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I am not quite sure where to start with this problem. I am new to polynomial rings and want to learn how to factor polynomials in polynomial rings made of fields.

Factor $X^4 + 3$ into irreducible factors in $F_7[X]$.

Travis Willse
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Display Name
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    same as $x^4 - 4$ – Will Jagy Feb 27 '19 at 04:01
  • @WillJagy I see how x^4-4 is the same as X^4+3 in the ring, and i factored that into (x^2+2)(x^2-2), but how do i know they are both irreducible? – Display Name Feb 27 '19 at 04:06
  • What do you mean by $F_7$ ? A field of what? @LawrdyLawrd – Crunchy Feb 27 '19 at 04:13
  • @Crunchy F7 is the finite field with 7 elements. F7[X] is the polynomial ring made from that field. – Display Name Feb 27 '19 at 04:16
  • Running my script :-) $-3=4$ is a third root of unity in $\Bbb{F}_7$. Therefore its square roots have orders $3$ and $6$ respectively. The square roots of those are roots of unity of orders $3$ and $6$, and two roots of unity of order $12$. The first two reside in the prime field because $6\mid 7-1$, and give rise to linear factors. The last two reside in a quadratic extension because $12\nmid7-1$ but $12\mid 7^2-1$. Thus they are zeros of a quadratic factor. – Jyrki Lahtonen Feb 27 '19 at 04:55
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    @Crunchy Because finite fields are uniquely determined by their order, up to isomorphism of course, the notation $\mathbb{F}_{p^n}$ is standard and it is widely used in field theory. – stressed out Feb 28 '19 at 00:58

2 Answers2

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You are working in $\Bbb F_7[x]$, the ring of polynomials with coefficients in the finite field $\Bbb F_7$. As Will Jagy also suggested, $3\equiv -4$ mod $7$, hence your polynomial is basically $x^4-4$.

$$x^4-4=(x^2+2)(x^2-2)$$

We have other two factors to check. Notice that $3^2=9\equiv 2$ mod $7$, hence the second factor splits in $(x+3)(x-3)$, that is

$$x^4+3=x^4-4=(x^2-2)(x^2+2)=(x+3)(x-3)(x^2+2).$$

Finally, consider $(x^2+2)$. You can check by hand that no element in $\Bbb F_7$ is a solution of $q(x)=x^2+2$; namely, for any $a\in \Bbb F_7,$ we have that $a^2+2\not\equiv 0$ mod $7$. Thus it is irreducible.

For these reasons, the total factorization of $x^4+3$ in $\Bbb F_7[x]$ is $$x^4+3=(x+3)(x-3)(x^2+2)$$

J. W. Tanner
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InsideOut
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As mentioned in the comments by Will Jagy, $x^2+3 \equiv x^2-4 $ in $\mathbb{F}_7$. The later one is decomposed as $(x^2+2)(x^2-2)$. Now since both of these polynomials are of degree less than $4$, you can use the following fact:

A polynomial $p(x)$ of degree at most three is irreducible if and only if it has no root.

Note that for polynomials of degree four, the statement above can fail. Consider $(x^2+1)^2=x^4+2x^2+1$ over the real numbers for example.

So, you just need to show that $x^2-2$ and $x^2+2$ have no root in $\mathbb{F}_7$. If you know quadratic reciprocity, you can use it here to determine if they are irreducible or not. It turns out that $x^2+2$ is irreducible while $x^2-2=(x+3)(x-3)$ as mentioned by InsideOut. If you don't know quadratic reciprocity, just check it by hand.

stressed out
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