4
$\begingroup$

I am trying to find the degree of the splitting field of this polynomial over these two fields. For the degree over $\mathbb{Q}((-3)^{1/2})$ I got 3. I am pretty sure this is correct. For $\mathbb{F}_5$ I am not so sure though. I guess I am intimidated by characteristic p fields. Here are my thoughts so far: In $\mathbb{F}_5, x^6-3=x*x^5-3=x^2-3$. I think that this polynomial is irreducible in $\mathbb{F}_5$ which would make the degree 2. However I am not sure how to show this.

Any help would be very much appreciated.

Thanks

|cite|improve this question
$\endgroup$
  • 13
    $\begingroup$ Note: The fact that $x^5 \equiv x\pmod{5}$ for all $x$ does not mean that the polynomial $x^6-3$ is equal to the polynomial $x^2-3$; the polynomial functions they define are the same, but the polynomials themselves are not. After all, the splitting field of $x^{25}-x$ over $\mathbb{F}_5$ is the field of $25$ elements, but using your argument we would conclude that $x^{25}=(x^5)^5 = x^5 = x$, so the polynomial would equal $0$ (it doesn't, though the polynomial function $a\mapsto a^5-a$ is identically zero). $\endgroup$ – Arturo Magidin Jul 21 '11 at 3:24
  • 1
    $\begingroup$ Have you figured out what the splitting field of $X^6 - 1$ over $\mathbf{F}_5$ is? That would be a good start. $\endgroup$ – Dylan Moreland Jul 21 '11 at 4:53
  • 2
    $\begingroup$ $x^6-1$ divides $x^{24}-1$ which divides $x^{25}-x=x^{5^2}-x$. So any root of $x^6-1$ must also be a root of $x^{5^2}-x$, but the roots of this polynomial are precisely the elements of $\mathbb{F}_{5^2}$. Since this is a degree 2 extension and $x^6-1$ doesn't split over $\mathbb{F}_5$ then this must be the splitting field... I think. $\endgroup$ – MJoszef Jul 21 '11 at 7:34
  • $\begingroup$ @MJoszef: Correct. $F_{25}$ is the splitting field of $x^6-1$ over $F_5$. $\endgroup$ – Jyrki Lahtonen Jul 21 '11 at 7:43
  • 1
    $\begingroup$ @Jyrki I read your post below. It was very helpful, thank you. Now that I have the splitting field of $x^6-1$ I would be finished if I could find a root of $x^2-2$ in either $\mathbb{F}_5$ or $\mathbb{F}_{25}$. Checking the values in $\mathbb{F}_5$ I see that there is no such root. If such a root is in $\mathbb{F}_{25}$ then it must also be a root of $x^{25}-x=x(x^{24}-1)$. To be a root of this polynomial it would suffice to be a root of $x^{24}-1=(x^2)^{12}-1$. But if it is a root of $x^2-2$ then we have $2^{12}-1=0$ (all of this mod 5 of course), and indeed it checks out. So the degree is 2. $\endgroup$ – MJoszef Jul 21 '11 at 8:27
4
$\begingroup$

The finite fields are often easier to deal with in this type of questions. A useful property of the finite fields is that all non-zero elements are roots of unity. The field $GF(5^m)$ contains all the roots of unity of order any factor of $5^m-1$. This is because the multiplicative group is cyclic. Another key fact to remember is that a given finite field has only one (up to isomorphism) extension of any given (finite) degree.

I hint at two solutions using these two facts. You are welcome to expand either or both of them to a full solution. Depending on how familiar you are with these properties (I cannot tell, because it depends how much of the basic properties of finite fields you have covered at this point).

A solution to this problem using the first fact would start as follows. Here we see that $3^2\equiv -1 \pmod 5$ and $3^4\equiv 1\pmod 5$, so $3$ is a primitive root of unity of order $4$. Therefore any sixth root of $3$ in any extension of $F_5$ is a root of unity of order $n$, where $n$ is a factor of $24$. Therefore all the zeros of $x^3-6$ are in a field containing the $24^{th}$ roots of unity, which is... A remaining question then is: could the splitting field be a proper subfield of that field?

One solution using the second fact is to look for factorizations of $x^6-3$ over $F_5$. We notice that $3=8=2^3$ in $F_5$, so $$ x^6-3=(x^2)^3-2^3=(x^2-2)(x^4+2x^2+4). $$ The factor $x^2-2$ is easy to deal with. The second factor is trickier (which is why I slightly prefer the first solution). We can write it in $F_5[x]$ as $$ x^4+2x^2+4=x^4+7x^2+9=(x^4+6x^2+9)+x^2=(x^2+3)^2-4x^2 $$ a difference of two squares, which leads to a factorization $$ x^6-3=(x^2-2)(x^2-2x+3)(x^2+2x+3). $$ But this is more than a bit ad hoc. Anyway, you can pick it up from here.

A third possibility (hinted at by Dylan Moreland) would be to combine these approaches, and determine the smallest extension field of $F_5$ that contains all the sixth roots of unity and one of the roots of $x^6-3$. Here I would use a zero of the easy factor $x^2-2$.

|cite|improve this answer
$\endgroup$
6
$\begingroup$

The roots of $x^6-3$ over the rationals are the numbers $\rho^m\root6\of3$ where $\rho$ is a primitive complex 6th root of unity and $m=0,1,\dots,5$, so the splitting field can be written $K={\bf Q}(\rho,\root6\of3)$. Now you've got a tower of fields ${\bf Q}\subset{\bf Q}(\rho)\subset K$. You ought to be able to show that ${\bf Q}(\rho)$ is ${\bf Q}(\sqrt{-3})$, and I think you'll find you got the wrong answer for the first question.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ I am confused, sorry. Here's what I'm thinking- since primitive sixth root of unity is $(-3)^{1/2}/2 + 1/2$ the two fields are equal. Can I say $x^6-3$ is irreducible over $\mathbb{Q}$ (by eisensteins) so the degree of $K$ over $\mathbb{Q}$ is 6. Hence the degree of $K$ over $\mathbb{Q}((-3)^{1/2})=6/\phi(6)=6/2=3$. $\endgroup$ – MJoszef Jul 21 '11 at 5:49
  • 1
    $\begingroup$ You can say the degree of ${\bf Q}(\root6\of3)$ over $\bf Q$ is 6, but ${\bf Q}(\root6\of3)$ is not the splitting field. $\endgroup$ – Gerry Myerson Jul 21 '11 at 6:02
  • $\begingroup$ The splitting field of $x^6-3$ would be $\mathbb{Q}(\rho,3^{1/6})$. But this is (after some messing around) equal to $\mathbb{Q}(3^{1/6},i)$. Hence the splitting field is of degree 12 over $\mathbb{Q}$. Thus the answer should be (hopefully) 6. I'm sorry I'm being so slow with this stuff, it's really been tripping me up. Thanks so much for helping. $\endgroup$ – MJoszef Jul 21 '11 at 6:36
  • $\begingroup$ Well done. No need to introduce $i$; since $\rho$ is not real, it's not in ${\bf Q}(\root6\of3)$, so it's of degree at least 2 over that field, but it's only of degree 2 over the rationals, so it's of degree exactly 2 over that other field. $\endgroup$ – Gerry Myerson Jul 21 '11 at 7:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.