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I need to prove that $K^{\mathbb N}$ is a K-vector space, where $K$ is a field, and each element of $K^{\mathbb N}$ is $(a_i)_{i \in \Bbb N}$, a sequence of elements of $K$,

The operation is $$\oplus :(a_i)_{i \in \Bbb N}\oplus(b_i)_{i \in \Bbb N}=(a_i+b_i)_{i \in \Bbb N}$$ and the action is $$\otimes: k \otimes(a_i)_{i \in \Bbb N}=(k \cdot a_i)_{i \in \Bbb N}$$

where $\cdot$ is the multiplication of $K$ and $+$ is the usual sum.

I have proven that $(V,\oplus)$ is an abelian group basically from the fact that $K$ is a field:

Since $K$ is a field, we have that $a_i$ has inverse additive, that there is a neutral element for addition, $e=0$, that $a_i+b_i=b_i+a_i$ and that $a_i \cdot b_i = b_i \cdot a_i$ for any index $i$.

Thus it follows that

$(1)$ $$\eqalign{ & {({a_i})_{i \in {\Bbb N}}} \oplus {(0)_{i \in {\Bbb N}}} = {({a_i} + 0)_{i \in {\Bbb N}}} = {({a_i})_{i \in {\Bbb N}}} \cr & {(0)_{i \in {\Bbb N}}} \oplus {({a_i})_{i \in {\Bbb N}}} = {(0 + {a_i})_{i \in {\Bbb N}}} = {({a_i})_{i \in {\Bbb N}}} \cr} $$

so $E= {(0)_{i \in {\Bbb N}}}$ is the neutral element.

$(2)$ $${({a_i})_{i \in {\Bbb N}}} \oplus {({b_i})_{i \in {\Bbb N}}} = {({a_i} + {b_i})_{i \in {\Bbb N}}} = {({b_i} + {a_i})_{i \in {\Bbb N}}} = {({b_i})_{i \in {\Bbb N}}} \oplus {({a_i})_{i \in {\Bbb N}}}$$

So the operation is commutative

$(3)$ $${\left( {{a_i}} \right)_{i \in {\Bbb N}}} \oplus {\left( { - {a_i}} \right)_{i \in {\Bbb N}}} = {\left( {{a_i} + {{\left( { - a_i} \right)}}} \right)_{i \in {\Bbb N}}} = {\left( 0 \right)_{i \in {\Bbb N}}}$$ so there is an inverse for any $\bf a$.

$(4)$ $$\left( {{{({a_i})}_{i \in {\Bbb N}}} \oplus {{({b_i})}_{i \in {\Bbb N}}}} \right) \oplus {({c_i})_{i \in {\Bbb N}}} = {({a_i})_{i \in {\Bbb N}}} \oplus \left( {{{({b_i})}_{i \in {\Bbb N}}} \oplus {{({c_i})}_{i \in {\Bbb N}}}} \right)$$ follows analogously.

The other conditions are not hard to prove

$$\tag {1} 1\otimes (a_i)_{i \in \Bbb N}=(a_i)_{i \in \Bbb N} $$ $$\tag {2} a \otimes \left((a_i)_{i \in \Bbb N}+(b_i)_{i \in \Bbb N}\right)=a \otimes(a_i)_{i \in \Bbb N}+a \otimes(b_i)_{i \in \Bbb N}$$ $$\tag {3} (a+b) \otimes(a_i)_{i \in \Bbb N}=a \otimes (a_i)_{i \in \Bbb N}+b \otimes (a_i)_{i \in \Bbb N}$$

I'm only missing associativity:

$$(a \cdot b)\otimes (a_i)_{i \in \Bbb N}=a \otimes (b \otimes (a_i)_{i \in \Bbb N})$$

Would this be legitimate?

$$(a \cdot b) \otimes {({a_i})_{i \in {\Bbb N}}} = {((a \cdot b) \cdot {a_i})_{i \in {\Bbb N}}} = {(a \cdot (b \cdot {a_i}))_{i \in {\Bbb N}}} = a \otimes {(b \cdot {a_i})_{i \in {\Bbb N}}} = a \otimes \left( {b\otimes {{({a_i})}_{i \in {\Bbb N}}}} \right)$$

and the inner manipulations follow from the fact that $K$ is a field.

I guess the others are allright, but let me know.

By the way, is there a faster way to prove the above? (Like the 3 propositions for a subspace to be one?)

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The argument for associativity is fine. Why wouldn’t it be legitimate? It’s pretty much the same sort of thing that you do to verify $(2)$ and $(3)$. –  Brian M. Scott Jun 25 '12 at 23:25
@BrianM.Scott It just happened to confuse me. –  Pedro Tamaroff Jun 25 '12 at 23:32
Your proof of $(a \cdot b)\otimes (a_i)_{i \in \Bbb N}=a \otimes (b \otimes (a_i)_{i \in \Bbb N})$ looks fine to me. The only point I wanted to mention is that this property, strictly speaking, is not associativity since the three elements $a$, $b$ and $(a_i)_{i \in \mathbb{N}}$ are not taken from the same set and the two operations $\cdot$ and $\otimes$ are different. –  Sayantan Jun 25 '12 at 23:33
I think your argument is fine. Why don't you get a bit more practice by writing out similar arguments for statements 1, 2, and 3? –  dfeuer Jun 25 '12 at 23:33
It would also be highly beneficial to practice rewriting this entire question to be easier to understand. Separate it into clear sections, define any necessary terms, etc. Also, you mention that $a_i\cdot b_i=b_i\cdot a_i$, but I don't see how this is relevant to anything. –  dfeuer Jun 25 '12 at 23:37

1 Answer 1

up vote 5 down vote accepted

Why wouldn't it be legitimate? Everything you defined works JUST fine. Absolutely no problem.

The only way you could prove those things faster is if you assumed some properties of direct products of abelian groups / vector spaces, since $K^{\mathbb N}$ is the same thing as the direct product of $|\mathbb N|$ $K$-vector spaces (note that the direct product and direct sum is different in this case). But proving this statement essentially comes up to moving around those trivial identities you did above. So the answer is no, I guess.

You defined things using the notation $(a_i)_{i \in I}$, and if you do exactly the same proofs you can show that $K^I$ is a $K$-vector space, i.e. the vector space of functions from $I$ to $K$. Also note that what you did is quite natural if you look at $K^{\mathbb N}$ as the vector space of sequences in $K$, because sequences add each other and multiply by scalars of $K$ quite naturally.

Hope that helps,

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