Constructing a splitting field for polynomial over F_5

[PsychonautQQ]

Homework Statement

f(x) = x^7 + 3x^6 + 3x^5 - x^3 - 3x^2 - 3x where the coefficients are elements of F_5. Show that this polynomial is divisible by x^5-x and construct a splitting field L for f over F_5 and computer [L:F_5]

Homework Equations

The Attempt at a Solution

So the first thing I did was turn all the negative coefficients positive so that f(x) = x^7 + 3x^6 + 3x^5 + 4x^3 + 2x^2 + 2x and we want to divide this polynomial by x^5+4x. Upon doing the long division, I get a quotient x^2 + 3x + 3 with remainder 4x^3... I checked my steps and I got the same answer with the same remainder, can anyone check my work here?

 

[fresh_42]

$x^2+3x+3$ is correct, without remainder. You don't need to convert the coefficients. The equations keep being true, even outside the standard representation of $\mathbb{F}_5.$ I only used it to simplify the roots of this last polynomial of degree $2,$ e.g. $\frac{1}{2}=3.$
After that one can write down all $7$ roots which makes the splitting field quite obvious.

 

[PsychonautQQ]

$x^2+3x+3$ is correct, without remainder. You don't need to convert the coefficients. The equations keep being true, even outside the standard representation of $\mathbb{F}_5.$ I only used it to simplify the roots of this last polynomial of degree $2,$ e.g. $\frac{1}{2}=3.$
After that one can write down all $7$ roots which makes the splitting field quite obvious.

Yes, excellent. And yet I am still having trouble finding all the roots. So the polynomial reduces to ##(x^2+3x+3)(x^5+4x)=(x^2+3x+3)(x^4+4)x. So the first three roots are the solutions to the quadratic and zero, are the other 4 roots c*i(4)^1/4 where c is a primitive fourth root of unity?

 

[fresh_42]

You can factor $x^4+4$. It would be more obvious if you had kept the sign: $x^4+4=x^4-1$. Now use $x^4=(x^2)^2$.
And $x^2+3x+3=0$ can be done by the usual formula. The only critical point in the formula is $\frac{1}{2}$ but this is simply $3$.

 

[PsychonautQQ]

You can factor $x^4+4$. It would be more obvious if you had kept the sign: $x^4+4=x^4-1$. Now use $x^4=(x^2)^2$.
And $x^2+3x+3=0$ can be done by the usual formula. The only critical point in the formula is $\frac{1}{2}$ but this is simply $3$.

are the solutions to $x^4-1=0$ the powers of a primitive fourth root of unity? i.e. the powers of i? even in $F_5$?

 

[fresh_42]

are the solutions to $x^4-1=0$ the powers of a primitive fourth root of unity? i.e. the powers of i? even in $F_5$?

Not so complicated. Just remember the formula $(x-a)(x+a)=x^2-a^2$. This gives you $(x^2-1)(x^2+1)$ in our case. The first factor splits again according to the same formula. To split the second, now write $x^2+1=x^2-4$ and consider $2\cdot 2=4$ even in $\mathbb{F}_5$.

Edit: The fourth roots of unity will work as well, but with the way above, it's easier to see, that $i=2$ in $\mathbb{F}_5$.