Hello Experts,
Please answer my 2 questions with details:
1) How do I prove that x^3 - x^2 + x + 2 is prime in Z[x]?
2) And if I know that is prime ideal means that k was prime? Vice versa too?
Please give me a step by step answer.
Thanks in advance!
Please answer my 2 questions with details:
1) How do I prove that x^3 - x^2 + x + 2 is prime in Z[x]?
2) And if I know that
Please give me a step by step answer.
Thanks in advance!
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1)
Be aware that prime and irreducible elements are not the same, in general abstract ring.
However, since Z[x] is a unique factorization domain (because Z is a unique factorization domain), irreducible and prime elements (i.e polynomials) in Z[x] coincide.
Hence p(x) = x³ – x² + x + 2 is prime in Z[x] iff p(x) = x³ – x² + x + 2 is irreducible in Z[x].
By way of contradiction, suppose that p(x) is reducible in Z[x]. Since p(x) is monic, it factorizes as product of monic polynomials of degree at most 2, and p(x) has (at least) one monic factor of degree 1, say p(x) = (x–b)q(x) with b in Z and q(x) in Z[x], and p(b) = b³ – b² + b + 2 = 0.
Now b(b² – b + 1) = –2 forces b ∈ {1, –1, 2, –2}, which is false, because each of p(1), p(–1), p(2), p(–2) are nonzero.
p(x) is irreducible in Z[x].
2)
If the ring A is commutative with identity 1 ≠ 0, then an element k ∈ A is prime if and only if is a nonzero prime ideal.
This assertion follows from definition of prime element:
k is prime iff k is a nonzero nonunit and k|ab → k|a or k|b
and the characterization of pime ideals in a commutative ring R:
an ideal P is prime iff P ≠ R, and ab ∈ P → a ∈ P or b ∈ P.
**
bye
Be aware that prime and irreducible elements are not the same, in general abstract ring.
However, since Z[x] is a unique factorization domain (because Z is a unique factorization domain), irreducible and prime elements (i.e polynomials) in Z[x] coincide.
Hence p(x) = x³ – x² + x + 2 is prime in Z[x] iff p(x) = x³ – x² + x + 2 is irreducible in Z[x].
By way of contradiction, suppose that p(x) is reducible in Z[x]. Since p(x) is monic, it factorizes as product of monic polynomials of degree at most 2, and p(x) has (at least) one monic factor of degree 1, say p(x) = (x–b)q(x) with b in Z and q(x) in Z[x], and p(b) = b³ – b² + b + 2 = 0.
Now b(b² – b + 1) = –2 forces b ∈ {1, –1, 2, –2}, which is false, because each of p(1), p(–1), p(2), p(–2) are nonzero.
p(x) is irreducible in Z[x].
2)
If the ring A is commutative with identity 1 ≠ 0, then an element k ∈ A is prime if and only if
This assertion follows from definition of prime element:
k is prime iff k is a nonzero nonunit and k|ab → k|a or k|b
and the characterization of pime ideals in a commutative ring R:
an ideal P is prime iff P ≠ R, and ab ∈ P → a ∈ P or b ∈ P.
**
bye
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For 1) To factorise f(x) = x^3 - x^2 + x + 2, you would write it as f(x) = (x + a) * (x^2 + b*x + c), where a, b, c are numbers to be found (normally you would say (ax + b) * (cx^2 + dx + e), but because the coefficient of x^3 is 1, you know that a and c would both be 1, so I have skipped that step). Multiplying this out gives x^3 + (a + b) x^2 + (c + ab) x + ac; comparing the coefficients gives that (a+b) = -1 and (c + ab) = 1 and ac = 2. Solving these simultaneous equations will give you a, b, c that cannot all be integers, and so the polynomial cannot be factorised in Z[x] (the field of polynomials in x with integer coefficients).
Can you remind me what means exactly? It's been a while...
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