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I assumed a root $x$ of a monic polynomial $p(t)\in R[t]$ where $R$ is a PID. I need to show that $x$ lies in $R$

I assumed $p(t)=a_{0}+...+a_{n-1}t^{n-1}+t^{n}$

Then we have $a_{0}=-x^{n}-...-a_{1}x$

As $a_{0}\in R$ Therefore, RHS belongs to the PID $R$

i.e. $(-x^{n-1}-...-a_{1})x\in R$

I can't see a way to proceed further. How do I utilize that every ideal of R is principal?

Uday Khanna
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  • First you need to make sure where you chose the root. It must come from the quotient field. Thus write $x = y/z$ with $y,z$ in $R$. Personally I find it more intuitive to show something more general namely that every unique factorization domain (also known as factorial domain) is int closed. – quid Jun 11 '20 at 11:07

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Write $x=\frac{p}{q}$ where $q\neq 0$ and $\gcd (p,q)=1$. You can do this since $R$ is a UFD. Now substituting this you get $$a_0+a_1\frac{p}{q}+a_2\left (\frac{p}{q} \right )^2+\dots +\left (\frac{p}{q} \right )^n=0$$Clearing the denominator gives you $$a_0q^n +a_1pq^{n-1}+\dots +p^n=0$$$$\implies p^n=-\left (a_0q^n +a_1pq^{n-1}+\dots +a_{n-1}p^{n-1}q\right )$$ This shows $q \mid \text{RHS} \implies q\ \mid\ \text{LHS}$, i.e. $q \mid p^n$ but $\gcd(p^n,q)=1$

This forces $q$ to be a unit in $R$ and hence $x\in R$.

Note: The above proof works for any UFD.

Kite
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Baidehi
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