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$S=\bigl\{\sum_{n}^{} a_nt^{n} | a_1=0 \bigr\}\subseteq \mathbb{Q[t]}$

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.

Bernard
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2 Answers2

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The formal proof isn't much different from what you wrote. There are not many ways in which you can factor these two polynomials. Since there are no polynomials of degree $1$ in $S$ we have to conclude that if $t^2$ is a product of some two polynomials $p,q\in S$ then one of them must have degree $2$ and the other degree $0$, hence one of them is invertible. So $t^2$ is irreducible. For $t^3$ it is a similar proof.

A hint for the second part: how can you write $t^6$ as a product of irreducible polynomials in $S$?

Mark
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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.

Mark Fischler
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