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We can say, that any field $\mathbb{K}$ -- $1$-dim vector space on itself: $\mathbb{K}_{\mathbb{K}}$. So any vector of one another finite-dimensional vector space $V_{\mathbb{K}}$, after choosing the some basis can be represented as the element of isomorphic space $\mathbb{K}_{\mathbb{K}}^{n} = \prod_{i=1}^n\mathbb{K}_i$, where $n$ -- dimension of $V_{\mathbb{K}}$. But we can determine operations on elements of $\mathbb{K}_{\mathbb{K}}^{n}$ like on the direct group product: $(x_1,\dots,x_n) + (x'_1, \dots, x'_n) = (x_1 + x'_1, \dots ,x_n + x'_n)$ and similar for second field operation: $(x_1,\dots,x_n) \times (x'_1, \dots, x'_n) = (x_1 \times x'_1, \dots ,x_n \times x'_n)$.

But usually we doing it only with one field operation $+$. Why?

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    $\begingroup$ We can consider the multiplication on $\Bbb K^n$, and that makes it a ring (with null divisors if $n\ge 2$). In ring theory these objects are studied. $\endgroup$
    – Berci
    Commented Apr 24, 2019 at 11:37
  • $\begingroup$ The formal answer is "you can define it if you want". The informal answer is "that's just useless, because it doesn't have any sensible invariances, etc." (so for instance it doesn't mean anything geometrically) $\endgroup$ Commented Apr 24, 2019 at 11:39

3 Answers 3

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On the vector space, there is an addition that does not depend on the choice of basis you make, sometimes called a "natural" addition.
However, you usually can't define a natural multiplication in your way, it would always depend on a basis and different bases will give different multiplications. Therefore, it is mostly not used when dealing with vector spaces.

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Because if $\mathbb K$ is a field, $\mathbb K^n$ remains a $\mathbb K$-vector space but may not be a field (with the same addition). For example, $\mathbb R$ is a field, but $\mathbb R^n$ can't have a field structure (with the classical addition) for $n=3$ and $n\geq 5$. Even when $n=4$ the field is not commutative. See for example the Wiki page on division $\mathbb R$-algebras

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  • $\begingroup$ Of course, note that for $n=2$, you do get a commutative field --- $\mathbb{C}$! $\endgroup$
    – Ehsaan
    Commented Apr 24, 2019 at 13:05
  • $\begingroup$ That said, OP's question was specifically about coordinate-wise multiplication. $\endgroup$
    – Ehsaan
    Commented Apr 24, 2019 at 13:06
  • $\begingroup$ Did I say anything about the case $n=2$ ? :) $\endgroup$
    – elidiot
    Commented Apr 24, 2019 at 13:28
  • $\begingroup$ No! Which is why I added it ;) $\endgroup$
    – Ehsaan
    Commented Apr 24, 2019 at 16:39
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The category of fields has no product. If it had, then the projection on one factor would be injective, which is clearly false!

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  • $\begingroup$ Why is it false? $\endgroup$
    – Just do it
    Commented Apr 24, 2019 at 13:24
  • $\begingroup$ When $A$ and $B$ are sets, the projection on $A$ is not injective if $A$ has at least one element and $B$ has at least two elements. Because $(a, b)$ and $(a, c)$ have the same projection but are different elements. The reason why the projection would be injective is that any morphism of fields is injective, in case you wondered. $\endgroup$
    – elidiot
    Commented Apr 24, 2019 at 13:30
  • $\begingroup$ Of course, the problem is exactly this! Morphisms of fields are necessarily injective and in a field 0 is different from 1. $\endgroup$ Commented Apr 24, 2019 at 20:13

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