Can $p(x)\in \mathbb{F}_{3}(x)$ with $p(x)=\frac{x²+x+1}{x+1}$ be expressed as a polynomial?

I tried it with different steps, like with polynomial long division:

$ (x^2 +x +1):(x+1)=x + \frac{1}{x+1} \\ -(x^2+x)\\ \quad \quad \quad\quad 1 $

So the division results to $ \frac{x^2 +x +1}{x+1}=x + \frac{1}{x+1}=x+(x+1)^{-1}$, which is not a polynomial because one exponent is not a natural number.

I'm not sure when I proofed that $p(x)$ can't be expressed as a polynomial. My intuition tells me that the polynomial long division is not enough, because then - I suppose- $p(x)$ couldn't be expressed as polynomial even if $p(x)\in K(x)$ for any field $K$ that I know* and I don't think that this is the case (although it would be possible).

What do you think? Is $p(x)$ not expressible as a polynomial for any of the fields I mentioned?

*|Since the result of the polynomial long division doesn't change for the fields $\mathbb{R},\mathbb{Q},\mathbb{C} $ and $\mathbb{F}_{p}$ (p is a prime number). |

I also tried to transform $p(x)$ using some of the properties of $\mathbb{F}_{3}$, but I haven't made much progress at this point: $$ \frac{x^2+x+1}{x+1} =\frac{x^2+4x+4}{x+1} =\frac{(x+2)^2}{x+1} =\ldots. $$

PS: I'm not used to write about math in english, please ask if something doesn't makes sense to you.

  • $\begingroup$ I'm happy about the answers, but I still have a question: The polynomial division results to $ \frac{x^2 +x +1}{x+1}=x + \frac{1}{x+1}=x+(x+1)^{-1}$, which is not a polynomial because one exponent is not a natural number. Since the result of the polynomial division doesn't change for $\mathbb{R},\mathbb{Q},\mathbb{C} $ and $\mathbb{F}_{p}$, $p(x)$ is not a polynomial for the given fields*. Would this argument also proof that $p(x)$ is not a polynomial for the given fields? If not, why? *I only consider the fields that I‘m familiar with. $\endgroup$ – CherryBlossom1878 Mar 1 at 18:25

If it could be expressed as a polynomial, then $$(x+1)(x+a)=x^2+x+1$$ for some $a$. But comparing coefficients on both sides yields $a=0$ and $a=1$, which is a contradiction over any field.

  • $\begingroup$ Why should it be a monic linear polynomial? $\endgroup$ – Servaes Mar 1 at 17:11
  • $\begingroup$ If not, the LHS would not have $x^2$ as monomial of highest degree. $\endgroup$ – Dietrich Burde Mar 1 at 17:16

Because every polynomial in $\Bbb{F}_3[x]$ defines a function on $\Bbb{F}_3$, whereas $p(x)=\frac{x^2+x+1}{x+1}$ does not; it is not defined for $x=-1$ because the denominator then equals zero but the numerator doesn't. The same problem occurs in every field.

  • $\begingroup$ But if $x^2+x+1$ over some field would factor as, say $(x+1)(x-a)$, then we could cancel $x+1$ and the argument with $x=-1$ would not work. $\endgroup$ – Dietrich Burde Mar 1 at 17:04
  • $\begingroup$ @DietrichBurde Over any field you have $(-1)^2+(-1)+1=1$, which doesn't vanish in any characteristic. $\endgroup$ – Servaes Mar 1 at 17:06
  • $\begingroup$ Yes exactly, this is missing. It exactly shows that this cannot factor like this. $\endgroup$ – Dietrich Burde Mar 1 at 17:07
  • $\begingroup$ @DietrichBurde I have made this argument more explicit. $\endgroup$ – Servaes Mar 1 at 17:09
  • $\begingroup$ @DietrichBurde In my opinion that's not necessary. $p$ is undefined at $-1$ regardless of whether any cancelling can be done. A polynomial cannot be undefined. $\endgroup$ – Arthur Mar 1 at 17:46

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