SOLUTION: please help me to solve that the polynomial {x^2-3} is irreducible over the field of rational numbers

Question 1084201: please help me to solve that the polynomial {x^2-3} is irreducible over the field of rational numbers
Found 2 solutions by rothauserc, ikleyn:
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
We use the fact that f(x) is irreducible over the field of rational numbers(Q) if and only if f(x+a) is irreducible for any a an element of Q.
:
We prove that the polynomial f(x+1) is irreducible
:
We have
:
(x+1)^2 - 3 = x^2 +2x + 1 - 3 = x^2 +2x -2
:
since the nonzero coefficient of highest degree is 1(x^2 term), f(x+1) is monic
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all the non-leading coefficients are divisible by the prime number 2(the 2x term)
:
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Eisenstein's Criterion
Suppose we have the following polynomial with integer coefficients.
P(x) = a(n)x^n + a(n−1)x(n−1) +...+ ⋯ + a(1)x + a(0)
If there exists a prime number p such that the following three conditions all apply:
p divides each a(i) for i not equal to n,
p does not divide a(n), and
p^2 does not divide a(0),
then Q is irreducible over the rational numbers
****************************************************************************
Since the constant term is not divisible by 2^2, Eisenstein�s criterion implies that the polynomial f(x+1) is irreducible over Q
:
Therefore by the fact stated above, the polynomial f(x) is also irreducible over Q.

Answer by ikleyn(52803) (Show Source): You can put this solution on YOUR website!
.
Had it be reducible over the field of rational numbers, it would be a product of two linear polynomials.

Then it would have rational roots.

But its roots +/- are irrational numbers. Contradiction which proves the base statement.

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