Doubt about Shilov’s linear Algebra: What exactly is isomorphism

linear algebra

I was reading Shilov's Linear Algebra and he started defining isomorphism, but I am having problems understanding it because of the language he uses.
He says

"Two fields $K$ and $K'$ are said to be isomorphic if we can set up a
one-to-one correspondence between $K$ and $K'$ such that the number
associated with every sum (or product) of numbers in $K$ is the sum
(or product) of the corresponding numbers in $K'$. The number
associated with every difference (or quotient) of numbers in $K$ will
then be the difference (or quotient) of the corresponding numbers in
$K'$."

I understood almost nothing.

My questions:

(1)Is not a one-to-one correspondence a correspondence where I associate one number of $K$ to one number of $K'$ without repeating, such as in a function, meaning that there is exactly the same quantity of numbers in $K$ as in $K'$?

(2)What does he mean by numbers associated with every sum? Does he mean that the result of the sum of two given elements of $K$ must be equal to the sum of the two elements of $K'$ that are associated with them?

(3) So does the same thing happens with the difference (or quotient)?

(4) When he writes "sum (or product)", is he saying that the two numbers, or the number (it will depende on the answer to my second question), must satisfy both sum case and product case or only have to satisfy one of them?(I know that or means that is either one thing or the other, not both, but… You know, I prefer to be sure).

Best Answer

For (1), you have to associate a unique, different element of $K$ to each element of $K'$, and vice versa. More importantly, this has to be the same map below that respects the sum and products. You're not just saying that $K$ and $K'$ have the same cardinality, but that the particular map the author is talking about is one-to-one.

For (2), yes.

For (3), yes, because this map $f$ has $f(-x) + f(x) = f(x + (-x)) = f(0) = 0$. (For the last part, note that $f(0) = f(0 + 0) = f(0) + f(0)$.)

I'm not sure what you mean in (4), but the idea is that $f(xy) = f(x)f(y)$.

The text in the quote is unclear and unnecessarily verbose; this is an instance where symbols make things much clearer. What the author means is that an isomorphism of fields $K$ and $K'$ is a function $f:K \to K'$ with the following properties:

  • $f$ is bijective: For any $x'\in K'$, there exists a unique $x\in K$ with $f(x) = x'$.
  • $f(x + y) = f(x) + f(y$) for any $x, y\in K$;
  • $f(xy) = f(x) f(y)$ for any $x, y\in K$.

There are some other properties, like $f(1) = 1$ and $f(x^{-1}) = f(x)^{-1}$ for nonzero $x$, that follow immediately from these properties; but that's the definition itself.

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