Testwiki:Reference desk/Archives/Mathematics/2012 September 11

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September 11

A countanle $\pi -weight$

Hi, Does anyone know what is a (countable) A countanle $\pi -weight$ for a topological space? And also, is there a difference between $\bar{A}$ to $\overline{int(\stackrel{‾}{A})}$. In other notation: is there a difference between $Cl(A)$ to $Cl(int(Cl(A)))$ ?. Thanks! Topologia clalit (talk) 11:14, 11 September 2012 (UTC)

Don't know about pi-weights. For the second question, let A in R be a closed interval union a point. Staecker (talk) 11:35, 11 September 2012 (UTC)

The π-weight of a space is a minimal cardinality of its π-base.—Emil J. 12:05, 11 September 2012 (UTC)

Thank you both! Topologia clalit (talk) 13:12, 12 September 2012 (UTC)

R[X]/(X^2-1)

If I have a Field (eg R) and I build the quotient as follows R[X]/(p(x)) with a polynomial p(x), then I get a field extension as usual if p(x) is a minimal polynomial. What happens if p(x) is reducible (ie. not a minimal polynomial)? For example in the cases: R[X]/(X^2-1)

or

R[X]/(X^2)

--helohe (talk) 19:45, 11 September 2012 (UTC)

The result will not be a field, but a ring with zero divisors. —Kusma (t·c) 20:13, 11 September 2012 (UTC)

in the case R[X]/(X^2) I guessed it will be something like a subset of R x Z/4Z . Does that make sense? --helohe (talk) 20:24, 11 September 2012 (UTC)

R[X]/(X²-1) is isomorphic to the group ring R[Z/2Z], but I don't know if you can phrase R[X]/(X²) in a similar way. Rckrone (talk) 00:08, 12 September 2012 (UTC)

R[X]/(X²-1) is also isomorphic to the ring of the Split-complex numbers, and R[X]/(X²) is isomorphic to the ring of dual numbers --84.229.150.202 (talk) 04:40, 12 September 2012 (UTC)

If the polynomial p is square-free, you can write it as a product $p=p_1{p}_2\cdots p_n$ of irreducible polynomials. Then the Chinese remainder theorem ensures that the ring $R[x]/(p(x))$ is isomorphic to the product of fields $(R[x]/(p_1(x)))\times \cdots \times (R[x]/(p_n(x)))$. In particular, if R is a field of characteristic other than 2, then $R[x]/(x^2-1)$ is just $R\times R$.—Emil J. 09:15, 12 September 2012 (UTC)