Krdytkyu

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 $0le R<1$, such that $k$ is $C_R$.

Recall the definition of $C_R$ property of a field $k$: $Rgeq 0$, for $forall$ $d,n in mathbb Z^{+}$, with $d^{R}<n$, $forall fin k[t_1,cdots,t_n]$ homogeneous, $deg f=d$ , $exists$ $0neq x in k^n$ 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 $xin k^{n}$ 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:

1. 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?




2. 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!