In the commutative case, this is equivalent to " R " being a reduced ring, since " R " has no nonzero nilpotent elements.
12.
But this would imply that a matrix ring over a field is not semisimple ( as there are nontrivial nilpotent elements ) while that ring is even simple . talk)
13.
For example, we can speak of " reduced ( i . e ., has no non-zero nilpotent elements ), then so are its localizations.
14.
An equivalent condition is that " S " / " PS " has a non-zero nilpotent element : it is not a product of finite fields.
15.
If " k " is not formally real, the fundamental ideal is the only prime ideal of " W " and consists precisely of the nilpotent elements;
16.
In algebraic geometry, differentials and other infinitesimal notions are handled in a very explicit way by accepting that the coordinate ring or structure sheaf of a space may contain nilpotent elements.
17.
:: : : If e squares to zero, it's what's called a nilpotent element and you can't divide by it in the ordinary way.
18.
One can show that the tensor product of fields of " L " with itself over " K " for this example has nilpotent elements that are non-zero.
19.
The nilpotent elements of a commutative ring " R " form an nilradical of " R "; therefore a commutative ring is reduced if and only if its nilradical is zero.
20.
This is an ideal because the sum of any two nilpotent elements is nilpotent ( by the binomial formula ), and the product of any element with a nilpotent element ( by commutativity ) is nilpotent.
