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?
How can a polynomial be irreducible over $\mathbb{Q}$ but reducible over $\mathbb{Z}$
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abstract-algebra
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0Think 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. – 2012-12-09
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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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2Concretely: $2x=2\cdot x$. 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. – 2012-12-09