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What are some examples of this? Say for $F_{4}$. I know this is a very simple question, but I can't find any info on it.

Edit: Yes, I was thinking of $F_{2}[x]/(x^2+x+1)$. I was confused.

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Maybe I'm missing something, but isn't it just $ax+by=c$ as usual? – anon Feb 13 '12 at 8:46
@ad423 you just write a normal equation of a line (as anon mentions), but now you replace numbers by polynomials over $\mathbb{F}_4$. That is, every constant is now allowed to be a polynomials and variables vary over polynomials.. The polynomials themselves should have coefficients in $\mathbb{F}_4$. – aelguindy Feb 13 '12 at 9:41
The title says that the field is a polynomial field. I automatically (mis-)read $F_4[x]$ in the question. I don't know which one OP means. – aelguindy Feb 13 '12 at 10:11
@ad423: Thanks for the edit! Makes more sense now! – Jyrki Lahtonen Feb 14 '12 at 6:36

Ok. So $F_4=\{0,1,\alpha,\alpha+1\}$, where $\alpha$ is a root of the equation $\alpha^2=\alpha+1$. Or, using the notation of quotient rings, $\alpha=x+ I\in F_2[x]/I$, where $I$ is the ideal generated by $x^2+x+1$.

About lines over $F_4$? The recipe from anon's comment applies to any field. A line is the set of solutions $(x,y)$ of a linear equation $$ ax+by=c, $$ where $a,b,c$ are constants from that field, and at least one of $a,b$ is non-zero. If here $b\neq0$, then we can solve $y$ from this equation, and express the line in the 'high-school form' $$ y=-\frac{a}{b}x+\frac{c}{b} $$ with the 'slope' $-a/b$ and 'intercept' $c/a$ visible. OTOH, if $b=0$, then we assumed that $a\neq0$, and (solving $x$) the equation is equivalent to $$ x=\frac{c}{a} $$ that represents a line parallel to the $y$-axis.

As an example of a line over $F_4$ let us look at the following equation $$ \alpha x+\alpha^2y=1. $$ Because $\alpha^2=\alpha+1\neq0$, we can solve for $y$. To that end we need the inverse of $\alpha^2$. Here we get $$\alpha^3=\alpha\cdot\alpha^2=\alpha(\alpha+1)=\alpha^2+\alpha=(\alpha+1)+\alpha=2\alpha+1=1$$ by repeatedly applying the equation $\alpha^2=\alpha+1$. So $\alpha^2\cdot\alpha=1$, and we can infer that $\alpha=(\alpha^2)^{-1}$ is the inverse. Let's multiply our equation by $\alpha$! We get $$ \alpha\alpha x+\alpha\alpha^2 y=\alpha 1\Leftrightarrow \alpha^2x+y=\alpha. $$ This gives us the equation of the line $$ y=-\alpha^2x+\alpha=\alpha^2x+\alpha=(\alpha+1)x+\alpha $$ in the usual 'slope and intercept' -form.

We can also list all the four points on this line. We get the $y$-coordinate of a point from this equation by plugging in all four elements of $F_4$ as the $x$. The points are (leaving this as an exercise in $F_4$-arithmetic: $$ (0,\alpha),\ (1,1),\ (\alpha,\alpha+1),\ (\alpha+1,0). $$

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