# Can $p(x)\in \mathbb{F}_{3}(x)$ with $p(x)=\frac{x²+x+1}{x+1}$ be expressed as a polynomial? Is it not possible for any of the given fields?

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.

• 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. – 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.
• If not, the LHS would not have $x^2$ as monomial of highest degree. – 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.
• 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. – Dietrich Burde Mar 1 at 17:04
• @DietrichBurde Over any field you have $(-1)^2+(-1)+1=1$, which doesn't vanish in any characteristic. – Servaes Mar 1 at 17:06
• @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. – Arthur Mar 1 at 17:46