I am confused about the following argument which I can't disprove
We know that if $V$ is a finite dimensional vector space over a field $\mathbb K$(algebraically closed,characteristic $0$) then symmetric algebra of $V$ is given by $S(V) = \mathbb K[x_1,....x_n]$,i.e a polynomial ring in $n$ indeterminates.In this situation we know that $V^*$ is also a finite dimensional vector space of dimension $n$ and hence $S(V^*) = \mathbb K[x_1,....x_n]$. Thus we have the following:
$\mathbb K[x_1,....x_n] = \oplus_{d \geq 0} (\mathbb K[x_1,....x_n])_d = \oplus_{d \geq 0} S^dV =\oplus_{d \geq 0} S^d(V^*) $, since the decomposition is unique we have $ (\mathbb K[x_1,....x_n])_d = S^dV = S^d(V^*), \forall d \geq 0 $
Now at this point I have the following questions(rather confusions):
$(1)$ In many sources it's said that $S(V^*)$ is the ring of polynomial functions on $V$ but $SV$ is not.But the above argument shows that both of them are the same.
$(2)$ Since we know that $S^d(V) = (\mathbb K[x_1,....x_n])_d$ and we can identify it with space of symmetric multilinear functionals on $V$,then we have space of symmetric multilinear functionals from $V \times V...\times V \to \mathbb K$($d$ copies)and from $V^* \times V^*...\times V^* \to \mathbb K$ ($d$ copies)are same.Is this true?
$(3)$ How to show that there exist a natural pairing $S^dV \otimes S^d(V^*) \to \mathbb K$(By the above argument $S^dV$ and $S^d(V^*)$ are same)
$(4)$ what does $(SV)^*$ look like?
I think I am missing some cardinal argument.Any help from anyone is welcome.