Let $I$ be the subset of $\mathbb{Q}[x]$ that consists of all the polynomials whose first five terms are 0.
I've proven that $I$ is an ideal (any polynomial multiplied by a polynomial in $I$ must be at least degree 5), but I'm unsure to how to prove that it is not a prime ideal. My intuition says that its not, because we can't use $(1)$ or $(x)$ as generators.
I know that $I$ is a prime ideal $\iff$ $R/I$ is an integral domain. Again, I'm a little confused on how represent $\mathbb{Q}[x]/I$
By definition, $I$ is a prime ideal if for all $f,g\in\mathbb{Q}[x]$ with $fg\in I$ you have $f\in I$ or $g\in I$. So to show that $I$ is not prime, it suffices to give a single counterexample. Look at $f = x^2$ and $g = x^3$...
You need only find two polynomials $p,q\in\Bbb Q[x]$ such that $p,q\notin I$ and $p\cdot q\in I$. Think about monomials.
Hint $\ $ For any prime ideal $\rm\,P\!:\,\ x^n\in P\:\Rightarrow\:x\in P.\:$ Thus $\rm\ x^5 \in I\,$ but $\rm\ x\not\in I\ \Rightarrow\ I\,$ is not prime.
Equivalently, $\rm\, R\ mod\ P\,$ has a nilpotent ($\Rightarrow$ zero-divisor): $\rm\, x^5\equiv 0,\ x\not\equiv 0,\,$ so it is not a domain.
Remark $\ $ Generally prime ideals can be generated by irreducible elements (in any domain where nonunits factor into irreducibles), since one can replace any reducible generator by some factor, then iterate till all generators are irreducible. In particular, in UFDs, where irreducibles are prime, prime ideals can be generated by prime elements. This property characterizes UFDs. Well-known is Kaplansky's case: a domain is a UFD iff every prime ideal $\!\ne\! 0$ contains a prime $\!\ne\! 0.$
Denote by $[f]$ the residue class of $f$ in $\Bbb{Q}[x]/I$. Then $[x],[x^4] \neq 0$, because neither $x \in I$ nor $x^4 \in I$. But what can you say about the product $$ [x] \cdot [x^4] $$ Can you conclude that $I$ is not a prime ideal?
