I'm currently learning Abstract Algebra using Gallian for the first time. I'm learning about factor rings and such right now. There is an Example where Gallian finds the order of the factor ring $R = Z_3[x]/ \langle x^2 + 1 \rangle$ where $Z_3[x]$ is the ring of polynomials with integer coefficients in modulo 3 and $I = \langle x^2 + 1 \rangle$ is the principle ideal generated by $x^2 + 1$ In trying to show that it has 9 elements, he begins the argument by saying
The fact that $x^2 + 1 + I = 0 + 1$ means that when we are dealing with coset representatives we may treat $x^2 + 1$ as equivalent to $0$ and therefore $x^2 = -1$. For example, the coset $2x^2 + x + 1 + I = -2 + x + 1 + I = x + 1 + I$.
The rest of the argument uses this idea that $x^2 = -1$ to show that all possible $P(x) + I$ degenerate into $ax + b + I \hspace{0.2cm} | \hspace{0.2cm} a, b \in Z_3$ and so it follows it is order 9.
What I was wondering is how this idea that $x^2 = -1$ is "valid?" Is there a proof allowing us to do this? Because it feels as though we are treating the coset representatives as a group under addition, even though it is the cosets themselves that are the group under addition. I'm guessing the ability to set $x^2 = -1$ somehow inherits from the factor ring operations $(a + I) + (b + I) = a + b + I$ and $(a + I)(b + I) = ab + I$ being well-defined, but I can't exactly fathom on how a proof would work as I don't really know what I'm setting out to prove. It's just that the move to suddenly say $x^2 = -1$ seems a bit.. hand-wavy or informal? Or perhaps I'm just missing out on the logic behind this.
If someone could help, or provide a proof / explanation of why this is valid, please let me know! And I would be grateful if this explanation could apply to the general case of this as well... i.e can we do algebra like this on all coset representatives?
Thanks!
The set $\mathbb Z_3[x]/I$ is the set of cosets $\{f+I\mid f\in\mathbb Z_3[x]\}$, and the ring structure is defined on these cosets.
(One does this by choosing representatives, applying the ring operations on the representatives, and then taking the corresponding coset. That is why one has to show that everything is well-defined, so didn’t depend on the original choices of representatives.)
For your question, we need to count how many distinct cosets there are. We have $I=(x^2+1)$, so $$ 0+I = I = \{(x^2+1)f\mid f\in\mathbb Z_3[x]\} $$ $$ x+I = \{x+(x^2+1)f\mid f\in\mathbb Z_3[x]\} $$ and so on. For $p,q\in\mathbb Z_3[x]$ we have $p+I=q+I$ if and only if $x^2+1$ divides their difference $p-q$. This clearly holds true for $p=x^2$ and $q=-1$, so we have equality of cosets $x^2+I=-1+I$.
Given $p\in\mathbb Z_3[x]$ polynomial division allows us to write $p=(x^2+1)f+(ax+b)$ for some unique $f\in\mathbb Z_3[x]$ and some unique $a,b\in\mathbb Z_3$. Thus every coset can be written uniquely as $(ax+b)+I$, and there are precisely 9 cosets.
Of course, once these ideas become more familiar, one abbreviates the notation, maybe writing $\bar p$ for the image of $p$ in $\mathbb Z_3[x]/I$, so for the coset $p+I$. Then maybe one becomes lazy and just writes something like ‘mod $I$ we have …’ or ‘… holds in the factor ring’, but with the explicit understanding that we are really dealing with cosets.