Fulton 1.10

Proposition 4. If a field $L$ is a ring finite over a subfield $K,$ then $L$ is module finite, and hence algebraic over $K.$

We provide a detailed proof below to clarify Fulton’s proof.

Suppose a field $L$ is ring finite over a subfield $K,$ that is, $L=K[v_1,\ldots,v_n]$ for $v_1,\ldots,v_n\in L.$ We proceed by induction on $n,$ the ring dimension of $L$ over $K.$

Suppose $n=1,$ that is $L=K[v].$ Let us define $\varphi:K[X]\to K[v]$ by $X\mapsto v.$ Since $K[X]$ is a PID, $\ker\varphi=(F)$ for some $F\in K[X].$ Since $K[X]/(F)=L$ is a field, $(F)$ is maximal. if $F=0,$ $K[X]\cong K[v]=L$ and therefore $K(X)\cong K[X],$ a contradiction by Problem 1.44. Thus $F\ne 0$ and $F(v)=0.$ So $v$ is algebraic over $K$ and $L=K[v]$ is module finite over $K$ by Proposition 3.

Suppose the proposition holds for all $m$ such that $1\le m <n.$ Observe that $L=K[v_1,\ldots,v_{n+1}]= K(v_{n+1})[v_1,\ldots,v_n].$ Thus by the inductive hypothesis, $L$ is module finite over $K(v_{n+1}).$ It suffices to show $v_{n+1}$ is algebraic over $K.$

Observe that for each $i\in\{1,\ldots,n\}$ there exists $F_i\in K(v_{n+1})[X],$ with $F_i(X)=X^{m_i}+a_{i,m_i-1}X^{m_i-1}+\ldots+a_{i,0},$ such that $F_i(v_i)=0.$ Observe that there exists $a\in K[v_{n+1}]$ such that $aa_{i,j}\in K[v_{n+1}]$ for all $i,j.$ For each $i$ note that $0=a^{m_i}F_i(v_i) =(av_i)^{m_i}+aa_{i,m_i-1}(av_i)^{m_i-1}+\ldots+a^{m_i}a_{i,0}=G(av_i).$ By our selection of $a,$ observe that $G\in K[v_{n+1}][X].$ So $av_i$ is integral over $K[v_{n+1}]$ for each $i.$ Therefore $K[av_1,\ldots,av_n,v_{n+1}]$ is integral over $K[v_{n+1}].$ Moreover, for each $z\in L=K[v_1,\ldots,v_{n+1}]$ there exists $m>0$ such that $a^mz\in K[av_1,\ldots,av_n,v_{n+1}].$ In particular this is true for $K(v_{n+1})\subset L.$

Suppose $v_{n+1}$ is transcendental over $K.$ Then $K(v_{n+1})\cong K(X).$ But the existence of $a$ is a contradiction to Problem 1.49(b). Thus $v_{n+1}$ is algebraic over $K$ and $L$ is module finite over $K.$

1.51.* Let $K$ be a field, $F\in K[X]$ an irreducible polynomial of degree $n>0.$

(a) Show that $L=K[X]/(F)$ is a field, and if $x$ is residue of $X$ in $L,$ then $F(x)=0.$

Since $K[X]$ is a PID, $(F)$ is maximal and $K[X]/(F)$ is a field. Let $\varphi$ be the homomorphism $K[X]\to K[X]/(F).$ Observe that $F(x)=F(\varphi(X))=\varphi(F(X))=0.$

(b) Suppose $L&39;$ is a field extension of $K,$ $y\in L&39;$ such that $F(y)=0.$ Show the homomorphism $\varphi:K[X]\to L&39;$ defined by $X\mapsto y$ induces an isomorphism $L\to K(y).$

Since $F(y)=0,$ $(F)\subset\ker\varphi.$ Thus by Lemma 1.1 there exists an induced homomorphism $\widetilde{\varphi}:L\to K(y).$ Now suppose $\ker\varphi\ne (F).$ Then $\ker\varphi=K[X]$ since $(F)$ is maximal, a contradiction. Thus $\ker\varphi=(F)$ and thus $\widetilde{\varphi}$ is an injective homomorphism. Moroever, $\widetilde{\varphi}(L)=\varphi(K[X])$ is a subfield of $K(y)$ containing $K$ and $y.$ So $\widetilde{\varphi}(L)=K(y)$ and $\widetilde{\varphi}$ is an isomorphism.

Observe that $G\in\ker\varphi=(F).$ So $F\mid G.$

(d) Show that $F=(X-x)F_1, F_1\in L[X].$

Observe that $X-x$ is irreducible in $L[X]$ and $F(x)=0.$ By (c), $X-x\mid F.$

1.52. Let $K$ be a field and $F\in K[X].$ Show there is a field $L$ containing $K$ such that $F=\prod_{i=1}^n(X-x_i)\in L[X].$

Assume $F$ is monic. Let us proceed by induction on $n,$ the degree of $F.$

If $n=1$ then $F=X-x,$ $x\in K.$

Suppose the statement is true for all monic polynomials in $K[X]$ of degree less than $n.$ If $F$ is reducible, the statement follows by the inductive hyptothesis. Suppose $F$ is irreducible. Then $L=K[X]/(F)$ is a field. Let $x$ be the image of $X$ in $L.$ Then $F(x)=0$ in $L,$ so $F=G(X-x)$ for some $G\in L[X]$ of degree less than $n.$ Thus by the inductive hypothesis, there exists an extension field $M$ of $L$ such that the statemenet holds. But since $L$ is an extension field of $K,$ $M$ is also an extension field of $K.$

1.53.* Suppose $K$ is a field of characteristic zero, and $F$ is an irreducible polynomial in $K[X]$ of degree $n>0.$ Let $L$ be a splitting field of $F,$ so $F=\prod_{i=1}^n(X-x_i),$ $x_i\in L.$ Show the $x_i$ are distinct.

Suppose some of the $x_i$ ’s are equal, that is, there exists $x\in L$ such that $(X-x)^m\mid F$ for some $m>1.$ Then $F=(X-x)^mG$ for some $G\in L[X].$ Observe $F_X=m(X-x)^{m-1}G+(X-x)^mG_X.$ Since $K$ is of characteristic zero, $F_X\ne 0.$ Observe $F_X(X)=0.$ By Problem 1.51(c), this implies $F\mid F_X,$ a contradiction since $0<\deg F_X<\deg F.$

1.54.* Let $R$ be a domain with quotient field $K,$ and let $L$ be a finite algebraic extension of $K.$

(a) For any $v\in L$ show there is a nonzero $a\in R$ such that $av$ is integral over $R.$

Let $v\in L.$ Then $v^n+a_{n-1}v^{n-1}+\ldots+a_0=0,$ for $a_i\in K.$ Let $a$ be the common denominator of $a_0,\ldots,a_{n-1}.$ Then $(av)^n+aa_{n-1}(av)^{n-1}+ \ldots+a^na_0=0.$ Thus $av$ is integral over $R.$

(b) Show that there is a basis $v_1,\ldots,v_n$ or $L$ over $K$ such that each $v_i$ is integral over $R.$

Let $w_1,\ldots,w_n$ be a basis for $L$ over $K.$ By (a) there exist $a_1,\ldots, a_n$ such that $a_1w_1,\ldots,a_nw_n$ are all integral over $R.$ Let $v\in L.$ Then $v=\sum_{i=1}^nb_iw_i,$ for $b_i\in K.$ So $v=\sum_{i=1}^nb_ia_i^{-1}(a_iw_i).$ So $a_1w_1,\ldots,a_nw_n$ spans $L$ and hence is a basis for $L$ over $K.$