A question about fields containing a copy of $\Bbb{Q}$

Question

When we say a field contains a copy of the field of rational numbers $\Bbb{Q}$, what does this really mean? Does it mean it contains a field isomorphic to $\Bbb{Q}$, or does it mean it contains $\Bbb{Q}$ itself?

Also, say the field contains $\Bbb{Q}$. Consider the element $a+a+a=3a$. Here, does $3$ belong to $\Bbb{Q}$? Or is $3a$ just a way of representing $a+a+a$?

Thanks

Answer 1

It contains a field isomorphic to $\mathbb{Q}$ (or there is an isomorphism from $\mathbb{Q}$ into that field). For example, $\mathbb{R}$ contains a copy of $\mathbb{Q}$. This usually comes up when you are constructing another (larger) field, like \mathbb{R}, from $\mathbb{Q}$ and then "embedding" $\mathbb{Q}$ in it. Similarly the field $\mathbb{Q}$ itself is constructed from the ring $\mathbb{Z}$, and then it is shown that $\mathbb{Q}$ contains a copy of $\mathbb{Z}$.

In fact, there is a very nice chain from $\mathbb{Z}$ to $\mathbb{C}$:
$\mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R} \subset \mathbb{C}$.
What is nice about this chain is that each of the last three structures in the sequence is constructed from the preceding one by way of adding certain properties. $Z$ is a ring and therefore does not have multiplicative inverses. So we build the field $\mathbb{Q}$ and embed $\mathbb{Z}$ in it. But $\mathbb{Q}$ is not "complete", in the sense that it does not have the least upper bound property. Construct a complete field from it and you get $\mathbb{R}$. But $\mathbb{R}$ is not algebraically closed! Construct an algebraically closed field containing $\mathbb{R}$ and you get $\mathbb{C}$.

"Does it mean it contains $\mathbb{Q}$ itself?" - The only meaningful way to interpret that is to say, as above, that it contains a field isomorphic to $\mathbb{Q}$. It doesn't matter what labels you use to represent the elements.

When you write $a + a + a$ as $3a$, $3$ does not represent the element corresponding to $3$ in the field's copy of $\mathbb{Q}$. It is just a notation. The same notation, for example, can be used to represent $a + a + a$ in a (commutative) group in additive notation (this is exactly the same as $a^3$ in multiplicative notation).

Edit: As Stahl explains in the comment, in a field of characteristic $0$, you can identify $3$ with the rational number $3$.

Answer 2

Every AFAIK, every field of characteristic zero contains a copy of $\mathbb Q$ in this sense: once you have $1$, you have $1+1+...$. Once you have $1+1$ , you have $1/2:=(1+1) ^{-1}$. Then , e.g., $3/2=1/2+1/2+1/2$ From this you see that every element of the form $a/b ; b\neq 0$ can be obtained in this way.

Answer 3

Does it mean it contains a field isomorphic to $\Bbb{Q}$, or does it mean it contains $\Bbb{Q}$ itself?

My answer to this would be "yes and yes."

The reason is that the whole idea of isomorphism means that we don't distinguish between isomorphic things. They have exactly the same properties, so they are the same. You can say they are different sets, but really it would be better to stop thinking of $\Bbb Q$ as some magical set that is different from its isomorphic counterparts, somehow.

For any ring with identity, there is always a homomorphism of $\Bbb Z$ Into the ring mapping 1 to the identity. If the characteristic is zero, then this map is an injection. If the ring your mapping into is a field with characteristic 0, this extends to a homomorphism from $\Bbb Q$ into your ring. All of the resulting images of elements of $\Bbb Q$ in your ring are exactly what you expect.