12,454 reputation
11332
bio website math.brown.edu/~lubinj
location Minnesota
age 77
visits member for 2 years, 9 months
seen 7 hours ago

Aged mathematician


Jan
18
comment Application of Hensel's lemma
Highly unconventional notation. But whatever version of Hensel you use, its hypotheses are not satisfied here. The polynomial is irreducible by Eisenstein, of course.
Jan
18
comment Multiply Complex number to form a real value
Are you wondering about the identity $(a+bi)(a-bi)=a^2+b^2$?
Jan
18
comment Find basis of $\mathbb{Q}(\sqrt{2}+\sqrt[3]{4})$
But isn’t it clear that the composite field is $\mathbb Q(2^{1/6})$? From this, it’s obvious that OP’s basis is correct, once you know that the singly-generated field is equal to the composite.
Jan
18
revised Find basis of $\mathbb{Q}(\sqrt{2}+\sqrt[3]{4})$
corrected a sign error in the denominator of the last fraction
Jan
17
answered Find basis of $\mathbb{Q}(\sqrt{2}+\sqrt[3]{4})$
Jan
17
comment Rota's “lure of the algorithm”?
Right. A spectacularly effective or ingenious algorithm may perhaps contribute nothing to understanding.
Jan
17
comment Simpler expression for $\sum_{k=1}^{n}{k!}$
It lacks the summation, but is hardly simpler than the given formula.
Jan
17
answered Real analyticity of $\sqrt{x}\coth(\sqrt{x})$ : function that is decreasing with resp. to derivation
Jan
17
answered Is the notation $X²$ an abuse of language in a polynomial ring.
Jan
17
comment Epsilon Neighborhoods of the Rationals
Chances are that for each $x\in\mathbb Q$, $\epsilon(x)$ is a small number so that the sum of all of them (countably many) is not very large. If that’s the case, I or someone else can explain in an answer what’s probably intended.
Jan
15
answered If $n$ is a positive integer and is not a perfect square
Jan
15
comment How to correctly do a division using a slash?
The notation is inherently ambiguous. Parenthesize, parenthesize!
Jan
14
comment Image of homomorphism from ideal is ideal
@user115654, if I had had my wits about me, I would have used the fact that the image of $J$ is the same as the image of $I+J$ (!).
Jan
14
comment Image of homomorphism from ideal is ideal
@user115654, yes, I seem to be totally confused today.
Jan
14
comment Solutions of elliptic curve in finite field
You’re trying to run before you’ve learned to walk. You must familiarize yourself with the principles and methods of computation in and over finite fields.
Jan
14
comment Image of homomorphism from ideal is ideal
Yes, you’re mistaken: take the homomorphism $f\colon k[x]\to k$ by $f(p)=p(0)$, the constant term of the polynomial $p$. Here, $k$ may be taken to be a field. The kernel is $(x)$. Now let $J$ be the set of all multiples of $x+1$.
Jan
12
comment polynomial with an irrational expression of the free term
Moving the constant over to the other side and factoring will never give you anything useful. Remember that the reason we factor the $f$ in $f(x)=0$ is to use the convenient fact that if a product is zero, one of the factors must be zero.
Jan
10
comment One to the power of i with two different results.
@MPW, that’s exactly right, but what I had in mind was another (and maybe worse) argument, namely that exponentiation with integer exponent is not an analytic operation, but purely algebraic.
Jan
10
answered One to the power of i with two different results.
Jan
9
comment Show that there are no onto homomorphisms from $\mathbb{Z}_{18}$ to $S_3$
Functions aren’t abelian, groups are.