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If a polynomial with integer coefficients cannot be factored into two polynomials of lower degree with rational coefficients, then certainly, you can't do it over Z either. So what am I missing here?

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Maybe it would be helpful to say why you think what you're saying is true? – Andrew Dec 9 '12 at 1:42
My source of confusion is's_criterion – Joseph Dec 9 '12 at 1:46
"It will also be irreducible over the integers" What? – Joseph Dec 9 '12 at 1:46
Think about what it means to "factor" something in a ring. That may be the issue. For instance, in $\mathbb{Z}[x]$, we have the factorization of $2x + 2 = 2(x+1)$. But we don't think of this as a factorization in $\mathbb{Q}[x]$, since 2 is invertible. – Zach L. Dec 9 '12 at 2:54

In $\mathbb Q$, which is a field, all non-zero elements are units, so there's no such thing as a common constant factor; factoring a polynomial over $\mathbb Q$ means factoring it into two polynomials of lower degree. By contrast, over $\mathbb Z$ you can factor a polynomial into a constant factor common to all coefficients and a polynomial of the same degree. Of course the same "factorization" also holds over $\mathbb Q$, but over $\mathbb Q$ the constant factor is a unit, so this doesn't count as factoring the polynomial.

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Concretely: $2x=2\cdot x$. Over $\Bbb{Q}$, $2$ is a unit, so this doesn't show that $2x$ is reducible (and in fact it isn't). Over $\Bbb{Z}$, this is a proof of reducibility. – Chris Eagle Dec 9 '12 at 1:54

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