I came across with the lemma below on Isaac's Algebra:

Lemma. Let $F \subseteq R$, where F is a field and R is a domain. If $|R:F|<\infty$ then R is a field.

As far as I know, $|E:F|$ denotes the degree of a field extension so here, $E$ is a field. In the book it is said that when "an integral domain $R$ contains a field $F$, this forces $R$ to be a unitary overring of $F$".

I would like to know how should I think $|R:F|$ as and what does it mean to be a unitary overring of some field?

  • $\begingroup$ This seems weird to me. Wouldn't $|M_{2\times 2}(\mathbb{R}):\mathbb{R}| <\infty$? $\endgroup$ – Stella Biderman Mar 31 '17 at 23:27
  • 2
    $\begingroup$ @StellaBiderman But it's not a domain... $\endgroup$ – Arnaud D. Mar 31 '17 at 23:28
  • $\begingroup$ @ArnaudD. Ah right, missed that word. $\endgroup$ – Stella Biderman Mar 31 '17 at 23:29
  • $\begingroup$ Presumably "domain" here implies commutative. $\endgroup$ – Dustan Levenstein Apr 1 '17 at 0:08
  • 1
    $\begingroup$ $S$ is an overring of $R$ when $R$ is a subring of $S$. The adjective unitary just means $R$ and $S$ have the same multiplicative identity. I'm not really sure how to answer your first question... It's just a vector space dimension. This kind of thing shows up repeatedly in Galois theory. $\endgroup$ – Dustan Levenstein Apr 1 '17 at 0:11

As already stated by B.A.: $[R:F]$ is the dimension of $R$ as a vector space over $F$. The fact that $R$ is a field if this dimension is finite follows from the dimension formula of linear algebra: multiplication with an element $r\in R\setminus 0$ yields an $F$-linear map $R\rightarrow R$, which is injective since $R$ is a domain. Hence it must be surjective, which implies that $r$ has an inverse.

| cite | improve this answer | |

You should think of $|R:F|$ as size of the basis needed so that the F-span of them is R (i.e. coefficients in F).

I think the comments answered some of your question so I will just answer the main part. (I assume that an integral domain is commutative in this context)

I will use some Galois theory (and so I hope you have been introduced to what I will be using).

Since $|R:F|< \infty$ (say $|R:F|=n$) you have for every $r\in R$ the set $\{1,r,r^2,\dots,r^n\}$ is linearly dependent and none of which are zero as we are in an integral domain. In particular every element in R is algebraic over F.

(note algebraic means there exists a polynomial such that r is a root)

So take the algebraic closure $\bar{F}$ of F (which is a field and it is the collection of all roots for all polynomials).

We get $ F \subseteq R \subseteq \bar{F}$ (note that being an integral domain means that the only thing that is not assured is the existence of inverses so the operation of R will not violate operations of $\bar{F}$. so we can have this embedding).

So the following claim should be enough

Let L be an algebraic extension of F. If $ F \subseteq R \subseteq L$ for some ring R. Then R is a field.

proof: All properties of being a field hold by virtue of being imbedded in L which is a field. The only nontrivial part is the existence of an inverse.

Let r be an element of R. r is algebraic so:

$\sum_{m=0}^{n}{a_mr^m}=0$ where $a_m \in F$ not all zero. Say j is the smallest integer such that $a_j \ne 0$

So we have $\sum_{m=j}^{n}{a_mr^m}=0$

$r\in L$ a field, so $r^{-1} \in L$. so by multiplying by $r^{-j-1}$ we get

$r^{-1}= -a_j^{-1}\sum_{m=0}^{n-j}{a_mr^m}$. However all powers of r are in R as R is a ring so we get the left hand side must be in R too.

| cite | improve this answer | |

Not the answer you're looking for? Browse other questions tagged or ask your own question.