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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3Maybe it would be helpful to say why you think what you're saying is true? – 2012-12-09
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0My source of confusion is here.http://en.wikipedia.org/wiki/Eisenstein's_criterion – 2012-12-09
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0"It will also be irreducible over the integers" What? – 2012-12-09
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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