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Let $F$ be a field, and suppose that $X^2 + X + 1$ is irreducible in $F[x]$. Let $K = F[x]/(X^2+X+1)F[X]$. Find a polynomial $q(X)$ of degree at most 1 satisfying $(q(X) + I) \cdot ((X+1) + I) = 1 + I$

So basically we want to find a $q(X)$ such that $q(X)(X+1) - 1 \in I \Longleftrightarrow q(X)(X+1) - 1 = f(X)(X^2+X+1)$ for some $f(X) \in F[x]$. The question is how do I find such a $q(X), f(X)$. And I looked through the previous duplicates and I still just don't get this. I don't get how you can just "treat $X^2 + X + 1$ as $0$" Like sure, they're equivalent in the sense they both exist in the ideal, but we can just replace? And everyone is saying to use the Division Algo, but on what? I'm so confused and have been stuck for hours.

beginner
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1 Answers1

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You need $q$ to be of degree at most $1$, so you just try a general polynomial of degree $1$ and see what it has to be. We have:

$(ax+b)(x+1)=ax^2+(a+b)x+b$

And we want it to satisfy an equality of the form $ax^2+(a+b)x+b=c(x)(x^2+x+1)+1$. By comparing degrees $c=c(x)$ has to be a constant. So we get:

$ax^2+(a+b)x+b=cx^2+cx+(c+1)$

Comparing coefficients gives $a=c, b=0, c=-1$. So the polynomial you are looking for is $q(x)=-x$.

Mark
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  • Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Feb 22 '23 at 02:58
  • @BillDubuque And how am I supposed to know this is a duplicate? The question looked legit, and gave me no reason to look at OP's history. It was closed after I wrote an answer. OP even mentioned that in none of these "duplicates" he got an answer. – Mark Feb 22 '23 at 13:58
  • This ssite is over 13 years old. Of course common questions like this has been asked many tens (if not hundreds) of times already. – Bill Dubuque Feb 22 '23 at 18:50