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Suppose $R=\mathbb{Q}[x]$ is a ring and define $T$ to be $T=\{f(x) \in R : f(0) \neq 0\}$

Prove that $\frac{x}{1}$ is the only irreducible element in $T^{-1}R$ (disregarding associates).

My approach: assume $\frac{x}{1}$ is reducible. So there exist non-unit elements $\frac{a}{b}, \frac{c}{d}$ in $T^{-1}R $ such that $\frac{x}{1}=\frac{a}{b} \frac{c}{d}$. So $x=ac$ and since the degree of $x$ is $1$ it follows that the degree of $a$ or $c$ needs to be $0$. So one of the elements $\frac{a}{b}, \frac{c}{d}$ is a unit and therefore our assumption that $\frac{x}{1}$ is reducible can't be true. Is my reasoning thus far correct?

Now I need to show that $\frac{x}{1}$ is the only irreducible element. But how can one prove that?

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Let $g(x)/h(x)\in T^{-1}R$. Notice that it is irreducible if and only if $g(x)$ is irreducible in $\mathbb Q[x]$, since it is associated to it. There are two cases:

  • $g(0)\neq 0$: in this case $g(x)\in T$ so it is invertible in $T^{-1}R$ and it can't be irreducible.
  • $g(0)=0$: then $g(x)=xg'(x)$ for some $g'(x)\in\mathbb Q[x]$. But if $g'(x)$ were not invertible in $T^{-1}R$, than $g(x)=xg'(x)$ is a non trivial factorization of $g(x)$ (since you already know that $x$ is irreducible). So $g(x)$ is not irreducible.

This proves that if $g(x)$ is irreducible, than it is associated to $x$.

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$T^{-1}R$ is the localization of $R$ at the prime ideal $(x)$, so it is a local ring, moreover it's a PID.

Each irreducible element generates a maximal ideal, and there is only one maximal ideal.

Alternatively, you can go on foot and consider something arbitrarily of the form $\frac{p}{s}$ where $p\in \mathbb Q[x]$. After factoring $p$ completely over $\mathbb Q[x]$, you are left with a product of irreducibles in $\mathbb Q[x]$, and they all have nonzero constant terms except for factors of $x$, if any exist. This means that $p=x^nq$ for some $q$ that is a unit in the localization, and that $\frac{p}{s}=\frac{q}{s}x^n$. This shows that the only irreducible present in any factorization of a nonzero nonunit element is $x$.

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  • $\begingroup$ I don't get why $p=x^nq$ and not $p=x^nq+r$.. $\endgroup$ Commented Nov 9, 2016 at 22:18
  • $\begingroup$ Everything that isn't a multiple of $x$ in the factorization becomes a unit in the localization $\endgroup$ Commented Nov 9, 2016 at 23:03

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