Answer to Question #271309 in Abstract Algebra for N. Kokom

Question

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

Answer 1

Solution:

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

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

We also notice that $3^2=9 \nmid 6=a_0$

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

Answer: $x^5+9x^4+12x^2+6$ is irreducible over Q[x].

(b)We apply the mod 2 test.

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

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

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

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

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

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

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