Set of Polynomials over Infinite Set has Same Cardinality

Theorem
Let $S$ be a set of infinite cardinality $\kappa$.

Let $S[x]$ be the set of polynomial forms over $S$ in the indeterminate $x$.

Then $S[x]$ has cardinality $\kappa$.

Proof
Since $S[x]$ contains a copy of $S$ as constant polynomials, we have an injection $S \to S[x]$.

We define an injection from $S[x]$ to the set $\mathcal F$ of finite sequences over $S$ as follows:

Each polynomial in $f \in S[x]$ is of the form $f = a_0 + a_1 x + a_2 x^2 + \dots + a_n x^n$ where $a_n$ is non-zero and each $a_i$ is in $S$.

We send each polynomial $f$ to the sequence of its coefficients $(a_0,\dots,a_n)$.

This is evidently injective by the definition of equality of polynomials.

Now the set of finite sequences over $S$ is a countable union of sets of cardinality $\kappa$.

Therefore, since such a union has cardinality $\kappa$, $\mathcal F$ has cardinality $\kappa$.

Therefore there is a bijection $\mathcal F \leftrightarrow S$.

Composing this with the injection $S[x] \to \mathcal F$, we have an injection $S[x] \to S$.

So by the Cantor-Bernstein-Schroeder Theorem there is a bijection $S[x] \leftrightarrow S$.

Hence, we have $|S[x]| = |S| = \kappa$.