Proving $x^4+2$ cannot be factored into $2$ degree polynomials

Question

My book says that it can't be because if I try to write $x⁴+10x³+15x²+5x+12$

as:

$$x^4+2$$

(which is $p(x) mod 5$)

then $x⁴+2$ is irreducible because:

$$x^4+2 = (ax²+bx+c)(a'x²+b'x+c)$$

is impossible over $\mathbb{Z}_5$

I used $a=1$ and did:

(x²+bx+c)(x²+dx+e)

which is equal to

$$x^4 + (b+d)x³+(bd+c+e)x²+(eb+cd)x+ec$$

then I tried to solve this system but as you can see, it's incomplete.

How do I prove that I can't solve this system over $\mathbb{Z}_5?$

Answer 1

Comparing coefficients of several $x^n$, you can get that

• $b = -d$ (look at $x^3$) and they are both non-zero, because otherwise $e = -c$ (look at $x^2$) and $-e^2 = 2$ which is impossible in $\mathbb{Z}_5$
• Assuming $ b = -d \not= 0$, you get that $e=c$ (by looking at the degree 1 coefficient) so $e^2 = 2$ which is also impossible.

So all cases are settled, and this polynomial is irreducible.

Answer 2

It could be useful to note that $2$ is not a quadratic residue in $\mathbb{F}_5$. It follows that the equation $x^2 - 2 = 0$ has no solutions in $\mathbb{F}_5$. Therefore, your polynomial can't be factored as $(x - a)P(x)$. Afterwards, you have to check if it can be factored as two monic irreducible polynomials.

Answer 3

Step 1: If a monic polynomial factors, it can factor into monic polynomials (you did this already).

Step 2: Either the degree 4 polynomial has a root (which this one doesn't), or it has two irreducible degree-2 monic factors.

Step 3: List all the irreducible quadratic monic polynomials mod 5. They are: $$x^2+2, x^2+3, x^2+x+1, x^2+x+2, x^2+2x+3, x^2+2x+4, x^2+3x+3, x^2+3x+4, x^2+4x+1, x^2+4x+2$$

Note: There are $10=\frac{5\cdot 4}{2}$ of them; see here.

Step 4: Check that no two of them multiply to $x^4+2$. There are not that many cases to check, as the constant coefficients must multiply to $2$ (mod 5). Hence they must be $\{1,2\}$ or $\{3,4\}$.

Answer 4

Hint Comparing degree $3$ coefficients gives $d = -b$. Then, comparing degree $1$ coefficients gives $d(c - e) = 0$.

We cannot have $c = e$, because comparing degree $0$ coefficients would give $c^2 = 2$, but $2$ is not a square in $\Bbb Z_5$. Hence $d(c - e) = 0$ implies $b = d = 0$. This leaves $x^4 + (c + e) x^2 + c e = x^4 + 2$. Now, $e = -c$, so $-c^2 = 2$ and hence $c^2 = -2$, but $-2$ is not a square in $\Bbb Z_5$ either, so there are no such factorizations.

Answer 5

$x^4+2$ can be factored into 2 degree polynomials (If you do not say explicitly with rational coefficients, then it can be factored in such a way); indeed

$$x^4+2=(x^2+\sqrt[4]8x+\sqrt 2)(x^2-\sqrt[4]8x+\sqrt 2)$$