Is $k\langle X \rangle \cong k[X]$

Question

Let $k$ be a field. Is $k\langle X \rangle \cong k[X]$? Here the left hand side denotes the free algebra generated by $X$, the right hand side are polynomials with coefficients in $X$.

I guess the answer is yes, but I want a quick sanity check.

Answer 1

The answer, as stated in the comments, is yes. Recall that the free $k$-algebra on a set of generators $X_1,\dots, X_n$ is the free $k$-module with basis consisting of all words in the elements $X_1,\dots, X_n,$ with multiplication given by concatenation of words. For example, if $a,b\in k,$ then $a X_1X_2 X_1 X_3$ and $b X_4X_1 X_2 X_4 X_5$ are both elements of $k\langle X_1,\dots, X_6\rangle$, and their product is $$ (a X_1X_2 X_1 X_3)(b X_4X_1 X_2 X_4 X_5) = ab X_1 X_2 X_1 X_3 X_4X_1 X_2 X_4 X_5. $$

However, if $n=1,$ words are simply just some number of $X$'s -- i.e., powers of $X$! So, the free $k$-algebra on a single generator $X$ is the free $k$-module with basis $1, X, XX = X^2, X^3,\dots$, and with multiplication given on basis elements by $(aX^i)(bX^j) = ab X^{i+j}.$ This is precisely the polynomial algebra $k[X]$ in one indeterminate.