# Computing the minimal polynomial over a field $K$ using a finite field extension $K\subseteq E$ by means of a $K$-linear map.

I can honestly say that this site has helped me out a great deal in the past between posting questions and/or researching questions posted by others. Typically, I turn to reaching out for help if every angle, attempt, etc., at solving problems by myself has failed. I will express the question in a moment, but let me first explain that this question was a homework problem that I got wrong, and I've been reworking it so that I completely understand what's going on (as opposed to just following along and/or ignoring my error). Further, the course I'm taking is a first course in Galois Theory to put everything into perspective, and we are on Spring break for the next week, so I'm re-working problems I didn't get correct. I've been working very hard on the material, and I'm doing very well so far (the previous homework assignment had extra credit which made up for my error on the problem stated below). Let me now provide the problem statement (exactly as it was given), in order to express what, specifically, I need help with - that being said, I would greatly appreciate as much detail as possible; in any case, however, I would greatly appreciate any aspect of help on this - I asked for the details since I've only got so far reworking through this:

**Question:**$~~~$Let $K\subseteq E$ be a field extension of degree $n$, and let $\alpha\in E$. Show that:

$(a)$ The map $M_{\alpha}:E\rightarrow E$, where $M_{\alpha}(e)=\alpha e$ is an invertible $K$-linear map and therefore it can be regarded as an element of $GL(n,K)$.

$(b)$ Show that $\alpha$ is a root of the characteristic polynomial det($X\cdot I_{n}-M_{\alpha}$) of $M_{\alpha}$

$(c)$ Use the above results to find a polynomial in $\mathbb{Q}(X)$ which has $\sqrt{2}+\sqrt{3}$ as a root. Find the minimal polynomial of $\sqrt{2}+\sqrt{3}$.

**Previous Work:**$~~~$ Upfront, I've already got some help from my professor, and we worked through part $(c)$ together once I got the graded homework back. His purpose was to try to show me the specific case in part $(c)$ so that I can go back and try to solve the general case in parts $(a)$ and $(b)$ on my own. To provide some of this work that we have carried out, first let $\alpha=\sqrt{2}+\sqrt{3}$. We then work in $\mathbb{Q}\subset\mathbb{Q}(\sqrt{2}+\sqrt{3})\subseteq\mathbb{Q}(\sqrt{2},\sqrt{3})$ (on a side note, I've already proven that $\mathbb{Q}(\sqrt{2}+\sqrt{3})=\mathbb{Q}(\sqrt{2},\sqrt{3})$, however I didn't want to stray too far from what I need help with - mainly I'm saying I already know the minimal polynomial of $\alpha=\sqrt{2}+\sqrt{3}~$ over $\mathbb{Q}$ is $x^{4}-10x^{2}+1$ to which I denote in particular as irr$(\alpha,\mathbb{Q})=x^{4}-10x^{2}+1$ and also the deg$(\alpha,\mathbb{Q})=\big[\mathbb{Q}(\alpha):\mathbb{Q}\big]=4~$ pertaining to the notation I use).

$~~~~~~~~$To try and quickly show part $(c)$ that me and the professor worked through, it can be shown that if we first adjoining $\sqrt{2}$ to $\mathbb{Q}$, and then adjoining $\sqrt{3}$ to $\mathbb{Q}(\sqrt{2})$, we have that the set $\big\{1,\sqrt{2},\sqrt{3},\sqrt{6}\big\}$ is a basis for $\mathbb{Q}(\sqrt{2},\sqrt{3})$ as a $\mathbb{Q}$-vector space. Hence, from $(a)$ above we define the map $M_{\alpha}:\mathbb{Q}(\sqrt{2},\sqrt{3})\rightarrow\mathbb{Q}(\sqrt{2},\sqrt{3})$ where we have that $1\mapsto\sqrt{2}+\sqrt{3}=\alpha$, $\sqrt{2}\mapsto 2+\sqrt{6}=\alpha^{2}$, $\sqrt{3}\mapsto 3+\sqrt{6}=\alpha^{3}$, and $\sqrt{6}\mapsto 2\sqrt{3}+3\sqrt{2}=\alpha^{4}$. This mapping gives us the matrix:

$\begin{bmatrix} 0 & 2 & 3 & 0 \\ 1 & 0 & 0 & 3 \\ 1 & 0 & 0 & 2 \\ 0 & 1 & 1 & 0 \\ \end{bmatrix}$;

and then we must compute the determinant of the matrix below giving the characteristic polynomial (which, in turn, will be the minimal polynomial for $\alpha$):

$\begin{bmatrix} x & 2 & 3 & 0 \\ 1 & x & 0 & 3 \\ 1 & 0 & x & 2 \\ 0 & 1 & 1 & x \\ \end{bmatrix}$.

I omit the details here to save space and time (sorry for the lengthy question), but the above computation yields that the determinant of the above matrix is $(x^{2}-1)^{2}-8x^{2}$. This polynomial simplifies to $x^{4}-10x^{2}+1$ which is exactly the minimal polynomial I stated above.

This brings me to my problem. I need help with part $(a)$ mainly; once I can work through $(a)$, I know I can work through $(b)$ (or at least I think I can, and, if not, I can always come back here). Anyways, concerning part $(a)$, I've shown already that the map $M_{\alpha}:E\rightarrow E$ is a $K$-linear map (which is very easy) by showing for $\varepsilon_{1},\varepsilon_{2}\in E$ and for $k\in K$ that:

$M_{\alpha}(k\varepsilon_{1}+\varepsilon_{2})=\alpha(k\varepsilon_{1}+\varepsilon_{2})=k(\alpha\varepsilon_{1})+\alpha\varepsilon_{2}=kM_{\alpha}(\varepsilon_{1})+M_{\alpha}(\varepsilon_{2})$;

where above holds by appealing to the field properties of $E$, hence the map $M_{\alpha}$ is linear. I know that this means there exists an $n\times n$ matrix, call it $A$, such that for $e\in E$ we have $M_{\alpha}(e)=Ae$. This is where I run into confusion about how to proceed, as I can't find the basis to use, as in the particular case for $(c)$. I've been bouncing back and forth between using bases for $E$. First I tried using the basis $\big\{e_{1},e_{2},...,e_{n}\big\}$ where $e_{i}$ are the standard basis vectors of $E$. But I then have trouble coming up with the matrix $A$ having elements in $K$, as I keep getting hung up trying to figure out the expression for $\alpha e_{i}\in E$, so I figured this wasn't the way to proceed? Furthermore, I then tried using a basis $\big\{1,\alpha,...,\alpha^{n-1}\big\}$, but I still run into the same problems; also I'm starting to think that this set is not a basis for $E$ as a $K$-vector space, where this set is a basis for $K(\alpha)$ instead working in $K\subseteq K(\alpha)\subseteq E$.

This is my major dilemma in terms of developing the matrix $A$ as mentioned above. I figured, once I can come up with this matrix, I can show that it is invertible by showing that its determinant is not zero. I've also thought about using the definition of an inverse, linear transformation which clearly shows part $(a)$ is true, but then, in this fashion, how would I solve $(b)$? If using the definition is indeed the way to proceed, then this would be the only circumstance where I would respectfully ask on some detail(s)/hint(s) on how to solve $(b)$. In any case, I again apologize for the lengthy question, as I realize things become quite time-consuming; as you can probably tell, I've been at this for a while now, and, since I have all of next week off from class due to Spring break, I figured I would reach out on this site. Overall, I greatly appreciate any insight, recommendations, suggestions, etc., and thank you greatly for you time!

• Field extensions do not have minimal poynomials. – Mariano Suárez-Álvarez Mar 7 '15 at 22:37
• Isn't the linear map $M_\alpha$ invertible simply because it has a two-sided inverse $M_{\alpha^{-1}}$? Or am I confused? – Gregory Grant Mar 7 '15 at 22:49
• Gregory, you are not confused. See, I know the map is invertible, but is there a way to show this by means of the matrix $A$ from the equation above being the linear transformation $M_{\alpha}(e)=Ae$ for $e\in E$??? – Procore Mar 7 '15 at 23:01
• The question as stated doesn't require any inspection of the matrix, it just asks to show it's invertible. Maybe you need to restate the problem to be specific to what you're trying to do. – Gregory Grant Mar 7 '15 at 23:05
• Mariano, thank you for the suggestion, and I edited the title as well as the problem. I neglected to look over the title after proofreading the body of the question. – Procore Mar 8 '15 at 0:21

Since $E$ is a field, $\alpha$ has an inverse $\alpha^{-1}\in E$. Therefore $M_{\alpha}$ has a two-sided inverse $M_{\alpha^{-1}}$. Therefore $M_\alpha$ is invertible, you don't have to calculate any determinants.
• Gregory, see the following link for the problem statement, as it is given exactly as it was given to me, and I didn't want to stray from the given problem statement. Using your answer/comment above, how would I go about calculating the determinant in (b), as I started working it with this in mind and I'm thinking maybe my misunderstanding is notational in regards to $M_{\alpha}$ in part $(b)$??? – Procore Mar 7 '15 at 23:27
• Well, the matrix $M_\alpha$ is given by the coefficients on the right hand side of the system of equations: $$\alpha u_1 = a_{11}u_1+\cdots+a_{1n}u_n$$ $$\vdots$$ $$\alpha u_n = a_{n1}u_1+\cdots+a_{nn}u_n$$ where $u_1,\dots,u_n$ is a basis. So if you subtract the left hand sides to the right you get a matrix that takes a non-zero vector to the zero vector and so must be singular and therefore have zero determinant. The matrix of this transformation is exactly $M_\alpha - \alpha\cdot I$. I think that's all that's behind part (b). – Gregory Grant Mar 7 '15 at 23:55
• I thank you very much on this, as I see exactly what you mean, and I agree completely! Yet again, I made things $much$ more harder on myself - thank you again. Moreover, when I first started this problem I had to go back and review linear transformations a little, and I reproved that if $f:V\rightarrow W$ is a linear transformation where dim$_{F}(V)=m$ with dim$_{F}(W)=n$ and $F$ is a field, then there is an $n\times m$ matrix $A$ such that for $u\in V$ we have $f(u)=Au$. The proof for this statement threw me off as I kept thinking I needed to define some kind of basis for $E$ in the problem. – Procore Mar 8 '15 at 0:29
• No problem, and thank you again as well - I just re-did the problem, except I tried it with $\mathbb{Q}(\sqrt{3},\sqrt{5})=E$ for kicks. I'm still kind of new to the structure of this site, but I continue to use it as I always find that things turn out exceptionally well when I need to look for help here as a last resort. I'm the type who likes to carry the world on my back, while Math is my passion in life, so making things harder is a bad habit of mine. I appreciate you setting me straight, as I'm finding Galois theory intriguing, so this will only help me understand things better! – Procore Mar 8 '15 at 1:11