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Ok so my math teacher told us that $\sqrt {(a+b)}=\sqrt a+\sqrt b $ for all reals, which is clearly false. However when I told this to another math teacher he told me this was true for fields of characteristic 2. replacing that statement by this one which is more general $a^2+b^2=(a+b)^2$Which I interpret as fields where $a+a=0$ for any a.

My new question is for rings in general, in which rings does it hold that $a^2+b^2=(a+b)^2$ or in other words in what rings is $ab=-ba$ for all $a,b$?

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  • $\begingroup$ $a^2=b^2+(a=b)^2$? Do you mean $(a+b)^2=a^2+b^2$? $\endgroup$ – Brian M. Scott Sep 5 '13 at 19:50
  • $\begingroup$ I sense there is an error made in your simplification but I can't tell because of the $=$ signs. In general a ring need not even have square roots of elements. $\endgroup$ – Alexander Sep 5 '13 at 19:50
  • $\begingroup$ Did my edit just now fix it to what you actually wanted? $\endgroup$ – Brian M. Scott Sep 5 '13 at 19:53
  • $\begingroup$ this is what I should have typed $\endgroup$ – Jorge Fernández Hidalgo Sep 5 '13 at 19:56
  • $\begingroup$ If the ring contains a unity, then $ab+ba =0$ implies $a+a = 0$ by setting $b = 1$, so the ring is of characteristic $2$. Conversely, any commutative unital ring of characteristic $2$ satisfies your property. For the remaining cases, I don't know... $\endgroup$ – Joel Cohen Sep 5 '13 at 20:02
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Let $A$ be a unital ring such that $(a+b)^2 = a^2 + b^2$ for all $a, b \in A$. This is equivalent to $ab + ba = 0$ for all $a, b \in A$. Setting $b = 1_A$, you get that the ring is of characteristic $2$. Now in characteristic $2$, $ab + ba = 0$ can be rewritten as $ab = ba$, which means the ring is commutative.

Conversely, any commutative unital ring of characteristic $2$ satisfies your property.

You can extend this proof for any ring $A$ such that for all $a\in A$, there exists $e_A \in A$ such that $a e_a = a e_a = a$. As for the general case, I have no idea.

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  • $\begingroup$ Srry why does ab+ba=0 imply ab=ba in characteristic 2? $\endgroup$ – Jorge Fernández Hidalgo Sep 5 '13 at 21:16
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    $\begingroup$ Add $ba$ on both sides of the equality, you get $ab + 2 ba = ba$. But since $2ba = 0$ (you're in characteristic $2$), then $ab = ba$. $\endgroup$ – Joel Cohen Sep 5 '13 at 21:17
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If we are working in a commutative ring $R$, $\sqrt{a+b}=\sqrt{a}+\sqrt{b}$ implies that $4ab=0$. If you want this latter equation to hold for all $a, b$, then you need $4=0$

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  • $\begingroup$ $4$ is $1+1+1+1$. I don't understand the first part of your question. $\endgroup$ – Alexander Sep 5 '13 at 19:59
  • $\begingroup$ I don't want to keep using the square root notation since not all elements in a ring must have a unique square root $\endgroup$ – Jorge Fernández Hidalgo Sep 5 '13 at 20:00
  • $\begingroup$ Right. Some algebra gives the statement $4ab=0$ which one can ask of any elements. Then if you let $a=b=1$, you have $4=0$. $\endgroup$ – Alexander Sep 5 '13 at 20:02
  • $\begingroup$ aren't you using commutatively for that? $\endgroup$ – Jorge Fernández Hidalgo Sep 5 '13 at 20:02
  • $\begingroup$ Ah ok, yes. I thought you were interested in commutative rings. Let me think about the non-commutative case. $\endgroup$ – Alexander Sep 5 '13 at 20:06
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If $V$ is a $2$-dimensional vector space over an arbitrary field $k$, choose a basis $\{e_1, e_2 \}$ for $V$. Then, the subring of the exterior algebra $\wedge^*V$ generated by $e_1, e_2, e_1 \wedge e_2$ is such a noncommutative ring.

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