I'm reading Dr. Pete Clark's paper Around Chevalley-Warning Theorem and I have a question about the proof of the following proposition on page 14:
Proposition: For a field $k$, the following are equivalent: (i) $k$ is algebraically closed.
(ii)$k$ is $C_0.$
(iii) There is some $0\le R<1$, such that $k$ is $C_R$.
Recall the definition of $C_R$ property of a field $k$: $R\geq 0$, for $\forall$ $d,n \in \mathbb Z^{+}$, with $d^{R}<n$, $\forall f\in k[t_1,\cdots,t_n]$ homogeneous, $\deg f=d$ , $\exists$ $0\neq x \in k^n$ such that $f(x)=0$.
Specifically, I don't quite understand the part of the proof from $(iii)\implies (i)$. The proof goes as following: we prove it by contrapositive, so we assume first k is not algebraically closed. Let $l/k$ be a field extension of degree d. Choose a basis $\alpha_1, \cdots, \alpha_d$ of $l/k$, and use this to identify $l$ with $k^{d}$, and view $\alpha \in l$ as giving a k-linear endomorphism of $k^{d}$ by multiplication. For indeterminates $t_1,\cdots,t_d$, let $p(t_1,\cdots,t_d)=det(\alpha_1 t_1+\cdots+ \alpha_d t_d)$. Thus P is a homogeneous polynomial of degree d with coefficients in k. Moreover, every nonzero $x\in k^{n}$ corresponds to a nonzero element of $l$, and since $l$ is a field, multiplication by $x$ has nonzero determinant. It follows that $P(x_1,\cdots,x_d)=0$ if and only if $x=0$. So whenever we have a degree d field extension , we have a degree d homogeneous polynomial in d variables which has only trivial solution, which means $k$ is not $C_R,_d$ for any $R<1$.
I'm not very familiar the concept of field extension but I have seen a few examples. I have the following questions regarding the proof:
I don't think I fully understand the meaning of the sentence: "view $\alpha \in l$ as giving a $k$-linear endomorphism of $k^{d}$ by multiplication." Does it mean we could multiply $\alpha$ with any element in $k^{d}$ and get a new element in $k^d$? Is it true that this endomorphism is an automorphism? Could anyone elaborate what this means?
How do you evaluate the determinant appeared in this equation: $p(t_1,\cdots,t_d)=det(\alpha_1 t_1+\cdots+ \alpha_d t_d)$? Isn't $\alpha_1 t_1+\cdots+ \alpha_d t_d$ a polynomial? How to calculate the determinant of a polynomial? Why $P$ is a homogeneous polynomial of degree $d$?
Thank you in advance for any feedback and advice!