How do I subtract in $\mathbb{Z}_{n}$?

I'm currently trying to understand polynomial division in abstract algebra. How is it possible for a polynomial with negative coefficients, say $f(x)=x^{4}-3x^{3}+2x^{3}+4x-1$ be a member of the set of polynomials in $\mathbb{Z}_{5}[x]$? This comes as a problem to me since as far as I know, negative numbers are not members of $\mathbb{Z}_{5}$, but if some arithmetic will result in negative numbers, it should undergo modular arithmetic. For example, note that 2 and 3 are elements of $\mathbb{Z}_{5}$. So for subtraction,

$2-3=-1=4\pmod 5=4\in\mathbb{Z}_{5}$.

But in Fraleigh's example for polynomial division, he left it as it is, i.e. $2x^{2}-3x^{2}=-x^{2}$. Shouldn't it be $4x^{2}$ since we're working on $\mathbb{Z}_{5}$?

• $-1\equiv_5 4$ as you've noted. Commented Apr 22, 2018 at 16:02
• $-x^2$ and $4x^2$ are, in your context, the same thing. Thing of it more akin to leaving something unsimplified (e.g. how in ordinary real numbers, $4/8 = 1/2$). Commented Apr 22, 2018 at 16:04
• You have two common options for how you define $\Bbb Z_n$. You could define it as though the elements are individual numbers, e.g. with $\Bbb Z_3=\{0,1,2\}$ with addition and multiplication accordingly, or you could define it as though the elements are sets, or in particular equivalence classes, e.g. with $\Bbb Z_3 = \{\overline{0},\overline{1},\overline{2}\}=\{\{\dots,-6,-3,0,3,6,\dots\},\{\dots,-5,-2,1,4,7,\dots\},\{\dots,-4,-1,2,5,8,\dots\}\}$. The latter is more common. It is also common to leave the overline or brackets off of an equivalence class and just use any representative Commented Apr 22, 2018 at 16:07
• So, in $\Bbb Z_5[x]$, you would have $\overline{2}x^2-\overline{3}x^2=\overline{-1}x^2=\overline{4}x^2$ since $\overline{-1}=\overline{4}$ in this context. As for "how do you subtract" just like in other scenarios the subtraction $x-y$ is defined as adding $x$ by the additive inverse of $y$, that is $x-y=x+(-y)$. In your case and using the first interpretation of $\Bbb Z_n$ I mentioned, this would be $2-3=2+(-3)=2+2=4$, noting that $3+2=0$ implies that the additive inverse of $3$ is $-3=2$. Commented Apr 22, 2018 at 16:09