# Does irreducibility in $\mathbb Q[X]$ always imply the irreducibility in $\mathbb Z[X]$? [duplicate]

I thought the irreducibility in $\mathbb Q[X]$ automatically implies the irreducibility in $\mathbb Z[X]$, because $\mathbb Z\subset \mathbb Q$.

Could someone give a counterexample of a polynomial $f\in \mathbb Z[X]$ which is irreducible over $\mathbb Q[X]$ and reducible over $\mathbb Z[X]$?

## marked as duplicate by Dietrich Burde, mrp, Peter Košinár, MathOverview, Namaste 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 12 '16 at 0:06

I know it might feel a little bit like cheating, but this is what they mean: The polynomial $5x + 5$ is reducible in $\Bbb Z[X]$, but not in $\Bbb Q[X]$. The reason is that in $\Bbb Q[x]$, the factor $5$ is invertible, and therefore "doesn't count", while in $\Bbb Z[X]$, $5$ is not invertible, and therefore $5\cdot (x+1)$ is a valid reduction.
The main idea to take with you is that "reducibility" is not a property of polynomials per se, but a property of elements in any ring. An element $r \in R$ is reducible if we can find non-invertible $s, t \in R$ such that $r = st$.
If you read that wikipedia article, they say exactly this: "(in which case $Q$ as integer polynomial will have some prime number, necessarily distinct from $p$, as an irreducible factor)"
$2 x$ is such an example. If you read the article carefully, the idea is already in there.