$S=\bigl\{\sum_{n}^{} a_nt^{n} | a_1=0 \bigr\}\subseteq \mathbb{Q[t]}$
This is the ring of polynomials with rational coefficients, where the coefficient of $t$ is always $0$. I want to show that $t^{2}$ and $t^3$ are irreducible in $S$, and that S is not a UFD.
My thoughts: the only way to factor $t^2$ and $t^3$ would be taking out factors of t, which isn't in S, so can't be done, but I don't know how to make this a mathematical proof? To show it isn't a UFD, I'm guessing there is an element which can be factorised in two different ways, but I don't know how to find this.
Assume $t^2$ can be factored without either factor being a unit; then the degree of each factor must be at least $1$ since polynomials of degree $0$ are units in this ring. But the subset we are considering has no polynomials of degree precisely $1$ (since those are of the form $m t + a$ with $m \neq 0$, which this subset excludes). So $t^2$ cannot be factored.
If $t^3$ can be factored without either factor being a unit, then one of the factors again would have to be of degree $1$ since the sum of the degrees is $3$ and neither factor is of degree $0$. So $t^3$ cannot be factored either.
The second part of the problem would be hard but the first part gives you a nice hint: If you can factor some polynomial into terms involving $t^3$, and also into terms involving $t^2$ but no $t^3$ instead, then since you hit a brick wall when trying to further factor each of those paths, you will get two inequivalent factorings. In fact, if you can factor down to a mix of $t^3$ and $t^2$ in two different ways, you have a polynomial without a unique factorization. Any polynomial of the form $t^n$ for $n \geq 8$ will have this property, as will $t^6$. Strangely enough, $t^7$ does have a unique factorization! But any one example means this is not a UFD.