Questions from students during the course are answered here anonymously.

Question 1: What are the conditions for the Bézout identity to be valid in an integral domain?

Answer: If $R$ is an integral domain, some equivalent conditions to the condition that the Bézout identity holds in $R$ are the following:

1. Every ideal in $R$ which is generated by two elements is principal.
2. Every finitely generated ideal in $R$ is principal.
3. The gcd of any two elements exists and every finitely generated ideal is invertible.

It should not be too difficult to see that (1) is very close to the Bézout identity. Also (2) is easily seen to be equivalent to (1) by induction. You maybe have not seen invertible ideals before. For this you consider an ideal $I$ of $R$ and the field of fractions $K$ of $R$. Then consider the subset $J$ of $K$ defined by $J=\{x \text{ in $$K$ such that $xI$ is a subset of $R$}\}$$. We say that $I$ is invertible, if the product $IJ$ is equal to $R$, where $IJ$ is defined in the same way as the product of two ideals.

Question 2: For this course, when we use that $R$ is a ring, are we always assuming that $R$ is a commutative ring?

Answer: Yes, in this course all rings are assumed to be commutative (and unital) unless stated otherwise.

Question 3: The solution of Problem 9 in Problem Set 1 uses Lemma 3.4(3) from the notes. However, I think that the whole lemma uses as a hypothesis that we are working in a field. $\mathbb{Z}$ is not a field, so, is the solution wrong or can we apply this lemma in a set of more general cases?

Answer: (The solution of this problem has now been updated) Indeed, we can not use Lemma 3.4(3) immediately since $\mathbb{Z}$ is not a field. We can use Lemma 3.6 to get over this problem. More specifically, Lemma 3.6 says that $f(x)$ is irreducible over $\mathbb{Z}$ if and only if it is primitive and irreducible over $\mathbb{Q}$. Since this specific polynomial is primitive, it is irreducible over $\mathbb{Z}$ if and only if it is irreducible over $\mathbb{Q}$. Hence, we may check its irreducibility over $\mathbb{Q}$ instead and so we can use Lemma 3.4(3).

Question 4: In Problem 14 in Problem Set 1 it is stated that $\mathbb{Z}[x]/(p(x)) = \mathbb{Z}_3[x + (p(x))]$. How can we deduce this?

Answer: This follows from Theorem 4.6. Let $\alpha=x+(p(x))$. Then $K=\mathbb{Z}_3[x]/(p(x))$ is isomorphic to $\mathbb{Z}_3[\alpha]$ (see the beginning of the proof of Theorem 4.6 in the notes for an explicit proof of this fact). Then by the statement of Theorem 4.6 we have that $K$ is isomorphic to $\mathbb{Z}_3[\alpha]=\mathbb{Z}_3(\alpha)$.

Question 5: In Problem 15 in Problem Set 1 why can we easily deduce that $\mathbb{Q}(2^{1/3},2^{1/3}e^{2\pi i/3},2^{1/3}e^{4\pi i/3}) = \mathbb{Q}(2^{1/3},e^{2\pi i/3})$?

Answer: Let me introduce some notation in this specific example. Let $x=2^{1/3}$ and $y=e^{2\pi i/3}$. Then we want to show that $\mathbb{Q}(x,xy,xy^2) = \mathbb{Q}(x,y)$. Notice first that $xy$ is in $\mathbb{Q}(x,y)$. Indeed, since $x$ and $y$ are in $\mathbb{Q}(x,y)$, and $\mathbb{Q}(x,y)$ is a field, we have that their product $xy$ is in $\mathbb{Q}(x,y)$. Similarly we have that $xy^2$ is in $\mathbb{Q}(x,y)$. Therefore, since all of $x$, $xy$ and $xy^2$ are in $\mathbb{Q}(x,y)$, and $\mathbb{Q}(x,y)$ is a field, we obtain that the smallest field that contains $x$, $xy$ and $xy^2$ is a subfield of $\mathbb{Q}(x,y)$. In other words, we have that $\mathbb{Q}(x,xy,xy^2)$ is a subfield of $\mathbb{Q}(x,y)$.

Now we show the other inclusion. We have that $x$ is in $\mathbb{Q}(x,xy,xy^2)$. Since $\mathbb{Q}(x,xy,xy^2)$ is a field, it follows that $x^{-1}$ is in $\mathbb{Q}(x,xy,xy^2)$. Then, since both $x^{-1}$ and $xy$ are in $\mathbb{Q}(x,xy,xy^2)$, we conclude that $x^{-1}xy=y$ is in $\mathbb{Q}(x,xy,xy^2)$. We have thus shown that both $x$ and $y$ are in $\mathbb{Q}(x,xy,xy^2)$ and so we conclude that $\mathbb{Q}(x,y)$ is a subfield of $\mathbb{Q}(x,xy,xy^2)$. Since both inclusions hold, we obtain that $\mathbb{Q}(x,xy,xy^2)=\mathbb{Q}(x,y)$.

In general, if we want to compare fields of the form $F(a_1,\ldots,a_k)$ and $F(b_1,\ldots,b_m)$ we need to show that each $a_i$ is in $F(b_1,\ldots,b_m)$ and each $b_j$ is in $F(a_1,\ldots,a_k)$. To do this, we have to write each $a_i$ using the elements of $F$, the elements $b_1,\ldots,b_k$ and the field operations, and similarly for each $b_j$.

Question 6: In the proof of Theorem 4.5 it is stated that the set $E=F[X]/(p(x))$ is a field extension of F. However, this does not make much sense. We know $E$ has equivalence classes of polynomials of $F$. Hence, I do not see how $F$ is included in $E$. The only way this is possible is having an embedding from $F$ to $E$. The only option I can find is having the function $f$ which is $f: F \rightarrow E$ with $f(a) = a + (p(x))$. This is a ring homomorphism and it seems to be an embedding. So, we can claim $E$ is a field extension of $F$ by considering the embedding $f$.

Answer: The situation is exactly as you describe it: we can view $E=F[x]/(p(x))$ as a field extension of $F$ by viewing an element $a\in F$ as the element $a+(p(x))\in F[x]/(p(x))$.

Question 7: In Problem 5 from Problem Set 2 it is used that the given $f(x)$ is the minimal polynomial of $b$ over $F$. Although this is true, I do not understand why we need to reason with the minimal polynomial and not just with the fact that $f(x)$ is irreducible, since Theorem 4.6 only requires an irreducible factor to state that $[F(b) : F] = deg(f(x))$.

Answer: You are right, the fact that $f(x)$ is the minimal polynomial of $b$ over $F$ is not required to show that $[F(b):F]=deg(f(x))$, it is sufficient to have that $f(x)$ is irreducible over $F$ and that $f(b)=0$.

Question 8: In Problem 9 from Problem Set 2 it is said that since $f(x) = x^3 + x + 1$ has two roots in $\mathbb{Z}_2(\alpha)$, it is straightforward that all its roots are in $\mathbb{Z}_2(\alpha)$ since the degree of $f(x)$ is $3$. Is this a consequence of applying the division algorithm two times with the two known roots? We know that the quotient used in each division is a polynomial that belongs to $\mathbb{Z}_2(\alpha)$, since the last division gives as quotient a polynomial of degree 1, thus its coefficients belongs to $\mathbb{Z}_2(\alpha)$, so the last root of $f(x)$ has to remain in $\mathbb{Z}_2(\alpha)$.

Answer: You have understood the situation correctly. In general, if a polynomial $p(x)$ over a field $F$ of degree $n$ has $n-1$ roots in a field $F$, then its last root is also in $F$ since we can divide it by a polynomial of degree $n-1$ with roots exactly the $n-1$ roots of $p(x)$ that we know are in $F$.

Question 9: In Problem 3 from Problem Set 3 it is said that a general element of $F(M_1 \cup M_2)$ is of the form $(\alpha_1 \beta_1 + \cdots + \alpha_k \beta_k)(\alpha_1' \beta_1' + \cdots + \alpha_m' \beta_m')^{-1}$. How do we get this?

Answer: First, $M_1$ and $M_2$ are both extensions fields of $F$. Then $F(M_1 \cup M_2)$ is the smallest field containing $M_1$ and $M_2$. First, note that the element described in the solution is an element in $F(M_1 \cup M_2)$ as it is given by using the field operations on elements of $M_1$ and $M_2$. For the other direction, pick an element $x$ of the field $F(M_1 \cup M_2)$. It is of the form $(a b^{-1})$ where both $a$ and $b$ are elements in the smallest ring generated by $M_1$ and $M_2$ (exactly as, for example, an element $x$ in $\mathbb{Q}$ is of the form $\tfrac{a}{b}$ where $a$ and $b$ are integers). So how do the elements of the smallest ring generated by $M_1$ and $M_2$ look like? Well, such an element is given by multiplying elements of $M_1$ with elements of $M_2$ and then adding the results (since $M_1$ and $M_2$ both include $F$, we do not need to think about elements of $F$). This gives us an element of the form $(\alpha_1 \beta_1 + \cdots + \alpha_k \beta_k)$ with $a_i$'s in $M_1$ and $b_j$'s in $M_2$. You may then say that you can now multiply this element again by elements of $M_1$ or $M_2$, or even worse multiply such an element by an element of a similar form. But note that if you multiply this with a similar element, you can use distributivity to end up with an element of exactly the same form and so that doesn't give you something new. Hence elements of the field $F(M_1 \cup M_2)$ are of the form $(a b^{-1})$ where $a$ is of the form $(\alpha_1 \beta_1 + \cdots + \alpha_k \beta_k)$ and $b$ is of the form $(\alpha_1' \beta_1' + \cdots + \alpha_m' \beta_m')$, as claimed.

Question 10: In Problem 9 from Problem Set 3 we used Fermat's little theorem. This theorem requires $p$ prime, however, the problem does not say anything about this. Is this a mistake from the formulation of the exercise?

Answer: You are right, it should be made clear that $p$ is a prime number.

Question 11: In Problem 3(a) from Problem Set 4, Example 5.5 from the lecture notes is used. However, we do not have $\sqrt{2}$, we are starting with $\sqrt{3}$. I do not understand why it is still possible to apply this result. Is this because we can use any order of the prime numbers when we apply this result, i.e., it is not necessary to start with: $2,3,5,7,\ldots$ and if we want we could start with, for example: $3,5,2,7,\ldots$?

Answer: Yes, that is correct. In Example 5.5 we say that $p_n$ is the $n$-th prime number. However, in the computations of this example, we only use the fact that $p_1,p_2,\ldots,p_n,\ldots$ is an ordering of the prime numbers, so you may choose any ordering of them.

Question 12: There seem to be a few typos in the solutions of Problem 6(b) from Problem Set 2, Problem 3(c) from Problem Set 4, Problem 8 from Problem Set 4 and Problem 5 from Problem Set 5. Is that correct?

Answer: Yes, there were a few typos there which are now corrected in the uploaded files.

Question 13: How can one obtain all subgroups of a certain (finite) group? Is it possible to know beforehand the number of subgroups that a certain group will have? Is there a faster way than checking its subset separately?

Answer: If you are given a large group, finding all of its subgroups is a hard problem. Essentially it boils down to checking all of its subsets and seeing if they form a subgroup (if the group is finite at least). You can imagine that this becomes tedious if you have even a relatively small group (say a group of order $8$). There are however a few results from group theory which may help save quite a bit of time, and make the problem of finding all subgroups of a small group manageable.

First of all, we have Lagrange's theorem which tells us that the order of a subgroup must divide the order of the group. Take for example $\mathbb{Z}_4$ and $\mathbb{Z}_2 \times \mathbb{Z}_2$ which both have order $4$, so any possible subgroup must have order $1$, $2$ or $4$. Of course, the subgroups of order $1$ and $4$ are always there and are unique (the trivial subgroup and the whole group). It remains to consider all subsets of $2$ elements and see if they form a subgroup. Note that each subgroup must contain the identity element. Hence you only need to consider all subsets which contain the identity element plus one more element from your group and see if they form a subgroup. In the case of $\mathbb{Z}_4$ you need to check $\{0,1\}$, $\{0,2\}$, $\{0,3\}$ and you see that $\{0,1\}$ and $\{0,3\}$ are not closed under addition while $\{0,2\}$ is, and hence the only subgroup of order $2$ is $\{0,2\}$. In the case of $\mathbb{Z}_2 \times \mathbb{Z}_2$ you need to check $\{(0,0),(1,0)\}$, $\{(0,0),(0,1)\}$, $\{(0,0),(1,1)\}$ and you see that all of them are closed under addition, and hence these are all the subgroups of order $2$ (notice that for a finite group we do not need to check that the inverse is in the group as long as the group is closed under the group operation).

The case of $\mathbb{Z}_4$ is especially interesting because $\mathbb{Z}_4$ is a cyclic group. For cyclic groups the following general result holds: a cyclic group with $n$ elements has exactly one subgroup for each divisor $d$ of $n$, and this is a cyclic subgroup itself. There are no other subgroups for cyclic groups.

PS: Some other useful results can be given by Cauchy's theorem (which states that whenever a prime number $p$ divides the order of a group, then that group has at least one subgroup of that order $p$) and Sylow's theorems. These are more advanced results in group theory which might be useful to find all subgroups of a group with a few more elements. In this course we will not need them for finding subgroups of a group.

Question 14: The solution of Problem 3(a) from Problem Set 5 claims that an element $\alpha$ of $G(\mathbb{Q}(\omega)/\mathbb{Q})$ is completely defined by its value on $\alpha(\omega)$. I understand the reason to say this. However, I do not understand how we calculate the possible values for $\alpha(\omega)$. Consider the following argument: since $\alpha$ must be the identity function when we restrict $alpha$ to $\mathbb{Q}$, we have that $\alpha(1) = 1$. Besides, since $\omega$ is a root of $x^p - 1$, we have that $\omega^p = 1$, so $\alpha(1) = \alpha(\omega^p) = (\alpha(\omega))^p$. Hence, $1 = \alpha(\omega)^p$ implies that $\alpha(\omega)$ is a $p$-th root of $1$, which gives $p$ different values for $\alpha(\omega)$. However, the solution deduces that $\alpha(\omega)$ is a root of $\Phi_p(x)$ and so there are $p -1$ different values for $\alpha(\omega)$. Where does my argument fail?

Answer: Your computation shows that $\alpha(\omega)$ is a root of $x^p-1$. It doesn't show that all roots of $x^p-1$ are values that $\alpha(\omega)$ does indeed take. But now notice that you know even more for $\omega$: you know that $\omega$ is a root of $\Phi_p(x)$. Hence by the same argument as the one you used, it follows that $\alpha(\omega)$ is a root of $\Phi_p(x)$. Notice that $1$ is not a root of $\Phi_p(x)$ and $x^p-1 = (x-1)\Phi_p(x)$. That is, while $\alpha(\omega)$ is indeed always a root of $x^p-1$, we have shown the even stronger property that it is a root of $Phi_p(x)$ and so it is not equal to $1$.

Of course, then you may ask how do we know that $\alpha(\omega)$ can then be any of the roots of $\Phi_p(x)$? It could be that again some of the roots of $\Phi_p(x)$ can not be attained as values of $\alpha(\omega)$, similarly as 1 can not be attained. But now we know that the Galois group $G(\mathbb{Q}(\omega)/\mathbb{Q})$ has order $p-1$ as seen in the exercise, and so $\alpha(\omega)$ must take $p-1$ different values (as any alpha in $G(\mathbb{Q}(\omega)/\mathbb{Q})$ is uniquely defined by its value on $\omega$). Since we know that all the values it can take are among the roots of $\Phi_p(x)$, and since there are exactly $p-1$ values there, it follows that $\alpha(\omega)$ can indeed be any of the roots of $\Phi_p(x)$.

Question 15: The last part of Problem 7 from Problem Set 5 asks for finding the fields $K$ such that $F\subseteq K\subseteq E$. To do that, we use the fact that $G(E/F)$ is isomorphic to $\mathbb{Z}_8$. My question is: is it true that each subgroup of $G(E/F)$ has to be mapped to a subgroup of $\mathbb{Z}_8$? More generally, is the number of subgroups of a group is invariant by isomorphism?

Answer: Yes, if you have two isomorphic groups, the number of subgroups they have is the same. Furthermore, not only the number is the same, but also the lattice of subgroups they have is the same and is preserved under any isomorphism of groups. This means that if $G$ is isomorphic to a group $G'$ via an isomorphism $f$, and if $H$ is a subgroup of $G$, then $f(H)$ is a subgroup of $G'$ and $H$ and $f(H)$ are isomorphic to each other (via $f$ again).

Question 16: In the solution of Problem 3(b) from Problem Set 6 it is stated that for a prime number $p$ we have that $\mathbb{Z}_p^{m}/\mathbb{Z}_p^{m-1}$ is isomorphic to $\mathbb{Z}_p$. Is this something that can be taken for granted or do we need to prove it?

Answer: You don't need to prove the stated group isomorphism unless asked to, so it is good to know how to do it! Here is a simple argument: $\mathbb{Z}_{p^m}$ has $p^m$ elements and $\mathbb{Z}_{p^{m-1}}$ has $p^{m-1}$ elements, so their quotient has $p^m/p^{m-1}=p$ elements (the quotient of finite groups has as many elements as the quotient of the number of elements between the two groups). The only group of prime order $p$ (up to isomorphism) is the cyclic group $\mathbb{Z}_p$, and so the quotient must be isomorphic to $\mathbb{Z}_p$.

Question 17: In the solution of Problem 11 from Problem Set 6 it is stated that "These circles intersect in two points: by Euclidean geometry,…". Which statement of Euclidean geometry is referred to here?

Answer: This is Proposition 10 in Book 1 of Euclid's Elements.

Question 18: The solution of Problem 12 from Problem Set 6 has two different parts where it is claimed that the construction done there creates a triangle with hypotenuse of length $1$. Why is this true?

Answer: For the first triangle: the points $O$, $X$ and $Y$ are different by assumption: $X$ and $Y$ lie on the segments $OQ$ and $OP$ respectively, and $OP$ and $OQ$ are not on the same line since the angle $\theta$ is greater than $0$. Hence the three points $O$, $X$ and $Y$ are not on the same line and so they form a triangle which lies on the first quadrant. Next, the line segment $OX$ has length $1$ by construction as $X$ lies on the circle with center at $O$ and radius $1$. Furthermore, the line segment $XY$ is orthogonal to the line going through $O$ and $Y$, again by construction. Hence the triangle formed by $O$, $X$ and $Y$ is an orthogonal triangle (with the right angle being the angle at $Y$) and the hypotenuse is the line segment $OX$ of length $1$.

Question 19: In the solution of Problem 3 from Problem Set 1, when solving for $x$,$y$,$z$,$w$ we proceed by dividing by $a+b\sqrt(-3)$. How is this possible when $\mathbb{Z}[\sqrt(-3)]$ is not a field?

Answer: Indeed, dividing by $a+b\sqrt(-3)$ is not possible in general as $\mathbb{Z}[\sqrt(-3)]$ is not a ring. However, in the solution of this problem we have made the assumption that $a+b\sqrt(-3)$ divides both $4$ and $2+2\sqrt(-3)$, and so we know that these specific divisions are possible, that is, they give us an element in the ring $\mathbb{Z}[\sqrt(-3)]$.

Question 20: At the end of the proof of the part (3)$\implies$(2) of Theorem 8.5, how do we conclude that $\beta = \phi^\star (\alpha)\in E$? How is Theorem 6.5 useful here?

Answer: First we want to apply Theorem 6.5. We have the embedding $\phi$ from $F(\alpha)$ to $\overline{F}$, the algebraic closure of $F$ which is an algebraically closed field. We also have an algebraic extension $F(\alpha)\subseteq E$ (the extension exists since $\alpha\in E$ and $F\subseteq E$, which gives that $F(\alpha)\subseteq E$, and the fact that it is algebraic follows since the extension $F\subseteq E$ is algebraic, as explained at the beginning of the proof). This shows that the requirements of Theorem 6.5 are satisfied (with the roles of $F$, $L$, $E$ and $\sigma$ in the theorem played by $F(\alpha)$, $\overline{F}$, $E$ and $\phi$ respectively in this proof), and so we obtain an embedding $\phi^{\ast}:E\to\overline{F}$ which restricts to identity on $F(\alpha)$. In particular, it restricts to identity on $F$. Since we have assumed that (3) holds, we obtain that this embedding restricts to an $F$-automorphism of $E$, that is it is a ring homomorphism from the field $E$ to the field $E$ which restricts to identity on $F$. In particular, $\phi^{\ast}(\alpha)$ is in $E$ (as is $\phi^{\ast}(x)$ for every $x\in E$). It remains to compute $\phi^{\ast}(\alpha)$. Notice that $\phi^{\ast}$ is an extension of $\phi$ and that $\phi$ is defined on $F(\alpha)$ via $\phi(b_0+b_1\alpha+\cdots+b_{n-1}\alpha^{n-1})=\phi(b_1+b_1\beta+\cdots+b_{n-1}\beta^{n-1})$. Then $\phi(\alpha)=\beta$ just by the definition of $\phi$ (that is, we set $b_0=b_2=\cdots=b_{n-1}=0$ and $b_1=1$). This concludes the proof.

Question 21: In the solution of Problem 5 from Problem Set 5, in the case $\text{char}(F)=p$, in the direction $\Rightarrow$, it is stated that "$(x-\alpha)^r$ has as constant term $-r\alpha$" which is false, and then this false statement is used to conclude that $-r\alpha\in F$. How can we conclude correctly that $-r\alpha\in F$? Perhaps we should look at the coefficient of $x^{r-1}$ instead? Moreover, how can we use $-r\alpha\in F$ to obtain that $\alpha\in F$?

Answer: (The solution of this problem has now been updated) You are right that the constant term of $(x-\alpha)^r$ is not $-r\alpha$, and you are right that we should look at the coefficient of $x^{r-1}$ instead. Notice that its coefficient is $-r\alpha$ by the binomial theorem. Since $(x-\alpha)^r$ is a polynomial in $F[x]$, this means that all of its coefficients are in $F$ and so $-r\alpha$ is in $F$. By multiplying by $-1$ we obtain that $r\alpha\in F$. Then we can of course add as many copies of this element in $F$ as we want and still get an element in $F$, that is $k(r\alpha)$ is in $F$ for every positive integer $k$. In particular, we have $k(r\alpha)$=$(kr)\alpha$ (remember that $r\alpha$ is just shorthand for $\underbrace{\alpha+\cdots+\alpha}_{r \text{ times}}$) and so $(kr)\alpha\in F$ for every positive integer $k$. Now pick $k$ such that $kr=1 \mod{p}$ (such a $k$ exists since $p$ is a prime). Since $F$ is a field of characteristic $p$, we have that $(kr)\alpha=\alpha\in F$ as required.

Question 22: At the end of the solution of Problem 8(b) from Problem Set 3, we have shown that $\beta$, an element of the quotient ring $F[x]/(f(x))$, is a root of both $f(x)$ and $g(x)$, where these are polynomials over $\mathrm{GF}(p)$ and $f(x)$ is irreducible. Then it is stated that, this implies that $f(x)$ divides $g(x)$ in $\mathrm{GF}(p)[x]$. Why is that?

Answer: This is a consequence of a more general and useful statement, so let us show that statement instead. Assume that we are working with polynomials over a field $F$ and that we have an extension field $K$ of $F$. If a polynomial $f(x)\in F[x]$ is irreducible and shares a root $\alpha\in K$ with another polynomial $g(x)\in F[x]$, then $f(x)$ divides $g(x)$. Here are two ways we can argue about this.

• First we can consider the minimal polynomial of $\alpha$ over $F$, say $q(x)$ (it exists, since $\alpha$ is algebraic over $F$ as it is the root of $f(x)$). Then we know that $q(x)$ divides $f(x)$ and that $q(x)$ divides $g(x)$ (by Theorem 5.2(2) in the notes). Therefore $f(x)=q(x)h(x)$ and $g(x)=q(x)z(x)$ for some polynomials $h(x)$ and $z(x)$ in $F[x]$. Since $f(x)$ is irreducible, one of $q(x)$ and $h(x)$ is a unit. Since $q(x)$ is irreducible, it is not a unit, and so $h(x)=c$ is a nonzero constant in $F$ (as these are the units of $F[x]$). But then $f(x)=c q(x)$ gives $q(x)=c^{-1} f(x)$. Replacing this in $g(x)=q(x)z(x)$ we obtain $g(x)=c^{-1}f(x)z(x)$ and so $f(x)$ divides $g(x)$.
• Another way is to consider the $\mathrm{gcd}$ of $f(x)$ and $g(x)$ (it exists since $F[x]$ is a UFD). Let $h(x)=\mathrm{gcd}(f(x),g(x))$. Since $h(x)$ divides the irreducible polynomial $f(x)$, it is either equal to $1$ or it is equal to $cf(x)$ for some nonzero constant $c\in F$. Then by Bézout identity there exist polynomials $p(x)$ and $q(x)$ such that $f(x)p(x)+g(x)q(x)=h(x)$. By plugging in $\alpha$ we have $f(\alpha)p(\alpha)+g(\alpha)q(\alpha)=h(\alpha)$ and the left-hand side is $0$ since $\alpha$ is a root of both $f(x)$ and $q(x)$. Hence $h(\alpha)=0$ and so $h(x)$ is not equal to $1$. Therefore $h(x)=cf(x)$. Since $h(x)$ divides $g(x)$ also (as it is a common divisor), we obtain that $cf(x)$ divides $g(x)$ and so $g(x)=r(x) cf(x)$ for some polynomial $r(x)$ in $F[x]$. This shows that $f(x)$ divides $g(x)$.