Caracterization of algebraic extensions

Question

An extension $L:K$ is algebraic If and only If every sub $K$-algebra of $L$ is a field.

Does anynone have the proof for this?

For proving it is an algebraic field i used that i can build a polynom $p(x) = -n + \sum_{i=1}^n a^{x_i} * (a^{x_i})^-1$ where the ring Will be $K$ united with $<a>$ therefore im assuming that at least one of these $a^{-1}$ is on $K$, but i don't know If i can hold that assumption

Answer 1

Suppose $K$ is not algebraic and take $b\in K$ transcendental over $L$; then $L[b]$ is not a field because…

Suppose $K$ is algebraic and take $b\in K$, $b\ne0$. Then $b$ satisfies a monic polynomial with coefficients in $L$, with nonzero constant term. Then…

Answer 2

Suppose $L:K$ is algebraic.

Let $R$ be a $K$-subalgebra of $L$.

We want to show $R$ is a field.

Since $R$ is a subring of $L$, $R$ is an integral domain.

To show $R$ is a field, it suffices to show that every nonzero element of $R$ is a unit in $R$.

Let $r\in R$, with $r\ne 0$.

Since $r\in L$, $r$ is algebraic over $K$.

Let $f\in K[x]$ be a monic polynomial of least degree, $n$ say, such that $f(r)=0$.

Write $f(x) = xg(x)+ a_0$, where $g\in K[x]$ is a monic polynomial with $\deg(g)=n-1$, and $a_0\in K$.

Suppose $a_0=0$.

Then from $f(r)=0$, we get $rg(r)=0$, hence, since $r\ne 0$, and $R$ has no zero divisors, it follows that $g(r)=0$, contradiction, since the degree of $g$ is less than the degree of $f$.

Therefore $a_0\ne 0$.

Then from $f(r)=0$, we get $$r\left(-\frac{g(r)}{a_0}\right)=1$$ hence, since $-{\large{\frac{g(r)}{a_0}}}$ is an element of $R$, it follows that $r$ is a unit of $R$.

Therefore, $R$ is a field.

For the converse, suppose $L$ is a field extension of the field $K$ such that every $K$-subalgebra of $L$ is a field.

Let $u\in L$.

Our goal is to show that $u$ is algebraic over $K$.

If $u\in K$, then $u$ satisfies the equation $x-u=0$, so $u$ is algebraic over $K$.

Next, assume $u\notin K$.

In particular, $u\ne 0$.

Let $R=K[u]$.

By hypothesis, $R$ is a field, hence, since $u\ne 0$, $u$ is a unit of $R$.

Then $uv=1$, for some $v\in R$.

Since $R=K[u]$, we can write $v=g(u)$, for some $g\in K[x]$.

Then, letting $f=xg(x)-1$, we have $f\in K[x]$, $f\ne 0$, and $$f(u)=ug(u)-1=uv-1=0$$ so $u$ is algebraic over $K$, as required.

Therefore $L$ is algebraic over $K$.