Show that the vector space of polynomials R[x] is isomorphic to a proper subspace of itself:
Vector Space Isomorphism exists when there exists a bijective (one-to-one and onto) linear mapping F:V $\rightarrow$U. the coefficient of the polynomials can be written as $(a_0,a_1,a_2...)$. But how to find the subspace?
What about instead to prove the dimension of the the two vector spaces is the same, which means isomorphic? But how to do it?
On
First, note that you can't do this with finite dimensional vector spaces. You have to come up with a linear transformation which is one-to-one but not onto. Any linear transformation is determined by what happens to the vectors in any basis. Use the standard basis for $R[x]$, that is, $x^n$ where $n\ge 0$. You have to map each $x^n$ to a polynomial such that the mapping is one-to-one. One way to do this is to map $x^n$ to a polynomial of degree $n+1$. You can check this ensures that only the zero polynomial maps to zero. You can check that it is not onto because, for example, $x^0$ is not the image of any polynomial. These two facts depend on looking at the leading term of polynomials and seeing that no nonzero polynomial is a linear combination of polynomials of lesser degrees.
Hint: Consider $p(x) \mapsto p(x^2)$.
This acts on a sequence of coefficients (which is all a polynomial is) by inserting zeros between them: $$(a_0,a_1,a_2,\dots,a_n,0,0,0,\dots) \mapsto (a_0,0,a_1,0,a_2,\dots,a_{n-1},0,a_n,0,0,0,\dots)$$ and so you can recover one from the other.