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In article <454672bc$0$150$edfadb0f@xxxxxxxxxxxxxxxxxxxx>,
Thunder <Thunder@com> wrote:
>Show that Q[x,y]/(x^2,xy) is not semisimple.
>The teacher wrote that it is because [x^2]=0
>why is that?
Commutative semi-simple rings with 1 have no nilpotent elements.
A ring is semisimple if and only if it is a direct sum of simple
ideals, if and only if it is a direct product of simple rings.
A ring is simple if and only if it is not trivial and its only
two-sided ideals are the trivial ones. If R is commutative, simple,
and has a 1, then it is a field: for given r in R, the ideal Rr is either
trivial (so in particular 1r = 0, so r=0), or Rr=R, in which case
there exists s in R such that sr=1, so r is a unit. Thus, R is a
field.
Now suppose that R = prod(R_i), with each R_i simple, and each R_i has
a 1. Then if (a_i)^n = (0) in R, then (a_i^n) = (0), so a_i^n=0 in
R_i. Since R_i is a field, the only possibility is a_i=0 for each i,
so (a_i)=(0). Thus, R has no nilpotent elements.
Your ring Q[x,y]/(x^2,xy) has a nilpotent element. If it were
semisimple, then it would be a product of simple rings. Since your
ring is commutative with 1, each of the factrs must be commutative
with 1, and so you get a contradiction.
--
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"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
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Arturo Magidin
magidin-at-member-ams-org
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