# Polynomial division over an extension [duplicate]

Suppose $$F \subseteq K$$ are fields. I'm trying to show that if $$P,Q \in F[X]$$ and $$P \mid Q$$ over $$K$$, then $$P \mid Q$$ over $$F$$ as well.

Denote the field of fractions of $$F$$ with $$F(X)$$. Then I have come to the point that it suffices to prove

$$F(X) \cap K[X] = F[X]$$

The inclusion $$\supseteq$$ is obvious. The other one should be straightforward, but I'm clearly forgetting something.

So, let $$P \in F(X) \cap K[X]$$, then $$P = \frac{F_1}{F_2}$$ with $$F_1,F_2 \in F[X]$$. I feel like this brings me back to where I started.

Other methods are appreciated as well.

• Doesn't Euclid's Algorithm only involve field operations with the coefficients of the starting polynomials? So that HCF is (sort of) independent of the field. – ancient mathematician May 30 '19 at 8:33
• I don't completely interstand your comment. Are you suggesting that the statement follows from the Euclidean algorithm? – user661541 May 30 '19 at 8:43
• Your reformulation of the question is clearly equivalent to the original one. The answer is very simple as long as you are aware of the uniqueness of quotient and remainder in the long division for polynomials; see here. Start with $Q=PF+R$ in $F[X]$ and compare with $Q=PH$ in $K[X]$. (Too simple to be true?) – user26857 Jun 1 '19 at 17:12
• @user26857. This is correct. Please post it as an answer! – user661541 Jun 1 '19 at 23:08

Let $$F$$ be a field and $$K$$ an extension field.

Let $$f,g\in F[X]$$. Then the highest common factors of $$f,g$$ in $$F[X]$$ and $$K[X]$$ coincide. This is because if we carry out the classical Euclidean Algorithm on $$f,g\in F[X]$$ the h.c.f. and all the intermediate polynomials lie in $$F[X]$$. But this calculation (thought of as in $$K[X]$$) is just the Euclidean Algorithm in $$K[X]$$, and yields the same h.c.f.

Now suppose $$p,q\in F[x]$$ and $$p|q$$ in $$K[X]$$. Then $$p$$ is the h.c.f. in $$K[X]$$ of $$p$$ and $$q$$. Hence $$p$$ is the h.c.f. in $$F[X]$$ of $$p$$ and $$q$$. So $$p|q$$ in $$F[X]$$.

A demonstration which illustrates exactly how the coefficients of the quotient in $$K[x]$$ lie in the base field $$F$$:

Given fields

$$F \subset K, \tag 1$$

with

$$P(x), Q(x) \in F[x] \tag 2$$

and

$$P(x) \mid Q(x) \; \text{in} \; K[x], \tag 3$$

then

$$\exists D(x) \in K[x], \; Q(x) = P(x) D(x) \; \text{in} \; K[x]; \tag 4$$

from this it follows that the term of $$Q(x)$$ of least degree $$q$$, that is

$$Q_q x^q, \; Q_q \ne 0, \tag 5$$

where

$$Q(x) = \displaystyle \sum_q^{\deg Q} Q_jx^j, Q_j \in F,\tag 6$$

satisfies

$$Q_qx^q = P_px^pD_dx^d = P_pD_dx^{p + d}, \tag 7$$

where again

$$P(x) = \displaystyle \sum_p^P P_jx^j, P_j \in F, P_p \ne 0, \tag 8$$

and

$$D(x) = \displaystyle \sum_d^{\deg D} D_jx^j, D_j \in K, D_d \ne 0; \tag 9$$

(7) is directly obtained by taking the product of (8) and (9) according to the usual rules of polynomial multiplication, and it affirms that

$$q = p + d \ge p, \tag{10}$$

since

$$d \ge 0. \tag{11}$$

At this point we pause to observe that, in light of (10), (7) yields

$$Q_q = P_qD_d, \tag{12}$$

whence

$$D_d = P_p^{-1} Q_q \in F; \tag{13}$$

we will extend this result to all $$D_j$$, $$d \le j \le D$$. Setting

$$Q_1(x) = \displaystyle \sum_q^{\deg Q} Q_jx^{j - q}, \tag{14}$$

we have

$$Q(x) = x^qQ_1(x), \tag{15}$$

and similarly setting

$$P_1(x) = \displaystyle \sum_p^P P_jx^{j - p}, \tag{16}$$

so that

$$P(x) = x^pP_1(x), \tag{17}$$

we write (4) as

$$x^qQ_1(x) = x^pP_1(x)D(x); \tag{18}$$

whence again invoking (10),

$$x^dQ_1(x) = P_1(x)D(x); \tag{19}$$

we perform this maneuver one last time with

$$D_1(x) = \displaystyle \sum_d^{\deg D} D_jx^{j - d}, \tag{20}$$

$$D(x) = x^dD_1(x); \tag{21}$$

then (19) yields

$$x^dQ_1(x) = x^dP_1(x)D_1(x), \tag{22}$$

or

$$Q_1(x) = P_1(x)D_1(x); \tag{23}$$

comparing the degree $$0$$ terms of either side of this equation once again gives (12); when we equate first-degree terms we obtain, via (14), (16), and (20),

$$Q_{q + 1} = P_{p + 1}D_d + P_pD_{d + 1}, \tag{24}$$

from which

$$D_{d + 1} = P_p^{-1} (Q_{q + 1} - P_{p + 1}D_d) \in F, \tag{25}$$so we now have

$$D_d, D_{d + 1} \in F. \tag{25}$$

Further scrutiny of (23), again in light of (14), (16) and (20), reveals that the coefficients of the powers of $$x$$ in general satisfy

$$Q_{q + k} = \displaystyle \sum_0^k P_{p +j}D_{d + (k - j)} = P_pD_{d + k} + \sum_1^k P_{p +j}D_{d + (k - j)}, \tag{26}$$

and thus

$$D_{d + k} = P_p^{-1} \left ( Q_{q + k} - \displaystyle \sum_1^k P_{p +j}D_{d + (k - j)} \right); \tag{27}$$

from this formula it is easily seen that $$D_{d + k}$$ depends on the $$D_{d +j}$$, $$0 \le j \le k - 1$$; therefore, since the all of the coefficients

$$P_j, Q_j \in F, \tag{28}$$

it follows from a simple inductive argument (the details of which are left to the reader) that

$$D_j \in F, \; d \le j \le \deg D, \tag{29}$$

that is,

$$D(x) \in F[x], \tag{30}$$

and hence that

$$P(x) \mid Q(x) \; \text{in} \; F[x]. \tag{31}$$

The equation $$Q = PU + V$$ of Euclidean division can be written as a linear system where the unknowns are the coefficients of $$U$$ and $$V$$ and the knowns are in $$F$$. By Cramer's rule, the solution, if any, must be in $$F$$.