Irreducible Polynomial Over the Ring of Polynomials Over Integral Domain Problem 333

Let $R$ be an integral domain and let $S=R[t]$ be the polynomial ring in $t$ over $R$. Let $n$ be a positive integer.

Prove that the polynomial
$f(x)=x^n-t$ in the ring $S[x]$ is irreducible in $S[x]$. Add to solve later

Proof.

Consider the principal ideal $(t)$ generated by $t$ in $S$.
Then the ideal $(t)$ is a prime ideal in $S$ since the quotient
$S/(t)=R[t]/(t)\cong R$ is an integral domain.

The only non-leading coefficient of $f(x)=x^n-t$ is $-t$, and $-t$ is in the ideal $(t)$ but not in the ideal $(t)^2$.
Then by Eisenstein’s criterion, the polynomial $f(x)$ is irreducible in $S[x]$.

(Remark that $S=R[t]$ is an integral domain since $R$ is an integral domain.) Add to solve later

More from my site

You may also like...

This site uses Akismet to reduce spam. Learn how your comment data is processed.

More in Ring theory Ring Homomorphisms from the Ring of Rational Numbers are Determined by the Values at Integers

Let $R$ be a ring with unity. Suppose that $f$ and $g$ are ring homomorphisms from $\Q$ to $R$ such...

Close