If a subring $B$ of a field $F$ is closed with respect to multiplicative inverses then $B$ is a field. ($B$ is then called a subfield of $F$)

I started out my proof like this:

Assume a subring $B$ of a field $F$ is closed with respect to multiplicative inverses. By definition, a subring is closed with respect to addition, negative (additive) and multiplication. Therefore $B$ contains all these properties.

To show that $B$ is a field then we must show that $B$ is a commutative ring with unity in which every nonzero element is invertible

$B$ is a ring since every subring is also a ring.

Since we are given that $B$ is closed with respect to multiplicative inverses, then it must contain a unity which shall be denoted $1$.

Also, since it is now a subring with unity and multiplicative inverse means that every element in $B$ is invertible or :

$$ xb=bx=1$$

where $x$ is the multiplicative inverse and $b$ is an element in $B$.

Now the only problem I have is how do I show that $B$ is commutative, the only thing I was thinking is to use the invertible part but i'm not sure.


1 Answer 1


By definition of a subring, we know that $B\subset F$. Moreover, by definition of a subring, the multiplication law in $B$ is the same as that of $F$, restricted to $B\times B$ (recall that multiplication is a map $\cdot :F\times F\rightarrow F$).

So since $a\cdot b=b\cdot a$ for all $(a,b)\in F\times F$ and $B\times B\subset F\times F$, we see that $a\cdot b=b\cdot a$ for all $(a,b)\in B\times B$, and so our subring is commutative.

Note that none of this depends on whether $B$ is closed under inverses or not. In general a subring of a commutative ring is also commutative, with the same proof as above.

This is the same way that $\cdot$ is shown to be associative. More generally, whatever functional laws hold in a ring also hold in a subring; for example, commutativity and associativity as above, but also the fact that $1$ is a multiplicative identity, that $0$ is an additive identity, and that $\cdot$ distributes over $+$.

  • $\begingroup$ Got it. thanks!!!!!! $\endgroup$
    – Emmie
    May 3, 2015 at 2:07

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