integral domain with a field as a subring

Asked Dec 18 '13 at 19:00

Active Dec 18 '13 at 19:00

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I would like to know if my solution to the following exercise is correct.

Let $A$ be an integral domain (with a unit) which has a field $\mathbb K$ as a subring and such that $A$ is a finite-dimensional vector space over $\mathbb K$. Show that $A$ is a field.

Here is what I do : let $n=dim_{\mathbb K}A$. Let $x$ be in $A$. Then $\{1,x,x^2,...,x^n\}$ is a linearly dependent family in A as a vector space. Therefore there exist $\beta_0,\beta_1,...,\beta_n \in \mathbb K$ such that $$\sum_{i=0}^{n}\beta_ix_i=0$$ Therefore $$\beta_1x+\beta_2x^2+...+\beta_{n-1}x^{n-1}+\beta_nx^n=-\beta_0$$

So $$x(\beta_1+\beta_2x+...+\beta_nx^{n-1})=-\beta_0$$

Since $\beta_0 \in \mathbb K$ it has an inverse so $$x \times (\beta_1+\beta_2x+...+\beta_nx^{n-1})(-\beta_0^{-1})=1$$

Therefore $x$ has an inverse, therefore $A$ is a field. Is there anything to add or to justify ? Where does the integrity of the ring come in ? Thanks in advance for your time.

asked Dec 18 '13 at 19:00

Asinus

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How do you know $\beta_0 \neq 0$? – bzc Dec 18 '13 at 19:03
You just made me understand why integrity is a necessary hypothesis. $x$ must be different from $0$. Then it works : If $\beta_0=0$, by integrity, $x \times something=0$. Since $x$ is not equal to $0$, that something must be $=0$. Then we use the same argument with $\beta_1$ – Asinus Dec 18 '13 at 19:08
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Here is a two-line proof. – Georges Elencwajg Dec 18 '13 at 19:12
Haha the guy has just the same proof ! But the 2-lines one is cooler. – Asinus Dec 18 '13 at 19:19

integral domain with a field as a subring

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