Which type of "numbers" will require order 3 factors for its irreducible polynomial factorizations?

Question

The complex numbers are famous for having the property of the fundamental theorem of algebra.

All polynomials have at least one root.

Together with being factor-able, this means we can factor every complex polynomial of degree n into a product of n linear factors. (For monic polynomials we can skip $k$) $$p(z) = k\cdot \prod_{i=1}^n (z-z_i)$$

For real valued polynomials this is not true. We will there also need to allow polynomial factors of degree 2 to reach some irreducible expression.

So it is clear that there exist two possibilities of numbers : those whose polynomials are guaranteed reducible down to 1 and 2 degrees. But what about 3? Which type of numbers will require us to allow also third-degree polynomial factors to factor "completely"?

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Edit > So to clarify, I am looking for a type of number so that for any polynomial with coefficients and variables in this set of numbers which I seek will be guaranteed to be able to factor it into factors of order 1,2 or 3 but never higher and that the case 3 by necessity happens for at least one polynomial.

We can always write : $$p(x) = \sum_{k=0}^N c_kx^k = K \prod_i (c_{i3}x^3+c_{i2}x^2+c_{i1}x+c_{i0})$$

So that there exists at least one p for which there is at least one i so that $c_{i3} \neq 0$

But for some p, all the $c_{i3}$ and also $c_{i2}$ may be $0$.

Answer 1

I'll follow @TorstenSchoeneberg's line of reasoning. You asked about numbers, so I'll write about a field $k$ with $\mathrm{char } k = 0$.

• Such a field is perfect, i.e. every algebraic extension of it is separable (this is one of equivalent definitions).
• For separable extensions, we have the well-known primitive element theorem: any finite separable extension $K/k$ is generated by a single element, $K = k(\gamma)$.
• Any element $\alpha \in \overline{k}$ has a minimal polynomial of degree at most $n$, as the extension $\overline{k}/k$ is algebraic, and a minimal polynomial of a field element must be irreducible.

Now let the degree of (the minimal polynomial of) $\alpha \in \overline{k}$ be $n$ and prove that $k(\alpha) = \overline{k}$. Indeed, for $\beta \in \overline{k}$ we have $k(\alpha, \beta) = k(\gamma)$ for some $\gamma$ by the primitive element theorem, so $|k(\gamma)/k| \le n$ (the dimension as a vector space), but $k(\alpha)$ is contained in $k(\gamma)$, so they are equal and $\gamma, \beta \in k(\alpha)$.

This allows us to use

Theorem 3.1 (Artin-Schreier). Let $C$ be algebraically closed with $F$ a subfield such that $1<[C:F]<\infty$. Then $C=F(i)$ where $i^2=−1$, and $F$ has characteristic $0$. Moreover, for $a\in F^\times$, exactly one of $a$ or $−a$ is a square in $F$, and every finite sum of nonzero squares in $F$ is again a nonzero square in $F$.

We conclude that a field of characteristic $0$ that has bounded degree $d$ of irreducible polynomials has $d \le 2$. The same holds without the a priori characteristic assumption (the bounded degree implies $\mathrm{char } k = 0$), but this is more difficult to prove.