, where $\alpha$ is declared to be a root of $x^4+x^3+1$. You can use $\alpha=x+\langle x^4+x^3+1\rangle$ if you really want to, but.... Anyway, the claim is that $\beta=\alpha^3+\alpha^2$ is a zero of $x^4+x+1$. All you need to do is to observe that $\beta\alpha=\alpha^4+\alpha^3=1$, so $\beta=1/\alpha$. Then use the fact that $x^4+x+1$ is the reciprocal polynomial of $x^4+x^3+1$. – Jyrki Lahtonen Oct 21 '24 at 18:28
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    (cont'd) In other words: $$\beta^4+\beta+1=(1/\alpha)^4+(1/\alpha)+1=\frac{1+\alpha^3+\alpha^4}{\alpha^4}=0.$$ – Jyrki Lahtonen Oct 21 '24 at 18:30
  • I used the very same pair of reciprocal polynomials at least here and here, albeit in a different context, so I should refrain from answering. Others are, of course, welcome to add their pieces – Jyrki Lahtonen Oct 21 '24 at 19:00
  • Wait, what is wrong with (1) @DietrichBurde? – Ben123 Oct 21 '24 at 19:29
  • Oh, wait, perhaps Sammy:s answer informs my mistake. – Ben123 Oct 21 '24 at 19:29
  • @JyrkiLahtonen Very nice! – Ben123 Oct 21 '24 at 21:08

1 Answers1

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This exercise is not stated very clearly. We are verifying that we have a root of a particular polynomial, and this polynomial has to be expressed in terms of an indeterminate. Any other symbol but $x$ would do. Here's a clearer statement of the exercise:

In the quotient ring $K = \mathbb{F}_2[x] / (x^4 + x^3 + 1)$, show that $x^3 + x^2$ is a root of $p(t) = t^4 + t + 1 \in K[t]$.

In this form, it's clear that we cannot reduce the polynomial in $t$ with coefficients in $K$ modulo the polynomial in $x$ that defines $K$.

How to work this out? One method that's methodical but can get a bit tedious for a larger quotient ring (field actually, but that's not so important right now) is to work out all the powers of $x$ back in terms of the basis $\{1, x, x^2, x^3\}$ via repeated application of the relation $x^4 = x^3 + 1$.

For instance, $$ x^5 = xx^4 = x(x^3 + 1) = x^4 + x = (x^3 + 1) + x = x^3 + x + 1. $$

The only monomials that you actually need in this particular calculation are
$$ \begin{aligned} x^{11} &= x^3 + x^2 + 1 \\ x^{14} &= x^3 + x^2 \\ x^{15} &= 1 \end{aligned} $$ Since $x^{15} = 1$ we only need to consider exponents modulo $15$: if $n = 15q + r$, then $$ x^n = x^{15q + r} = (x^{15})^q x^r = (1)^q x^r = x^r. $$ Thus, $$ \begin{aligned} p(x^{14}) &= (x^{14})^4 + (x^{14}) + 1 \\ &= x^{56} + x^{14} + 1 \\ &= x^{11} + x^{14} + 1 \\ &= (x^3 + x^2 + 1) + (x^3 + x^2) + 1 \\ &= 0. \end{aligned} $$

Sammy Black
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  • Ah, thanks for the clarification; indeed how you formluated it makes it clearer what we are trying to show... hm. Can you say more about "its clear that we cannot reduce the polynomial in $t$ with coeffcients in $K$ modulo the polynomial in $x$ that defines $K$". You are saying because $t^i$ is independent of $x$, and indeed $K[t]$ is an infinite-dimensional vector space with basis ${1,t,t^2,\ldots}$ so that we can't reduce $p(t)$ mod $x$. Or well, it is not entirely true since e.g. $t^4 = (\alpha^4+\alpha^3)t^4$. So we can reduce coefficients, but not variables $t^i$ itself. Do you agree? – Ben123 Oct 21 '24 at 22:13