Example of a certain vector space

Question

Let $k$ be a field. I am looking for an example of infinite dimensional $k$-vector space $V$ such that $V\cong V\oplus V$. Also kindly let me know in brief how you've thought about it . Thanks in advance!

Answer 1

As Rhys Steele mentioned, two vector spaces over $k$ are isomorphic iff they have the same dimension. For an infinite dimensional $k$-vector space $V$, $\dim(V) + \dim(V)= \dim(V)$ holds (see this question). Therefore $V \cong V \oplus V$.

Let's pick our favourite infinite-dimensional vector space and do this. Mine is $k[X]$, the vector space of polynomials in one variable over $k$. If you add it with itself, you obtain $k[X] \oplus k[Y] \cong k[X,Y]$, the vector space of polynomials in two variables. All those vector spaces have dimension $\mathbb N$, so they are abstractly isomorphic. But where is the isomorphism? $k[X]$ and $k[X,Y]$ look very different, don't they?

We can choose a basis of $k[X]$ (e.g. the monomial basis) and $k[X,Y]$ (e.g. the basis of all monomial products $1,X,Y,XY,X^2,Y^2,XY^2,X^2Y\dots$). Choosing any bijection between those two basis (which exists, since $\mathbb N \times \mathbb N$ and $\mathbb N$ have the same cardinality, i.e. are both countably infinite) provides a vector space isomorphism, by linear extension.

You might feel a bit cheated now. Aren't $k[X]$ and $k[X,Y]$ different things? They are the same as vector spaces though. The difference is the additional structure on those two vector spaces, since you can multiply polynomials. The multiplication turns $k[X]$ and $k[X,Y]$ into algebras (even graded algebras), and the isomorphism above won't preserve this structure. If you want to fix this, you obtain an example which feels more natural:

Let $V=k[X_1,\dots,X_n,\ldots] = \bigoplus_{i=1}^\infty k[X_i]$ the vector space of polynomials with coefficients in $k$ and countably infinitely many unknowns. Like before, if you take the direct sum of $V$ with itself, you just "double" the unknowns. So $V\oplus V$ is the vector space over twice as many unknowns (which could be called $X_1,X_2,X_3,\ldots$ and $Y_1,Y_2,Y_3,\ldots$). But that's still the same amount of unknowns. A bijection of the set $\{\,X_1,X_2,X_3,\ldots\}$ and $\{\,X_1,Y_1,X_2,Y_2,X_3,Y_3,\ldots\,\}$ gives rise to an algebra-isomorphism (which is in particular a linear isomorphism).