# Field extensions described as a matrix

I'm reading Abstract Algebra and there is an exercise I don't understand :

The exercise is $Q19.b$ : I just don't understand what the matrix the book talks about ? I can think of something like to coefficients of some minimal polynomial, but it's gust my intuition. What is the matrix the question talks about ?

From reading $Q20$ I see that there is a relation between that matrix in $Q19.b$ and with calculating minimal polynomials, but what is the relation ?

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Consider for example $F=\mathbb{Q}$ and $K=\mathbb{Q}(\sqrt{2},\sqrt{3})$. A basis for $K$ as an $F$ vector space is $1$, $\sqrt{2}$, $\sqrt{3}$, $\sqrt{6}$. Now, take an element of $K$, say, $\alpha = 1-\sqrt{2}$. Then multiplication by $\alpha$ is an $F$-linear transformation on $K$ ($\alpha(k_1+k_2) = \alpha k_1 + \alpha k_2$ for $k_1,k_2\in K$, and $\alpha(rk) = r(\alpha k)$ for any $r\in F$ and $k\in K$), so it has a matrix representation relative to $\beta=[1,\sqrt{2},\sqrt{3},\sqrt{6}]$. Computing, we have: \begin{align*} \alpha 1 & = (1-\sqrt{2})(1) \\ &= 1 - \sqrt{2}+0\sqrt{3}+0\sqrt{6};\\ \alpha\sqrt{2} &= (1-\sqrt{2})\sqrt{2}\\ &= -2(1) + \sqrt{2} + 0\sqrt{3} + 0\sqrt{6};\\ \alpha\sqrt{3} &= (1-\sqrt{2})\sqrt{3}\\ &= 0(1) + 0\sqrt{2} + \sqrt{3} - \sqrt{6};\\ \alpha\sqrt{6} &= (1-\sqrt{2})\sqrt{6}\\ &= 0(1) + 0\sqrt{2} -2\sqrt{3} + \sqrt{6}. \end{align*} So the matrix of "multiplication by $\alpha$" relative to the given basis is: $$\left(\begin{array}{rrrr} 1 & -2 & 0 & 0\\ -1 & 1 & 0 & 0\\ 0 &0 & 1 & -2\\ 0 & 0 & -1 & 1 \end{array}\right).$$ Now, we can do this with any elements of $K$; you should check that if you look at the matrices you get by doing that, then what you'll get is isomorphic to $K$ (under addition and multiplication of matrices).
Thanks for your explanation. However I am still confused by question 20 above in the question. For example are they saying with this method we can compute a polynomial of degree 4 with $1 - \sqrt{2}$ as a root, such a polynomial with coefficients in $\Bbb{Q}$? – user38268 Mar 30 '12 at 6:13
In 19a, you show multiplication by $\alpha$ is a linear transformation. Well, every linear transformation can be seen as matrix multiplication, once you've picked a basis for $K$ over $F$. So the element $\alpha$ corresponds to the matrix $A$ which has the property that multiplication by $\alpha$ is the linear transformation whose matrix is $A$.
Maybe it would help if you worked through a simple example, like multiplication by $\sqrt2$.
Wouldn't this just be a diagonal matrix with $\alpha$ on the diagonal? I was confused by this exercise a few days ago too. – user38268 Mar 29 '12 at 8:35
We're talking about an $F$-linear transformation, so the matrix entries have to be in $F$, not in $K$. But just do it: find a basis for ${\bf Q}(\sqrt2)$ over $\bf Q$, and find the matrix, with respect to that basis, representing multiplication by $\sqrt2$. – Gerry Myerson Mar 29 '12 at 11:36