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# a condition of algebraic extension

Theorem. A field extension $L/K$ is algebraic if and only if any subring of the extension field $L$ containing the base field $K$ is a field.

Proof. Assume first that $L/K$ is algebraic. Let $R$ be a subring of $L$ containing $K$. For any non-zero element $r$ of $R$, naturally $K[r]\subseteq R$, and since $r$ is an algebraic element over $K$, the ring $K[r]$ coincides with the field $K(r)$. Therefore we have $r^{{-1}}\in K[r]\subseteq R$, and $R$ must be a field.

Assume then that each subring of $L$ which contains $K$ is a field. Let $a$ be any non-zero element of $L$. Accordingly, the subring $K[a]$ of $L$ contains $K$ and is a field. So we have $a^{{-1}}\in K[a]$. This means that there is a polynomial $f(x)$ in the polynomial ring $K[x]$ such that $a^{{-1}}=f(a)$. Because $af(a)-1=0$, the element $a$ is a zero of the polynomial $xf(x)-1$ of $K[x]$, i.e. is algebraic over $K$. Thus every element of $L$ is algebraic over $K$.

# References

- 1 David M. Burton: A first course in rings and ideals. Addison-Wesley Publishing Company. Reading, Menlo Park, London, Don Mills (1970).

## Mathematics Subject Classification

12F05*no label found*

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