In following all schemes $X$
will be considered as separated, of finite type over
an algebraically closed field $K$ of characteristic $0$.
Recall that a Hilbert function $HF_S: \mathbb{N}_0 \to \mathbb{N}_0$ associates to a
finitely generated graded commutative algebra $S= \bigoplus_{i \ge 0} S_i$
over a field $K$, which is finitely generated by elements of positive degree
the value
$$ HF_S(n) := \dim_k S_n $$
(the $k$-dimension of $n$-graded piece). One can show that there exist a unique polynomial $P_S(t) \in \mathbb{Q}[t]$ and a $n_0(S) \in \mathbb{N}$, which depends on $S$, such that for all $n \ge n_0(S)$ we have
$$ HF_S(n) = P_S(n) $$
This $P_S$ is unique and called the Hilbert polynomial of $S$.
Consider following proposition and proof from "3264 & All That" by Eisenbud & Harris (p 204):
Proposition 6.7. A subscheme $Y \subset \mathbb{P}^n$ is a linearly embedded $k$-plane (ie isomorphic to $\mathbb{P}^k$) if and only if the Hilbert polynomial of $Y$ is
$$ P(d) = \binom{d+k}{k} = \frac{(d+k)(d+k-1)... (d+1)}{k(k-1)...1} $$
Proof: Since the dimension of the $d$-th graded component of a polynomial ring on $k+1$ variables is $\binom{d+k}{k} $, the Hilbert polynomial of a linearly embedded $k$-plane is $P(d)$.
Conversely, suppose that $Y$ has Hilbert polynomial $P$. From the degree and leading coefficient of $P$ we see that $Y$ is a scheme of dimension $k$ and degree $1$. Thus $L :=Y_{red} \to Y$ is a linearly embedded $k$-plane. (why?) This inclusion induces a surjection of homogeneous coordinate rings $S_Y \to S_L$, and the equality of Hilbert polynomials shows that it is an isomorphism in high degrees. Since the inclusions $L \subset Y \subset \mathbb{P}^n$ can be recovered as $\operatorname{Proj}(S_L) \subset \operatorname{Proj}(S_Y) \subset \operatorname{Proj}(S) = \mathbb{P}^n$, where $S=K[x_0,..., x_n]$ is the homogeneous coordinate ring of $\mathbb{P}^n$, and since $\operatorname{Proj}(S_Y)$ depends only on the high degree part of $S_Y$ , this shows $L \subset Y$ is actually an equality.
Question: Why it's immediate clear that $Y_{red}$ is a linearly embedded $k$-plane? Or equivalently, that a reduced $k$-dimensional subscheme $X \subset \mathbb{P}^n$ of degree $1$ isomorphic to a $k$-plane $\mathbb{P}^k$? So the proof suggests that reducedness is the key point here making the latter statement "trivial". Any ideas which arguments the authors use here?
My ideas: We know that Hilbert functions and therefore Hilbert polynomials are additive with respect exact sequences. The homogeneous coordinate rings of $Y$ and $Y_{red} $ are related by following exact sequence of homogeneous $S_Y$-modules:
$$ 0 \to \mathcal{Nil}(S_Y) \to S_Y \to S_Y/(\mathcal{Nil}(S_Y)) = S_{Y_{red}} \to 0 $$
Therefore $P_{S_Y}= P_{Y_{red}}+P_{\mathcal{Nil}(S_Y)}$. Do we know here more about $P_{\mathcal{Nil}(S_Y)}$? Since $\dim(Y) = \dim(Y_{red})$ and the dimension of the scheme equals to the degree of the associated Hilbert polynomial to it's coordinate ring, $P_{Y}$ and $P_{Y_{red}}$ have same degrees.
It is known that the leading coefficient $a_k$ of the Hilbert polynomial $P_X$ of a $k$-dimensional subscheme $X \subset \mathbb{P}^n$ of degree $d = \operatorname{deg}(X)$ equals $a_k = \frac{d}{k!}$. The degree is a positive integer, so the lowest possible value which a leading coefficient of a Hilbert polynomial could have is $\frac{1}{k!}$ and $P_{Y}$ has this coefficent. Therefore $P_{Y_{red}} $ and $P_{\mathcal{Nil}(S_Y)}$ cannot simultaneosly have degree $k$. But $\dim(Y) = \dim(Y_{red})$, so $P_{Y_{red}} $ has degree $k$, and it's leading coefficient coincides with $\frac{1}{k!}$, the leading coefficient of $P_{S_Y}$. Furthermore, since the degree of $Y$ equals to $a_n \cdot n!$, where $a_n$ is the leading coefficient of $P_{S_Y}$, as the proof states we have that $ Y_{red}$ has the the same degree as $Y$, namely $1$.
Why this implies that $Y_{red}$ is is a linearly embedded $k$-plane? What we can easily deduce is that if $Y_{red}$ decomposes as $X_1 \cup X_2 \cup ... \sup X_m$ in irreducible components then there is only one $X_i$ with $\dim X_i= \dim Y_{red}$, again because of additivity of Hilbert polynomial $P_{Y_{red}}= P_{X_1} + P_{X_2} + ... + P_{X_m}$, fact that the leading coefficient of Hilbert polynomial is always positive and that $P_{Y_{red}}$ has the lowest possible leading coefficient $\frac{1}{k!}$ by considerations above.
Assume now wlog $X_i=X_1$. Since $Y_{red}$ reduced, $X_i$ is reduced and irreducible, therefore integral. Conjecture: $Y_{red} = X_1$.
So the original problem becomes: Why is an integral (reduced, irreducible) $k$-dimensional subscheme
$X \subset \mathbb{P}^n$ of degree $1$ isomorphic to a $k$-plane $\mathbb{P}^k$?