Verifying F_5[x]/(x^4+1) is a Field

[T-O7]

So my job is to find a field with $$5^4$$ elements, and I know i can construct one by considering something of the form: $$F_5[x]/(x^4+bx^3+cx^2+1)$$. So I thought i'd just consider this one:

$$F_5[x]/(x^4+1)$$

The problem is I'm not sure how to verify that this is indeed a field, i.e. I'm having trouble showing that a general non-zero element (which is of the form $$a+ b\alpha+c\alpha^2+d\alpha^3$$) has an inverse. Does anyone know what to do?

 

[shmoe]

Your polynomial is not irreducible in $$F_{5}[x]$$, you might want to try another.

Also, why are you forcing the constant term of your polynomial to be 1? Was that a typo?

 

[HallsofIvy]

I was under the impression that every finite field was isomorphic to Z_p for some prime p and so must have a prime number of elements.

 

[shmoe]

Every finite field has prime power order and for every prime power there's a unique finite field of that order.

One way to show the is a finite field of order p^n is to consider the splitting field of $$x^{p^n}-x$$ over $$F_{p}$$, uniqueness follows pretty easily. This doesn't give much info on the structure of the field though, so if you want to actually do some computations in it, you want a more explicit construction, such as finding an irreducible polynomial of degree n in $$F_{p}[x]$$. The corresponding ideal will be prime, so the quotient group will be an integral domain and hence a field since it's finite.

 

[T-O7]

Okay, thanks a lot. I was originally trying to use some proposition in my book involving monic polynomials, which is why i had a 1 leading coefficient in the ideal. So now my problem reduces down to finding a quartic polynomial that is irreducible over $$F_5[x]$$. Now, you said earlier that $$x^4+1$$ is not irreducible over $$F_5[x]$$. Why not? The polynomial doesn't have any zeros over this field, does it?

 

[shmoe]

Having no zeros only implies irreducible if it's degree 3 or 2. Your quartic could still factor into irredcible quadratics. Can you factor $$y^2+1$$ over $$F_5$$?

 

[HallsofIvy]

Ah- a power of a prime. I knew I was forgetting something!

 

[T-O7]

Hmm...so how exactly can I find an irreducible quartic polynomial over $$F_5[x]$$?

 

[shmoe]

My query about the 1 in $$(x^4+bx^3+cx^2+1)$$ was a hint. Starting with the simplest 4th degree polynomials was a fine idea, and you know you need a non-zero constant term, so try $$x^4+d$$ for the various options of d.

First check your selected polynomial has no roots. If this is the case, start trying to divide it by the monic irreducible quadratic polynomials over $$F_5$$. There are 25 monic quadratics, but many of them have zeros, so they can be ruled out as possible factors.

 

[T-O7]

Okay, so I've gone through all the quadratic polynomials of $$F_5[x]$$ and after some hard scrutiny, i think there are only six irreducible ones:
$$x^2+x+1, x^2+4x+2, x^2+2x+3, x^2+3x+3, x^2+2x+4, x^2+3x+4$$
Now, I've tried dividing $$x^4+1$$ by each of these, and in each case i find that none of them divide $$x^4+1$$, they each give a remainder in $$F_5[x]$$. Have I miscalculated somewhere?

 

[shmoe]

There are 25 monic quadratics. The reducible ones have the form $$(x+a)(x+b)$$ for some a and b, 5 choices with a=b, 5*4/2=10 with a and b distinct, so 15 total. There are 10 irreducible quadratics here, you've missed 4.

 

[T-O7]

Arghh...yes, you were right. I had overlooked these four:
$$x^2+2, x^2+3, x^2+x+2, x^2+4x+1$$

And you were right again, $$x^2+1$$ did divide $$x^4+1$$. Well, now I guess it's either: $$x^4+2$$ or $$x^4+3$$. One of them's got to work right?

 

[shmoe]

Yep, at least one of those will work. It might be worth pointing out that in general you are NOT guaranteed a simple irreducible of degree n of the form $$x^n+d$$. If you were trying to find a field with $$5^5$$ elements for example (n=5), you wouldn't have such a nice irreducible.