A month ago I posed the problem of showing that a monic $\\newcommand{\\Z}{\\mathbb{Z}}f\\in\\Z[x]$ is squarefree over $\\newcommand{\\C}{\\mathbb{C}}\\C$ if and only if it is squarefree over $\\Z$.<\/p>\n
One direction is straightforward, although the easiest way to see it is to consider the contrapositive. If $f$ is not squarefree over $\\Z$ then its factorization over $\\Z$ is of the form $hg^2$ where $h$, $g\\in\\Z[x]$ and $g$ is nonconstant. Since $f$ can only factor farther<\/em> in $\\C$ it follows that $f$ is also not squarefree over $\\C$; in particular any root $\\alpha\\in\\C$ of $g$ will have multiplicity at least $2$ in the factorization of $f$ in $\\C$.<\/p>\n Thus if $f$ is squarefree over $\\C$ it is also squarefree over $\\Z$. Conversely, if $f$ is not squarefree over $\\C$ then it has some multiple root $\\alpha\\in\\C$ and its factorization is of the form $k\\cdot(x-\\alpha)^2$, so $f’=k'(x-\\alpha)^2+2(x-\\alpha)k$ and $\\alpha$ is also a root of $f’$.<\/p>\n Since $f$ and $f’$ are polynomials with integer coefficients with $\\alpha$ as a root, the minimal polynomial $g$ of $\\alpha$ over $\\newcommand{\\Q}{\\mathbb{Q}}\\Q$ divides both $f$ and $f’$. In particular, there must be some $h\\in\\Q[x]$ such that $f=gh$. Then $f’=g’h+h’g$ and since $g\\mid f’$ and $g\\mid h’g$ we must have $g\\mid g’h$. However, $g\\nmid g’$ since $g’$ has a smaller degree than $g$ and is $g’$ is nonzero (as the characteristic of $\\Q$ is $0$).<\/p>\n Being a minimal polynomial, $g$ is irreducible, and it is also prime as $\\Q[x]$ is a UFD. Thus from $g\\mid g’h$ and $g\\nmid g’$ we must have $g\\mid h$, i.e., $g^2\\mid f$ and $f$ is not squarefree over $\\Q$. By Gauss’ Lemma<\/a>\u00a0the factorization $f=g^2(h\/g)$ over $\\Q$ may actually be taken to be over $\\Z$ (by replacing $g$ and $h$ with their primitive parts), so $f$ is not squarefree over $\\Z$, as required.<\/p>\n Note that the requirement that $f$ be monic is necessary (at least, the content of $f$ should be squarefree). For example, if $f:=4$ then $f=2\\cdot2$ is not squarefree over $\\Z$, but is<\/em> technically squarefree over $\\C$, since $4$ is a unit in $\\C$!<\/p>\n","protected":false},"excerpt":{"rendered":" A month ago I posed the problem of showing that a monic $\\newcommand{\\Z}{\\mathbb{Z}}f\\in\\Z[x]$ is squarefree over $\\newcommand{\\C}{\\mathbb{C}}\\C$ if and only if it is squarefree over $\\Z$. One direction is straightforward, although the easiest way to see it is to consider … Continue reading