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Question: In $\mathbb{F}_3[x]$ write, if possible, the polynomial $1$ in the form: $$f(x)p(x)+g(x)q(x)=1$$ Where $p(x)=x^3+x^2+x+2$ and $q(x)=x^3+2x^2+2$.

This is a question from Concrete Introduction to Higher Algebra (Lay) assigned in my Classical Algebra class.

I first jumped to using the EEA matrix, but ran into issues.

I then used Euclid's algorithm/long division to find that $\gcd(p(x),q(x))=-x^2+x \neq 1$ so am I correct in saying that there is no way to write the polynomial $1$ in the form stated above?

My other question was, does the fact that these polynomials are in $\mathbb{F}_3[x]$ make any tangible difference in solving this problem?

pyridoxal_trigeminus
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    $x$ divides $-x^2+x$ but neither of $p(x)$ and $q(x)$, so you got the wrong gcd. Sometimes it made a difference. E.g. the polynomials $x^2+2$ and $x+1$ are coprime in $\mathbb Q[x]$, but in $\mathbb F_3[x]$, $2=-1$, hence the first poly $=x^2-1$ and is divisible by $x+1$, so $\gcd(x^2+2, x+1)=x+1$ in $\mathbb F_3[x]$. – Just a user Oct 26 '21 at 04:09
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    I redid my Euclid's algorithm and after reducing mod 3 I got the gcd to be 2. Does that sound right? If so, does that mean there is no solution to the above equation for 1? – pyridoxal_trigeminus Oct 26 '21 at 05:01
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    Yes, that's correct. And remember in $\mathbb{F}_3[x]$, $2 = -1$, and since $-1$ is a unit, you might as well say the GCD is $1$. – Sammy Black Oct 26 '21 at 05:08
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    @SammyBlack You mean yes correct on the gcd 2 part right? But not on whether 1 can be obtained on right side of OP equation? [Because if 2 an be obtained then since 2=-1 then -1 can be, and thn finally can get 1.] – coffeemath Oct 26 '21 at 05:12
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    In this ring, $2 = -1$, so you can always change signs in any equation by multiplying by $2$. – Sammy Black Oct 26 '21 at 05:16

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Euclidean algorithm: $$ \begin{array}{rcrll} x^3+x^2+x+2 &=& 1 &\cdot& (x^3+2x^2+2) &{}+ (2x^2+x) \\ x^3+2x^2+2 &=& 2x &\cdot& (2x^2+x) &{}+ 2 \\ 2x^2+x &=& (x^2+2x) &\cdot& 2 & \end{array} $$ This shows that $2 = -1$ is the GCD, but it's only defined up to a unit, so it's equivalent to the fact that $1 = 2 \cdot 2$ is the GCD, i.e. the polynomials are relatively prime.

Now, the backwards phase of the extended Euclidean algorithm will express $2$ as a $\mathbb{F}_3[x]$-linear combination of $p(x)$ and $q(x)$, which we can negate to get the Bézout identity. Begin with the penultimate line, and substitute back up the chain of equations (using $2 = -1$ freely): $$ \begin{array}{rcrcl} 2 &=& (x^3+2x^2+2) &{}+{}& x \cdot (2x^2+x) \\ &=& (x^3+2x^2+2) &{}+{}& x \cdot \bigl( (x^3+x^2+x+2) + 2(x^3+2x^2+2) \bigr) \\ &=& x \cdot (x^3+x^2+x+2) &{}+{}& (2x+1) \cdot (x^3+2x^2+2). \end{array} $$ So, $$ x \cdot p(x) + (2x+1) \cdot q(x) = 2, $$ which we can multiply by $2$ to get $$ 2x \cdot p(x) + (x+2) \cdot q(x) = 1. $$

Sammy Black
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