Is free algebra a PI-algebra?

Question

We take a nonempty set $\mathbf{X}$ and write $\mathbf{X}=\{X_i\mid i\in I\}$. A finite sequence of elements from $\mathbf{X}$ will be denoted by $X_{i_1}X_{i_2}\cdots X_{i_m}$ and called a word. Note that the empty sequence is not excluded, and we denote it by $1$ and call it the empty word. Let us define multiplication by juxtaposition, that is, $$(X_{i_1}X_{i_2}\cdots X_{i_m})\cdot(X_{j_1}X_{j_2}\cdots X_{j_n})=X_{i_1}X_{i_2}\cdots X_{i_m}X_{j_1}X_{j_2}\cdots X_{j_n},$$ and the set of all words becomes a monoid with unity $1$, which we denote by $\mathbf{X}^\ast$ called the free monoid on $X$. For convenience, we can write $X_i^n$ for $X_iX_i\cdots X_i$ ($n$ times). Accordingly, every $w\neq1$ in $\mathbf{X}^\ast$ can be written as $w=X_{i_1}^{k_1}X_{i_2}^{k_2}\cdots X_{i_r}^{k_r}$, in which $i_j\in I$ and $k_j\in\mathbb{N}$. The free algebra on $\mathbf{X}$ over $F$, denoted by $F\langle \mathbf{X}\rangle$, is the monoid algebra of $\mathbf{X}^\ast$ over $F$. The elements in $\mathbf{X}$ and $F\langle \mathbf{X}\rangle$ are called indeterminates and noncommutative polynomials, respectively. A nonzero noncommutative polynomial $$f=f(X_1,X_2,\cdots,X_n)\in F\langle X_1,X_2,\cdots\rangle$$ is called polynomial identity of an $F$-algebra $A$ if $$f(a_1,a_2,\cdots,a_n)=0$$ for all $a_1,a_2,\cdots,a_n\in A$, and $A$ is called PI-algebra.

Question. Is free algebra a PI-algebra?

I do not even find an answer to this question. I also wonder if we can prove by contradiction. However, I get no results. Please help me.

Answer 1

I assume that $F$ is a field here.

Edit: Apologies for the inaccurate initial answer, you're completely right. The real answer is that the free algebra is a PI-algebra when $n = 1$ (since then it's commutative and satisfies $xy - yx = 0$) but not when $n \ge 2$. To prove this it suffices to observe that

• if an algebra contains a copy of the free algebra on $k$ generators then it can't satisfy any nontrivial polynomial identity in $k$ variables, and
• $F \langle X, Y \rangle$ contains a copy of the free algebra on $k$ generators for every $k$, since the free monoid on $X, Y$ contains a copy of the free monoid on $k$ generators: it suffices to take the submonoid generated by any $k$ distinct words in $X, Y$ all of the same length $\ell$ (we can take $\ell = \lceil \log_2 k \rceil$), since any product of such words factors uniquely into subwords of length $\ell$. For example $\{ XX, XY, YY \}$ generates a free monoid on $3$ generators and $\{ XX, XY, YX, YY \}$ generates a free monoid on $4$ generators. Alternatively we can find a copy of the free monoid on countably many generators, namely $\{ XY^i, i \ge 0 \}$; this generates the submonoid of words starting with $X$ and any such word again has a unique factorization into these generators by considering the locations of the $X$es.