Question #271309

Determine which of the polynomials below is (are) irreducible over Q. a. x5+9x4+12x2+6 b. x4+x+1


1
Expert's answer
2021-11-26T05:10:17-0500

Solution:

(a) We can check whether x5+9x4+12x2+6x^5+9x^4+12x^2+6 is reducible over Q or not by Eisenstein's criterion.

Let p=3p=3 (a prime number). We know that ana_n is the coefficient of highest power. Now, we notice that 3an=13 \nmid a_n=1, but since 39,312,363|9, 3|12, 3|6 and 303|0, i.e., 33 divides all the other coefficients of given polynomial.

We also notice that 32=96=a03^2=9 \nmid 6=a_0

Thus, Eisenstein's criterion is satisfied and so, x5+9x4+12x2+6x^5+9x^4+12x^2+6 is irreducible over Q[x].

Answerx5+9x4+12x2+6x^5+9x^4+12x^2+6 is irreducible over Q[x].


(b)We apply the mod 2 test.

x4+x+1Z[x]x^4+x+1\in Z[x] is primitive. Thus x4+x+1Q[x]x^4+x+1\in Q[x] is irreducible if and only if x4+x+1Z[x]x^4+x+1\in Z[x] is irreducible.

The mod 2 reduction of x4+x+1Z[x]x^4+x+1\in Z[x] is f(x)=x4+x+1Z2[x]f(x)=x^4+x+1\in Z_2[x] . Since

f(a)=10f(a)=1 \neq0 for all aZ2 a\in Z2~ it follows that f(x)f(x) has no linear factors.

Suppose that f(x)f(x) is reducible. Then it must be the product of quadratic factors. There are 3 quadratic reducible polynomials in Z2[x]Z_2[x].Therefore there is 1 irreducible quadratic in Z2[x]Z_2[x] which is x2+x+1x^2 + x + 1 since this polynomial has no roots in Z2[x]Z_2[x]. Therefore f(x)=(x2+x+1)2=x4+x2+1f(x)=(x^2+x+1)^2=x^4+x^2+1 which is not the case.

We have shown f(x)Z2[x]f(x)\in Z_2[x] is irreducible. Thus x4+x+1Z[x]x^4+x+1\in Z[x] is irreducible

which means x4+x+1Q[x]x^4+x+1\in Q[x] is also.

Answer: x4+x+1x^4+x+1 irreducible over Q[x].



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