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I've got a problem from book Integer, Polynomials, and Rings by Ronald S. Irving, page 227.

The question:

Is there a value of the coefficient $c$ in the field $\mathbb{R}$ that makes $x^4+3x^3+2x+1 \equiv 3x^4+x^2+cx \pmod{x^2+x+2}$ a valid congruence in $\mathbb{R}[x]$? If so, what is it; if not, why?

So first, I want to find the difference of the two polynomials.

$(x^4+3x^3+2x+1)-(3x^4+x^2+cx)$

$=(-2x^4+3x^3-x^2+(2-c)x+1)$

$=(x^2+x+2) U $

But how to find the value of $c$? Is there a method that I should be use? I appreciate any help. Thank you.

liz_chan
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1 Answers1

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We use polynomial division. Note that $$-2x^4 + 3x^3 - x^2 = (x^2+x+1)q(x) + r(x)$$ for $r$ a linear polynomial and $q$ a quadratic. We solve for $q, r$ now.

$$(x^2+x+1)(ax^2 + bx + c) + dx + e = -2x^4 + 3x^3 - x^2$$ implies $$ax^4 + (a+b)x^3 + (a+b+c)x^2 + (b+c+d)x + (c+e) = -2x^4 + 3x^3 - x^2.$$

Therefore $a=-2, b=5, c=-4$ is our quotient, and $r(x) = -1x + 4$ is our remainder.

The only way this congruence could hold is if $-2x^4 + 3x^3 - x^2 =(x^2+x+1)q(x) + (c-2)x-1.$ Hence the fact that our remainder has constant term 4 implies there's no $c$ making this true.

  • Thank you for your answer. Sorry I am not really smart. I want to check my understanding about this with the value of $c$ if in $\mathbb{Z}_5[x]$. I got $c=4$. Since the difference of the two polynomials by substituting $c=4$ is $(-2x^4+3x^3-x^2-2x+1)=(-2x^2+5x-2)(x^2+x+2)+(5-10x)$ so the remainder is $0$ in $\mathbb{Z}_5[x]$. Thus $c=4$ makes a valid congruence in $\mathbb{Z}_5[x]$? Is that true? – liz_chan May 20 '21 at 05:39
  • @liz_chan Well, you're smarter than most people : ) What you wrote seems to wrok to me. –  May 20 '21 at 06:13