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Irreducible polynomial in Q[x] but redicuble in Z[x]

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Homework Statement


Give an example of a polynomial irreducible in Q[x], but reducible in Z[x]


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The Attempt at a Solution



I think there is no example of this. The coefficients of a reducible polynomial with integer coefficients will be rational, or am I mistaken?
 

Answers and Replies

  • #2
jbunniii
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The question makes no sense. If a polynomial can be factored in Z[x], then it can be factored (with the same factors) in Q[x].

Also, I'm not sure what you mean by this question:
The coefficients of a reducible polynomial with integer coefficients will be rational, or am I mistaken?
Since integers are rational numbers, it's (trivially) true that integer coefficients are rational coefficients. But what does this buy you?
 
  • #3
Hurkyl
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There is an example.

Recall the definition of irreducible:
A non-unit x is irreducible if and only if it cannot be written as yz where y and z are non-units​

Now, as is very often a useful, we simply translate the question according to the definition:

We seek polynomials f,g,h with integer coefficients such that:

  • f=gh, and none of f,g,h are units in Z[x]
  • There do not exist non-units u,v in Q[x] such that f=uv

A more symmetric phrasing might be useful, to spot what you're missing:

We seek a polynomial f with integer coefficients such that

  • There exist non-units u,v of Z[x] such that f=uv
  • There do not exist non-units u,v of Q[x] such that f=uv
 
  • #4
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The question is nonsense as written. If a polynomial f(x) in Q[x] is reducible (spelling!) in Z[x], then f(x) = g(x)h(x) were g(x) and h(x) are in Z[x]. We assume that the degrees of g(x) and h(x) are > 0. This means that g(x) and h(x) have integer coefficients. When multiplied, the product polynomial, f(x), must have integer coefficients in the first place.
g(x) and h(x) are also in Q[x], so f(x) IS reducible in Q[x].

The non-trivial result is that if a polynomial in Z[x] is irreducible, then it is also irreducible in Q[x]. This takes a bit of proof and is usually given in a first course in so-called abstract algebra.
 
  • #5
Hurkyl
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The question is nonsense as written.
Not only does the question make sense, the request can be satisfied.

We assume that the degrees of g(x) and h(x) are > 0. This means that g(x) and h(x) have integer coefficients. When multiplied, the product polynomial, f(x), must have integer coefficients in the first place.
g(x) and h(x) are also in Q[x], so f(x) IS reducible in Q[x].
And thus, we conclude that any example must violate the assumption....
 
  • #6
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I am still stumped. Do you have another hint to try and make me see what I am missing?
 
  • #7
HallsofIvy
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If a polynomial is reducible in the rational numbers, it can be written as a fraction (1 over the least common multiple of the denominators of the coefficients) times a polynomial that is reducible over the integers. That does NOT mean the original poynomial was reducible over the integers.
 
  • #8
Hurkyl
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I am still stumped. Do you have another hint to try and make me see what I am missing?
Have you noticed that I've already said:
  • If f=uv is an example, then at least one of u and v are units of Q[x],
  • If f=uv is an example, then at least one of u and v have degree 0?
?

If not then be aware that your problem is not algebra, but logic and its application to problem solving.
 
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