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Feb 1, 2019 · However, Serre is dealing with finite-dimensional complex representations of finite groups, and in that case, yes, every indecomposable representation is irreducible.
Oct 9, 2015 · A decomposable representation is one which is not indecomposable. The relationship between the notions of irreducibility and complete reducibility is quite different …
Why an absolutely irreducible representation is irreducible under all field extensions? Ask Question Asked 9 years, 3 months ago Modified 4 years, 2 months ago
What is an irreducible matrix? Ask Question Asked 6 years, 4 months ago Modified 6 years, 4 months ago
The implication 'irreducible implies prime' is true in integral domains in which any two non-zero elements have a greatest common divisor. This is for instance the case of unique factorization …
The condition of being irreducible if it doesn't have any roots is false. Consider, for example, the polynomial $$ x^4 + 4 x^2 + 3 = (x^2 + 1) (x^2 + 3) \in \mathbb {R} [x] $$ When the coefficient …
Mar 1, 2013 · Proving that a polynomial is irreducible over a field Ask Question Asked 13 years, 11 months ago Modified 4 years, 5 months ago
This is impossible: any polynomial ring over a field is a U.F.D. In such domains, irreducible elements are prime. The simplest example is the ring of quadratic integers $\;\mathbf Z [i\sqrt …
Oct 9, 2021 · Given a polynomial over a field, what are the methods to see it is irreducible? Only two comes to my mind now. First is Eisenstein criterion. Another is that if a polynomial is …
May 13, 2011 · From my reading I get the feeling that an irreducible representation is a matrix (in the case of SO (3) at least, though it seems that in general they are always tensors), is this …
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