Are there any rings that are not polynomial identity rings? I'm currently studying polynomial rings and I'm struggling to find any examples of rings that are not polynomial identity (PI) rings. I was wondering if someone could point me in the general direction? I've messed around with the Weyl algebra and quaternions but I can't prove definitively whether they're PI rings or not.
Any information is greatly appreciated.
 A: A simple example is the ring $R = \mathbb{Z}\langle X_1, X_2, \ldots\rangle$, which is the free ring on $\mathbb{N}$ generators.
To see this, note that for all $N$, the obvious inclusion $f : \mathbb{Z}\langle X_1, X_2, \ldots, X_N \rangle \to \mathbb{Z}\langle X_1, X_2, \ldots\rangle$ has a left inverse $g$ - for example, the one sending all $X_j$ to $0$ for $j > N$. So if we had $P \in \mathbb{Z}\langle X_1, X_2, \ldots, X_N \rangle$ such that for all $a_1, \ldots, a_N \in R$, $P(a_1, \ldots, a_N) = 0$, then we would have $f(P) = P(x_1, \ldots, x_N)= 0$. Then we’d have $g(f(P)) = P = 0$.
A: 
I've messed around with the Weyl algebra and quaternions but I can't prove definitively whether they're PI rings or not.

From theorem 1 in Kaplansky, Irving. "Rings with a polynomial identity." Selected Papers and Other Writings (1995): 59., a right primitive ring which satisfies a polynomial identity has to be finite dimensional over its center. The Weyl algebra is right primitive but is infinite dimensional over its center.  That is why it does not satisfy a polynomial identity.
On the other hand, there is the famous embedding of $\mathbb H$ into $M_2(\mathbb C)$, and of course, this matrix ring satisfies the standard identity so of course the subring does too.
