Calculate terms modulo a polynomial in algebras more efficiently
If we have an algebra that is modulo some polynomial, say $\mathbb{Z}_5 [x]_{x^2 + 1}$, it often is the case that we perform some operation in this algebra and afterwards have to take our result modulo said polynomial, so modulo $x^2 + 1$ for example.
In this case, a simple trick can be very helpful. Remember that by the definition of modulo,
(Of course you’d adapt the polynomial accordingly to the one you are given.) Then, we can simply say
Now, in our example, $-1 = 4$ since we are working in $\mathbb{Z}_5$. Thus
And that’s it already. Now we can reap the benefits of this simple transformation: If we have a polynomial like $2x^2 + 3x + 1$ that you want to take modulo $x^2 + 1$, you can simply substitute $x^2$ with $4$, turning this into $2 \cdot 4 + 3x + 1 = 3x + 9 = 3x + 4$. Especially if you have to do such modulo operations a few times, that’s much quicker and needs less notation than using long division.