Show $f(x) = x^4-10x^2+1$ is irreducible over $\mathbb{Q}$

Question

How do I show that $x^4-10x^2+1$ is irreducible over $\mathbb{Q}$? Someone says I should use the rational root test, but I don't exactly know how that applies. Thanks for any input.

Answer 1

We have that $p(x)=x^4-10x^2+1$ splits as $(x^2-2)(x^2+2)$ over $\mathbb{F}_5$, hence $p(x)$ has no linear factors over $\mathbb{Q}$ and by assuming it splits over $\mathbb{Q}$, since $p(x)=p(-x)$, it must satisfy: $$ x^4-10x^2+1 = (x^2+ax+b)(x^2-ax+b) $$ with $b^2=1$ and $a^2-2b=10$, contradiction.

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With a little reverse-engineering, I guess you just want to show that your polynomial is the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb{Q}$. We have that $\mathbb{Q}[x,y]/(x^2-2,y^2-3)$ is a vector space over $\mathbb{Q}$ with dimension $4$ and a base given by $1,x,y,xy$. With respect to such a base: $$\begin{pmatrix}1 \\ (x+y) \\ (x+y)^2 \\ (x+y)^3 \\ (x+y)^4\end{pmatrix}= \begin{pmatrix}1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 5 & 0 & 0 & 2 \\ 0 & 9 & 11 & 0\\49& 0 & 0 & 20 \end{pmatrix}\begin{pmatrix}1 \\ x \\ y \\ xy \end{pmatrix}$$ and by applying Gaussian elimination we get that the fifth row of the last matrix can be expressed as a linear combination of the previous four rows, so that $z^4-10z^2+1$ is a polynomial in $\mathbb{Q}[z]$ that vanishes at $z=\sqrt{2}+\sqrt{3}$. Since the rank of such a matrix is four (easy to check), the last polynomial is also the minimal polynomial of $\sqrt{2}+\sqrt{3}$, so it is irreducible.

Answer 2

There are so many ways to see the irreducibility of the polynomial $p(x)=x^4-10x^2+1$. Adding yet another way, using Eisenstein and trickery.

We see that $$ r(x):=p(x+1)=x^4+4x^3-4x^2-16x-8. $$ At this point users familiar with $2$-adics and Newton's polygon would already see a way. A more elementary route is to observe that $r(x)$ is irreducible over the rationals if and only if $r(2x)$ is. But $$ r(2x)=16x^4+32x^3-16x^2+32x-8=8 q(x) $$ with $$ q(x)=2x^4+4x^3-2x^2+4x-1. $$ The other trick we need is to use the fact that a polynomial (with a non-zero constant term) is irreducible iff its reciprocal polynomial $$ \tilde{q}(x)=x^4q(\frac1x)=-x^4+4x^3-2x^2+4x+2 $$ is irreducible. Here Eisenstein with $p=2$ kicks in.

Answer 3

Hint: If $p(x)$ is a factor, $p(-x)$ is a factor. Show that there can be no linear factor, and no quadratic factor of the form $x^2-a$. Then try to solve $x^4-10x^2+1 = (x^2+ax+b)(x^2-ax+b)$.

Answer 4

Perhaps it is worth noting that $x^{2n} -10 x^n + 1$ is irreducible over $\mathbb{Q}$ for every natural $n\ge 1$. Indeed, for $n=1$ the polynomial $$x^2 - 10 x + 1 = (x-(5 + 2\sqrt{6})\,)(x-(5-2\sqrt{6})\,)$$ is irreducible over $\mathbb{Q}$. Now, it is enough to check that in the decomposition $$x^{2n}-10 x^n + 1 =(x^n-(5 + 2\sqrt{6})\,)(x^n-(5-2\sqrt{6})\,)$$ both of the factors are irreducible in $\mathbb{Q}(\sqrt{6})[x]$.

Let's consider the general case of a binomial $x^n - \alpha$ over some field $F$. Decompose into linear factors over some extension $x^n - \alpha = (x-\gamma_1)\cdot \ldots \cdot (x-\gamma_n)$. Assume that $x^n -\alpha$ is reducible over $F$. Then we can find $k$ factors ($1\le k \le n-1$) whose product is in $F[x]$. Assume wlog that $$(x-\gamma_1)\cdot \ldots \cdot (x-\gamma_k)\in F[x]$$ We conclude $\gamma_1 \cdot \ldots \cdot \gamma_k = \beta \in F$. Now raise to the power $n$ and get $\beta^n = \gamma_1^n\cdot \ldots \cdot \gamma_k^n = \alpha^k$. Conclusion: If $x^n - \alpha$ is reducible in $F[x]$ then there exists $1\le k \le n-1$ and $\beta \in F$ so that $$\alpha^k = \beta^n$$

Back to our problem. It is not hard to check that the group of units of the ring of integers of the field $\mathbb{Q}(\sqrt{6})$ is generated by $5 + 2\sqrt{6}$ and $-1$. Therefore, for $\alpha = 5\pm 2 \sqrt{6}$ we cannot have an equality as above.

Answer 5

It's $\!\!\overbrace{(\color{#c00}3,\color{#0a0}{x^2\!+\!1})}^{\text{aka Schönemann}}\!\!$ Eisenstein: $\:\!f = (\color{#0a0}{x^2\!+\!1})^2\! - \color{#c00}3(\color{#0af}{4x^2}),\,$ prime $\:\!\color{#0a0}{x^2\!+\!1}\nmid \color{#0af}{4x^2}$ in $\:\!\Bbb Z_{\:\!\large \color{#c00}3}[x],\:\!$ so irreducible.

Answer 6

You can use the Generalized Arthur Cohn's irreducibility criterion (no other answer has mentioned it yet). But the solution is ugly -- you have to prove the primality of an integer larger than $10^9$.

In this case, $x^4-10x^2+1$ has a term with a negative coefficient, so the criterion couldn't be directly applied.

If a term with a negative coefficient had an odd exponent, then the substitution $x=-y$ might help. In this case, the exponent is even.

Trying substitutions $x=\pm y\pm 1$, $x=\pm y\pm 2$, $x=\pm y\pm 3$ doesn't work ($12$ tries overall, or $6$, because $|y+1|=|-y-1|$, $|y-1|=|-y+1|$, etc., and the exponents are all even) for giving a polynomial with only positive coefficients. Finally, $x=y+4$ does work, giving us the polynomial $y^4+16 y^3+86 y^2+176 y+97$.

If $y>176$ is an odd integer, then the polynomial is even and larger than $2$, so not prime. So $y$ must be even if we want the polynomial to be prime.

Trying $y=178$ doesn't give a prime number (you could prove it's divisible by $23$), but $y=180$ does, so we're done. Clearly proving the primality by hand would be a problem, so it's an ugly solution.

By a Gauss's Lemma for polynomials irreducibility over $\mathbb Z[x]$ for a polynomial with integer coefficients implies irreducibility over $\mathbb Q[x]$ (the Wikipedia page for Cohn's criterion currently uses $\mathbb Z[x]$).

Answer 7

This can be seen from the following fact:

If $P\in{\bf Z}[X]$ is monic with $d=\deg(P)\geqslant1$ and takes values in the set $\{\pm 1\}\cup\{\pm p\mid p\text{ prime}\}$ in at least $2d+1$ distinct integers $a_1,\dots,a_{2d+1}\in{\bf Z}$, then it is irreducible over ${\bf Q}$.

For the proof, see Theorem 5.1. The example of this polynomial is also given, where the integers $a_1,\dots,a_9$ can be taken to be $0,\pm2,\pm4,\pm6,\pm8$.

As a side note, this polynomial is reducible mod every prime (adding it because the question comment's link is dead).

Answer 8

If we substitute $x+8$ as x in $x^4-10x^2+1$, we get $x^4+32x^3+374x^2+1888x+3457$. If we prove that this polynomial is irreducible over the rational numbers, $x^4-10x^2+1$ is also irreducible over the rational numbers. We first prove that $x^4+32x^3+374x^2+1888x+3457$ is irreducible over the integers using Ostrowski's Criterion. Notice that the absolute value of the constant term, 3457, is prime. We also take the sum of the absolute values of the non-constant coefficients of that polynomial and get $1+32+374+1888=2295$. Since $3457>2295$, this polynomial satisfies Ostrowski’s Criterion, so it is irreducible over the integers. By Gauss' Lemma, a monic polynomial that is irreducible over the integers is irreducible over the rational numbers, so $x^4+32x^3+374x^2+1888x+3457$ is irreducible over the rational numbers. Therefore, we have proven that $x^4-10x^2+1$ is irreducible over $\mathbb{Q}$.