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Simply, if there is a polynomial $f$, in noncommuting variables, which vanishes under substitutions from ring $R$, the ring will be called a PI ring (Polynomial Identity ring). For example, commutative rings always satisfy the polynomial $f(x,y) = xy - yx$.

Is $M_{2}(K)$, the ring of all $2 \times2$ matrices over a field $K$, a PI ring? I have tried to construct a polynomial, assuming that this ring is a PI ring, but couldn't find any. Thanks.

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3 Answers 3

up vote 4 down vote accepted

See the Amitsur-Levitzki theorem.

Any four $2\times 2$ matrices $A_1,A_2,A_3,A_4$ satisfy $$\sum_{\sigma\in S_4} \mathrm{sgn}(\sigma)A_{\sigma(1)}A_{\sigma(2)}A_{\sigma(3)}A_{\sigma(4)}=0.$$ Hence $M_2(K)$ satisfies the polynomial identity $$\sum_{\sigma\in S_4} \mathrm{sgn}(\sigma) x_{\sigma(1)}x_{\sigma(2)}x_{\sigma(3)}x_{\sigma(4)}$$ of degree $4$ in four variables. This generalizes to $M_{n}(K)$.

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Arturo, I was going to post a link, before seeing that this had been tagged as homework. What are we supposed to do on MSE for such cases? – user16299 Jan 15 '12 at 8:03
@Yemon: He still has to prove the result... – Arturo Magidin Jan 15 '12 at 8:04
@ArturoMagidin: Is there any restrictions on the degree of this polynomial? I mean, when you gave above polynomial of degree 4, so I shouldn't expect to find any of degree 3. – Basil R Jan 15 '12 at 8:23
@Arturo, I got my last question's answer. It will satisfy a polynomial of degree exactly 4. Thanks. – Basil R Jan 15 '12 at 8:28

Besides to first pervasive answer, suppose $A$ and $B$ be such two matrices. The Cayley–Hamilton theorem implies that if $x$= $(AB-BA)$ then $(AB-BA)^2$ commutes with any other $2\times 2$ matrix in $M_{2}(K)$. It leads you to $f(x,y,z)=(xy-yx)^2z-z(xy-yx)^2$.

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If A and B are arbitrary matrices then your claim that $(AB-BA)^2$ lies in the centre of $M_2$ is false. I am not sure what you are trying to say or how your answer relates to the original question. – user16299 Jan 15 '12 at 22:32
Yemon: If tr$(AB-BA)$=0, so according to C-H Theorem, $(AB-BA)^2$ would be a scalar matrix which commutes with any other trivial$ 2\times 2$ matrix. Now, take the latter matrix as $z$. – Babak S. Jan 16 '12 at 14:14
Oh, I see where I was going wrong, you are of course right. I can't cancel my downvote, but if you edit your answer (just add a space or something) then that should unlock it so I can cancel my wrong-headed downvote. – user16299 Jan 16 '12 at 16:19
+1 Great first answer at Math.SE! – amWhy Mar 3 '13 at 18:03
@amWhy: Thanks. yes it was almost the first. It is old now like one year child. an infant :D – Babak S. Mar 3 '13 at 18:06

Matrix rings over a given commutative ring (in particular, over any field) are indeed PI, and in some sense are the motivating prototypes.

Since you say this is homework, I am reluctant to give an explicit reference for this fact, at least until you give some more context. Were you given any hints in the course? Is this an optional question for interest, or are you expected to work out the answer using things mentioned in class?

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Thanks for hints. I expected to examine and find this ploynomial for the given statement in the class. – Basil R Jan 15 '12 at 8:15

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