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Here's the setup: $k$ is a field extension of degree $d$ over $\mathbb{Q}$. So there are $d$ field embeddings $\sigma_1,\dotsc,\sigma_d:k \to \mathbb{C}$. Let $\{\alpha_1,\dotsc,\alpha_d\}$ be a collection of elements of $k$. The text I'm reading is (I think) trying to motivate the definition of the discriminant by posing the question of when $\{\alpha_i\}$ is a basis for $k$ as a vector space over $\mathbb{Q}$.

Of course, linear independence means considering the linear equation $$x_1\alpha_1+\dotsb+x_d\alpha_d=0$$ with coefficients $x_i \in \mathbb{Q}$. This is an equation in the abstract field $k$ so embed it in $\mathbb{C}$, I guess to make it more concrete? $$x_1\sigma_i(\alpha_1)+\dotsb+x_d\sigma_i(\alpha_d)=0$$ At this point, the text says

Thus one readily deduces that the set $\{\alpha_1,\dotsb,\alpha_d\}$ is a basis for $k$ if and only if $\det[\sigma_i(\alpha_j)] \not= 0$.

I do not readily deduce this at all to be frank. Seems to me that writing out a matrix implies a choice of basis already. What basis are they referring to? Are the $\sigma_i$ somehow acting as some sort of component functions, giving coordinates of the $\alpha_j$?

After this I'm willing to accept that it's a good idea to define the discriminant $\operatorname{disc}(\{\alpha_1,\dotsc,\alpha_d\})=\det[\sigma_i(\alpha_j)]^2$.

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  • $\begingroup$ In 3d, the magnitude of the determinant is the volume of the parallelpiped defined by the 3 vectors. If the volume is zero the 3 vectors lie in a plane, so they are not linearly independent. A basis must be linearly independent, so its determinant cannot be zero. The converse is also true. Similar analysis applies in 4d and higher dimensions. $\endgroup$ – Peter Sep 3 '20 at 1:24
  • $\begingroup$ The construction you are referring to fixes a basis, writes the vectors in question in components in that basis, and then takes the determinant of the resulting matrix. What is the basis of the field $k$ here? How does this matrix relate to the components of the field elements $\alpha_i$? $\endgroup$ – Ethan Dlugie Sep 3 '20 at 1:46
  • $\begingroup$ Your last comment is exactly right: the $\sigma_i$ are acting as component functions, giving coordinates of the $\alpha_i$. You can see how this works in simple examples where you can compute everything explicitly, say $k = \mathbb{Q}(i)$. $\endgroup$ – Qiaochu Yuan Sep 3 '20 at 1:59
  • $\begingroup$ @QiaochuYuan but coordinates with respect to what basis? For instance in the example you suggest $k=\mathbb{Q}(\sqrt{-1})$, the two embeddings are $\sigma_{\pm}:\sqrt{-1} \mapsto \pm i$. Then $(\sigma_+,\sigma_-)(3+2\sqrt{-1})=(3+2i,3-2i)$. I'm not really understanding how these coordinates represent $3+2\sqrt{-1}$, or even why you would need both coordinates to specify the field element. $\endgroup$ – Ethan Dlugie Sep 3 '20 at 3:37
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A homogeneous system of $n$ linear equations wth $n$ unknowns has a nontrivial solution iff the determinant of the system is $0$. That is the statement used in the OP.

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  • $\begingroup$ But the equations are essentially the same. That is, $\sum x_j\alpha_j = 0$ will have a nontrivial solution in the $x_j$ if and only if $\sum x_j\sigma_i(\alpha_j) = 0$ has a nontrivial solution. Why would we need all $n$ of the equations? $\endgroup$ – Ethan Dlugie Sep 3 '20 at 1:48
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    $\begingroup$ The equations have unknowns $x_1,...,x_n$ and coefficients $\sigma_i(\alpha_j)$. The equations are different. Every equation restricts possible choices of $x_i$, it is good to have $n$ of them to pin-point the solution. Fortunately you have enough ($n$ ) equations. $\endgroup$ – Mark Sapir Sep 3 '20 at 1:52
  • $\begingroup$ I see that the equations are literally different, but I fail to see why they don't all give the same result. If we have a linear dependence like $2\alpha_1+3\alpha_2-4\alpha_3=0$ in $k$, then applying any field embedding $\sigma:k \hookrightarrow \mathbb{C}$ which is the identity on $\mathbb{Q}$ will give an identical linear independence $2\sigma(\alpha_1)+3\sigma(\alpha_2)-4\sigma(\alpha_3)=0$. Why do the $n$ equations give any new information that wasn't evident in the original equation? $\endgroup$ – Ethan Dlugie Sep 3 '20 at 3:55
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    $\begingroup$ Just consider some example. Say, $\alpha_1=1+\sqrt{2}, \alpha_2=2+3\sqrt{2}$ and the isomorphism $\sqrt{2}\to -\sqrt{2}$. Then the system is $$x_1(1+\sqrt{2})+x_2(2+3\sqrt{2})=0$$ $$x_1(1-sqrt{2})+x_2(2-3\sqrt{2}) =0$$ $\endgroup$ – Mark Sapir Sep 3 '20 at 4:00
  • $\begingroup$ Hm I think I see. If I just started with the first equation you give, it is possible to rearrange it and get a polynomial equation over $x_1$ and $x_2$ with rational coefficients. But it may be difficult to determine whether such an equation has nonzero rational solutions for $x_1$ and $x_2$. Otoh, the linear system is much easier to check for solutions. Is that a fair explanation? $\endgroup$ – Ethan Dlugie Sep 3 '20 at 4:13

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