Polynomial factorization in $\mathbb{R}$ and $\mathbb{Z}_{[n]}$

Question

I've the following polynomial:

$$ a(x) = x^6 + x^5 + 2x^3 - 3x^2 +x -2 \in \mathbb{K}[x] $$

Set $\mathbb{K} = \mathbb{R}$.

A factorization of $a(x)$ is:

$$ a(x) = (x^2 + 1)^2(x-2)(x+1) $$

Now set $\mathbb{K} = \mathbb{Z}_{[5]}$.

If I'm asked to factorize $a(x)$ can I consider the factorization done in $\mathbb{R}$?

A factorization in a field $X$ is valid in a field $Y \subseteq X$?

EDIT: the right factorization of $a(x)$ in $\mathbb{R}$ is

$$ (x^2 + 1)^2(x+2)(x-1) $$

Answer 1

In fact, the factorisation over $\Bbb Q$ is given by $$ a(x)=(x^2+1)^2(x+2)(x-1). $$ This is also the factorisation over $\Bbb R$. Over $\Bbb F_5$ the Berlekamp algorithm gives $$ a(x)=(x + 4)(x + 3)^2(x + 2)^3 $$ This can also be seen from the factorisation over $\Bbb Z$ above, where $x^2+1=(x+2)(x+3)$ over $\Bbb F_5$.

Answer 2

You can and should use the factorization in $\Bbb R$ as a starting point. But notice that in $\Bbb Z/5 \Bbb Z, x^2+1=x^2-4$ has linear factors, so in fact your polynomial $a(x)$ will end up having $6$ linear factors over $\Bbb Z/5 \Bbb Z: (x+2)^3(x-2)^2(x-1)$.

Answer 3

In fact, over $\;\Bbb R\;$ , we have

$$x^6+x^5+2x^3-3x^2+x-2=(x-1)(x+2)(x^2+1)^2$$

so as all the polynomials above are integer ones, we can reduce each modulo $\;5\;$ :

$$a(x)=(x-1)(x-3)(x^2+1)^2\pmod5$$

But ...Observe that we also have $\;2^2=-1\pmod 5\;$ , so $\;x^2+1=(x-2)(x-3)\pmod5\;$ , and thus:

$$a(x)=(x-1)(x-3)(x-2)^2(x-3)^2=(x-1)(x-2)^2(x-3)^3$$

Note Added: Observe that the above polynomial modulo $\;5\;$ is exactly the same as the one given in Dietrich's answer...can you see why?

Answer 4

If you multiply out $$(x^2+1)^2(x+2)(x-1)$$ over $\Bbb Z_5$, you see that you still get $x^6 + x^5 + 2x^3 - 3x^2 +x -2$. So the factorisation is valid, although it is not a complete factorisation as over $\Bbb Z_5$ we have $x^2+1=(x-2)(x+2)$.

In general, for fields (or commutative rings, really) $Y\subseteq X$, a factorisation over $Y$ is a valid (but potentially not complete) factorisation over $X$. A factorisation over $X$ will be a valid factorisation over $Y$ if all coefficients of the factorisation are in $Y$. If you generalize from inclusion to homomorphism $Y\to X$, then a factorisation over $Y$ still gives a factorisation over $X$, but you can't go back from $X$ to $Y$ in any way way.

The relationship between $\Bbb R$ and $\Bbb Z_5$ is a combination of the above. Your factorisation over $\Bbb R$ is a valid factorisation over $\Bbb Z$, as all coefficients are integers. And then by the canonical map $\Bbb Z\to\Bbb Z_5$, this makes the factorisation valid over $\Bbb Z_5$.