# Minimal polynomial of $ab$, when those of $a$ and $b$ are known

Let $k \subseteq L$ be a finite separable field extension (not necessarily Galois) of prime degree $P \geq 3$.

Assume that $L=k(a)=k(b)$, for some $a,b \in L$.

Clearly, $k \subseteq k(ab) \subseteq k(a)=k(b)$, so from $[k(ab):k][L:k(ab)]=[L:k]=P$, we get that $k(ab)=k$ or $k(ab)=L$.

Further assume that $L=k(ab)$, so the degree of the minimal polynomial of $ab$ over $k$ is $P$.

Denote the minimal polynomial of $a$ over $k$ by $A(t)=t^P+a_{P-1}t^{P_1}+\cdots+a_1t+a_0$ and the minimal polynomial of $b$ over $k$ by $B(t)=t^P+b_{P-1}t^{P_1}+\cdots+b_1t+b_0$, where $a_{P-1},\ldots,a_1,a_0,b_{P-1},\ldots,b_1,b_0 \in k$.

What is the minimal polynomial of $ab$ over $k$ in term of the coefficients of $A(t)$ and $B(t)$?

Notice that it is not possible to get help from this question, since here $[k(a):k]=[k(b):k]=P \geq 3$, not $1$.

Remark: I think that it is easier to answer my question when $k \subseteq L$ is Galois, since, if I am not wrong, it is cyclic, etc. (I explained in a comment the answer to the Galois case).

Please see the following three relevant questions: i, ii (especially the answer of awllower) and iii (especially the answer of Lubin).

Thank you very much!

• I do not think there is a "universal" minimal polynomial (to see this, take $b=a+t$), but there is a "universal" polynomial satisfied by $ab$ in all cases. Why do you suppose $k(a)=k(b)$ ? Jan 17, 2018 at 23:32
• The question doesn't necessarily have a single answer: e.g. if $A(t)$ has a second root $a' \in L$ then an answer would also be a minimal polynomial for $a'b$ which should be different. Jan 17, 2018 at 23:33
• @DuchampGérardH.E., thank you. Please, can you elaborate on your example $b=a+t$? (Why not assume $k(a)=k(b)$?). Jan 17, 2018 at 23:43
• @Doris, thank you for your comment. (It is fine with me that there will be a different answer for $a'b$). Do you have an answer for fixed $a$ and $b$ as above? (What if $k \subseteq L$ is Galois?). Jan 17, 2018 at 23:49
• If $k \subseteq L$ is Galois, then it is cyclic (of degree $P$) Galois, and by Kummer theorem $a^P \in k$, $b^P \in k$ and $a^Pb^P=(ab)^P \in k$. Therefore, if we write $a^P=\alpha \in k$ and $b^P=\beta \in k$, then the minimal polynomial of $ab$ over $k$ is $t^P-\alpha \beta$ (and the minimal polynomials of $a$ over $k$ and of $b$ over $k$ are $t^P-\alpha$ and $t^P-\beta$, respectively). Jan 18, 2018 at 0:45

The question as stated does not have a well-formed answer. That is, the minimal polynomial of $ab$ is not a function of the minimal polynomials of $a$ and $b$. To see why, take for example the case $k=\mathbb{Q}$, $a=\sqrt{2}$, $b=a+1$. Then $ab = a^2+a = 2+a$ has minimal polynomial $(t-2)^2-2$. Now $a'=-a$ has the same minimal polynomial as $a$, so the question implies that the minimal polynomial for $a'b$ should be the same as the minimal polynomial for $ab$, but it is not, it is $(t+2)^2-2$.

On the other hand, polynomial resultants can be used to get other related polynomials. The important fact is that when $f,g \in F[x]$ are monic, where $F$ is some field, then $$\operatorname{Res}_x(f(x),g(x)) = \prod_{f(a)=0} g(\alpha) \in F$$ where $a$ runs over the roots of $f$. Also important is that the resultant is an explicit polynomial expression in the coefficients of $f$ and $g$.

• So for example $$\operatorname{Res}_x(A(x),y-bx) = \prod_{A(a')=0}(y - ba') \in k(b)[y]$$ is the polynomial over $k(b)$ in $y$ of degree $P$ whose roots are $a'b$ with $a'$ running over the roots of $A$.

• Similarly $$\operatorname{Res}_y(B(y),\operatorname{Res}_x(A(x),z-xy)) = \prod_{A(a')=0,B(b')=0}(z-a'b') \in k[z]$$ is the polynomial over $k$ in $z$ of degree $P^2$ whose roots are $a'b'$ with $a'$ running over the roots of $A$ and $b'$ running over the roots of $B$.

• If, instead of the minimal polynomial $B(x)$ of $b$ you actually have a polynomial $C(x) \in k[x]$ such that $b=C(a)$ (i.e. you have expressed $b$ in terms of $a$), then $$\operatorname{Res}_x(A(x),y-C(x)) = \prod_{A(a')=0}(y - C(a')) \in k[x]$$ is the polynomial over $k$ of degree $P$ whose roots are $C(a')$ where $a'$ runs over the roots of $A$. In particular, $b=C(a)$ is one of these roots so this is the minimal polynomial for $b$ assuming $k(a)=k(b)$.

• Taking this a little further, then if we use $xC(x)$ instead, then $aC(a)=ab$ and therefore $$\operatorname{Res}_x(A(x),y-xC(x)) \in k[y]$$ is the polynomial whose roots are $a'C(a')$ with $a'$ running over the roots of $A$, and in particular it has $ab$ as a root, and so it is the minimal polynomial of $ab$ if $k(ab)=k(a)$.

• This is (exactly) the answer I was looking for. (I have mentioned in one of the former versions of my question: "probably the resultant will help", but did not know how to get the minimal polynomial of $ab$, but just a polynomial for which $ab$ is a root). Jan 18, 2018 at 12:31