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

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 $$F_7[X]$$.

• same as $x^4 - 4$ Feb 27 '19 at 4:01
• @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? Feb 27 '19 at 4:06
• What do you mean by $F_7$ ? A field of what? @LawrdyLawrd Feb 27 '19 at 4:13
• @Crunchy F7 is the finite field with 7 elements. F7[X] is the polynomial ring made from that field. Feb 27 '19 at 4:16
• @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. Feb 28 '19 at 0:58

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

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.