Question

Not every polynomial that is irreducible over Q[x] is irreducible over Z[x] . State whether the statement is true or false, justify with reason.

Answer 1

Answer on Question #80554 - Math - Abstract Algebra

Question. Not every polynomial that is irreducible over $\mathbb{Q}[x]$ is irreducible over $\mathbb{Z}[x]$. State whether the statement is true or false, justify with reason.

Answer. True. The statement is equivalent to saying that there is a polynomial that is irreducible in $\mathbb{Q}[x]$ and is not irreducible in $\mathbb{Z}[x]$. A polynomial $p = 3x + 6$ is an example of such a polynomial.

In $\mathbb{Z}[x]$, $p = 3(x + 2)$. As $\mathbb{Z}$ is an integral domain, the set of units in $\mathbb{Z}[x]$ is the same as in $\mathbb{Z}$, namely, $\{-1,1\}$. Hence neither of the factors of $p$ is a unit, and $p$ is not irreducible.

Assume that $p$ factorizes as $q_{1}q_{2}$ in $\mathbb{Q}[x]$. As $\mathbb{Q}$ is an integral domain, the set of units in $\mathbb{Q}[x]$ is the set of all non-zero elements of $\mathbb{Q}$. As $p$ is not zero, both $q_{1}$ and $q_{2}$ are not zero. As $\mathbb{Q}$ is an integral domain, the sum of the degrees of $q_{1}$ and $q_{2}$ is the degree of $p$ which is 1. Hence for some $i \in \{1,2\}$, $q_{i}$ has degree 0. Hence $q_{i}$ is a unit in $\mathbb{Q}[x]$. Also $p$ is not a unit. We conclude that $p$ is irreducible in $\mathbb{Q}[x]$.

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