converse of reduction criterion? [duplicate]

By the reduction criterion, I mean the following test for the irreducibility of polynomial with Dedekind domain coefficients.

Let $\mathfrak{m}$ be maximal in Dedekind domain A and $f(X)\in A[X]$. If $f$ reduced modulo $\mathfrak{m}$ is irreducible in $A/\mathfrak{m}$, then $f$ is irreducible in $A$.

I know that the converse doesn't generally hold. There are irreducible polynomials with integer coefficients that is reducible when reduced modulo $p$ for some prime $p$.

So I conjecture the following for polynomials with integer coefficients.

For monic polynomial $f(X)\in \mathbb{Z}[X]$, if $f(X)$ modulo $p$ is reducible for all prime $p$, then $f(X)$ is reducible in $\mathbb{Z}[X]$.

I have thought of some galois theoretic approach but couldn't quite reach the conclusion. How would I prove or disprove this statement?

marked as duplicate by Dietrich Burde abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Oct 13 '18 at 18:37

• See here or here or here (or possibly elsewhere). Since I have been involved with a few of those, it is not kosher for me to pick a duplicate target. +1s to all. Our search engine is challenging to use. Particularly to new users who haven't learned the tricks – Jyrki Lahtonen Feb 25 '18 at 5:40

Corollary 16. The irreducible polynomial $x^4+1\in \mathbb{Z}[x]$ is reducible modulo every prime $p$.
Sketch of proof. Claim. $x^{4}+1$ is irreducible in $\mathbb{Z}[x]$.
Suppose it is reducible. It does not have roots in $\mathbb{Z}$. So it factors to two polynomials of degree two. We can check this cannot happen.
Claim. $x^{4}+1$ is reducible in $\mathbb{F}_{p}[x]$ for all $p$ prime.
If $p=2$, then $x^{4}+1=(x+1)^{4}$. Suppose $x^{4}+1$ is irreducible in $\mathbb{F}_{p}[x]$ for some odd prime $p$. Then $8|p^{2}-1$ (let $p=2k+1$). So $$x^{4}+1|x^{8}-1|x^{p^{2}-1}-1|x^{p^{2}}-x.$$ Let $\alpha$ be a root of $x^{4}+1\in \mathbb{F}_{p}[x]$. Then $\alpha\in \mathbb{F}_{p^{2}}$. So we have $\mathbb{F}_{p}\subset \mathbb{F}_{p}(\alpha)\subset \mathbb{F}_{p^{2}}$. But $[\mathbb{F}_{p}(\alpha):\mathbb{F}_{p}]=4$ and $[\mathbb{F}_{p^{2}}:\mathbb{F}_{p}]=2$, contradiction.