I am reading Shafarevich's Basic Algebraic Geometry 1(Third Edition). I met some trouble when I read the proof of Theorem1.14 in page 62. The theorem is

Theorem 1.14 If $f:X\to Y$ is a regular map and $f(X)$ is dense in $Y$, then $f(X)$ contains an open set of $Y$.

The proof assumes $X$ is irreducible and affine, $r$ is the transcendence degreee of the field extension $k(X)/k(Y)$. Choose $r$ elements $u_1, \dots, u_r\in k[X]$ that are algebraically independent over $K(Y)$. Then $$ K[X]\supset k[Y][u_1,\dots,u_r]\supset k[Y] \quad\text{and}\quad k[Y][u_1.\dots,u_r]=k[Y\times\mathbb{A}^r]. $$

In the seventh line of page 63, it says:

Any element $v\in k[X]$ is algebraic over $k[Y\times \mathbb{A}^r]$, hence there exists an element $a\in k[Y\times \mathbb{A}^r]$ such that $av$ is integral over $k[Y\times \mathbb{A}^r]$.

I don't know how to prove this statement.

  • 2
    $\begingroup$ If an element $x$ is algebraic, then there exists some elements $a_0,\dots,a_n$ such that $a_nx^n+\cdots+a_0=0$. By multiplying this equation with $a_n^{n-1}$ notice that $a_nx$ is integral. $\endgroup$
    – user26857
    Commented May 27, 2018 at 20:26
  • $\begingroup$ @user26857 nice answer, thank you! $\endgroup$
    – yahoo
    Commented May 27, 2018 at 23:19

1 Answer 1


That is a general fact, we are going to prove it:

Let be $R$ a domain, $K=q.f(R)$ and let be $L$ a finite and algebraic extension over $K$. Then:

i) $\forall v\in L$ $\exists$ $a\in R-\{0\}:$ $av$ is integral over $R$.

ii) Exists a basis of $L$ over $K$ such that their elements are integral over $R$.

Proof of i)

Let be $v\in L$, we have $L/K$ is algebraic then exists $a_n,\ldots,a_1,a_0\in K$ such that $$a_nv^{n}+a_{n-1}v^{n-1}+\cdots+a_1v+a_0=0$$ We have $K=q.f(R)$ so $$\forall\hspace{0.05cm}i\in\{0,1\ldots,n\}\hspace{0.05cm} \exists\hspace{0.05cm}b_i\in R,c_i\in R-\{0\}: a_i=\frac{b_i}{c_i}$$ Let be $\gamma:=\prod_{i=0}^{n}c_i$, $\gamma_j=\prod_{0\leq i\leq n,i\neq j}c_i$ and $r_j:=\gamma_{j}b_{j}\in R$, then $$a_nv^{n}+a_{n-1}v^{n-1}+\cdots+a_1v+a_0=0 \Rightarrow \gamma_{n}b_nv^{n}+\gamma_{n-1}b_{n-1}v^{n-1}+\cdots+\gamma_{1}b_1v+\gamma_{0}b_0=0\Rightarrow$$ $$r_nv^{n}+r_{n-1}v^{n-1}+\cdots+r_1v+r_0=0$$ Now we multiply by $r_n^{n-1}$ the last expression : $$0=r_n^{n-1}r_nv^{n}+r_n^{n-1}r_{n-1}v^{n-1}+\cdots+r_n^{n-1}r_1v+r_n^{n-1}r_0=$$ $$(r_nv)^n + r_{n-1}(r_nv)^{n-1}+\cdots+r_1r_n^{n-2}(r_nv)+r_{n}^{n-1}r_0$$ Thus the polynomial $P(x) = x^n + r_{n-1}x^{n-1}+\cdots+r_1r_n^{n-2}x+r_{n}^{n-1}r_0$ has a zero in $r_nv$, it is monic and belong to $R[X]$, so $r_nv$ is integral over $R$

Proof of ii):

We have $L/K$ is finite then $$m:=[L:K]=dim_{K}(L)<\infty$$ Thus exists $\{v_1,\ldots,v_m\}$ basis of $L$. If we use i) we have $\forall\hspace{0.05cm}i\in\{1\ldots,m\}\hspace{0.05cm} \exists\hspace{0.05cm}a_i\in R-\{0\}$ such that $a_iv_i$ is integral over $R$. We have to show $\{a_1v_1,\ldots,a_mv_m\}$ is lineary independent over $K$:

Let be $\lambda_1,\ldots,\lambda_n\in K$ such that $$\sum_{i=1}^{m}\lambda_i(a_iv_i)=0$$ Now we use $\{v_1,\ldots,v_m\}$ is lineary independent: $$0=\sum_{i=1}^{m}\lambda_i(a_iv_i) = \sum_{i=1}^{m}(\lambda_ia_i)v_i \Rightarrow 0=\lambda_ia_i\hspace{0.15cm}\forall\hspace{0.05cm}i\in\{1\ldots,m\}$$ We have $a_i\neq 0$ for all $i\in\{1\ldots,m\}$ then $\lambda_i=0$ for all $i\in\{1\ldots,m\}$. Thus $\{a_1v_1,\ldots,a_mv_m\}\subset L$ is lineary indepent over $K$ and the set $\{a_1v_1,\ldots,a_mv_m\}$ is a basis because $m=dim_K(L))$.


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