Good evening,
I would like to ask the following exercise :
"Determine the zero divisors, nilpotents and the non-invertibles in $ \mathbb{Q}[x] /(x^{4}-2x^{3}+3x^{2}-4x+2)$"
I saw that the polynomial can be factorized as $(x-1)^{2}(x^2+2)$ if i am not mistaken, but what really i am not fully understanding is that i always worked with finite fields or ring, like (for example) $ \mathbb{Z}/_{5 \mathbb{Z}} $ where i could at least used that $ \mathbb{Z}/_{5 \mathbb{Z}} $ was the union of the Invertibles and the Zero-divisors.
Or maybe use the fact that a generic Invertible polynomial $ h(x) $ is Invertible $\iff (f(x),h(x))=1 $, where $f(x)=(x-1)^{2}(x^2+2)$.
(And so on for Zero-divisors and Nilpotents).
Here my guess is that there are infinite polynomials because of $|\mathbb{Q}| = +\infty $, but i don't know (if my guess is true) how to prove or how to caracterize those classes.
Any help or tips would be appreciated,
Thank you all
Let $g(x)$ be a zero divisor. Then $\exists h,k \not = 0 \in \mathbb{Q}[x]$ s.t. $g(x)h(x) = 0 = k(x)(x-1)^2(x^2+2)$. Now note that niether $g$ nor $h$ can have both $(x-1)^2$ and $x^2+2$ as factors, as otherwise they would be zero. Now going through all the combinations yields that $g(x)$ is of the forms:
$$(x-1)p(x)\quad \quad (x-1)^2p(x) \quad \quad (x^2+2)p(x) \quad \quad(x-1)(x^2+2)p(x)$$
where $\gcd(p(x),x^{4}-2x^{3}+3x^{2}-4x+2) = 1$
To find the nilpotent element using the unique factorization in $\mathbb{Q}[x]$ we can conclude that both factors of $x^{4}-2x^{3}+3x^{2}-4x+2$ must appear in the prime factorization of a nilpotent element. Thus all nilpotent elements are of the form:
$$(x-1)^k(x^2+2)^np(x)$$
where again $\gcd(p(x),x^{4}-2x^{3}+3x^{2}-4x+2) = 1$
Finally to find the invertibles you can use the Euclidean Algorithm that says that $g(x)$ is invertible iff $\gcd(g(x),x^{4}-2x^{3}+3x^{2}-4x+2) = 1$