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18h
comment Decryption of RSA
Why does that seem inappropriate? What you need is an inverse $d$ to $e$ modulo $\varphi(n)$. To find an inverse modulo $\varphi(n)$, you need to know what $\varphi(n)$ is. To find $\varphi(n)$, you need to know the prime factorization of $n$.
1d
comment Inverse of multivariated polynomials over finite fields
It took 5 votes to close; that means 4 other people agreed your question needed clarification.
1d
comment Inverse of multivariated polynomials over finite fields
@Roger I'm going to stop giving this question any attention, on the grounds that I'm pretty sure you're trolling. It is not my responsibility, nor anyone else's, to try and second-guess what your question is; it is your responsibility to make it clear what your question is.
2d
comment If $\sigma\in S_p$ with $|\sigma| = p$, why is $\sigma$ a $p-$cycle, and why is $|\sigma^r| = p$?
How many $p$-cycles can $\sigma$ have?
2d
comment Is there an “internal” definition of the tensor product?
The "cone" of simple tensors is given by the bilinear map $V \times W \to V \otimes W$. So what you probably want is some conditions under which a bilinear map $V \times W \to S$ induces an isomorphism $V \otimes W \simeq S$?
2d
comment Inverse of multivariated polynomials over finite fields
If you define the polynomial given by $f(x,y) = x+1$, with dummy variable $y$, then $(f \circ f)(x,y) = f(f(x,y)) = f(x+1) = $.... It's just not grammatical. You need two inputs to $f$, so in order for the notion that $f$ have an inverse to be sensible, $f$ needs to have two outputs. What you want in this case is $f(x,y) = (x+1, y)$ - a dummy variable that $f$ does nothing to, not one that $f$ forgets.
Apr
24
comment Inverse of multivariated polynomials over finite fields
Well, first thing is to make sure your question is well-formed. In order for the notion of an "inverse" under composition to be sensible, you need to consider polynomials with the same number of outputs as inputs, i.e., a pair of polynomials $f(x,y), g(x,y)$. Was that your intention?
Apr
23
comment Inverse of multivariated polynomials over finite fields
$GF(2)$ is $\mathbb F_2 = \mathbb Z/2\mathbb Z$... What do you mean by $f(x,y)$? A polynomial? A rational function?
Apr
23
comment modification of Dedekind cuts
You can get a non-nilpotent infinitesimal by approximating the sequence $(1/\ln(n+1))$ by rationals.
Apr
23
comment modification of Dedekind cuts
yes, well, there is the system of dual numbers, where an infinitesimal is defined to have the property that it squares to $0$: en.wikipedia.org/wiki/Dual_number. Your system of numbers is more complicated than that one.
Apr
23
answered modification of Dedekind cuts
Apr
23
comment modification of Dedekind cuts
Dedekind cuts and Cauchy sequences are distinct definitions of the real numbers. This question is about Cauchy sequences.
Apr
23
comment Show that a polynomial over $\mathbb{Z}_{2}$ is irreducible
@DiegoRobayo That's a common notational confusion. I've known professors who insist that one should never use $\mathbb Z_p$ to mean $\mathbb Z/p\mathbb Z$, but not everyone follows that rule, so I think you just kind of have to accept it.
Apr
23
comment Show that a polynomial over $\mathbb{Z}_{2}$ is irreducible
No, that only shows that $p$ has no roots. There's still the possibility that $p$ factors into two irreducible quadratics.
Apr
23
comment What are all the integral domains that are not division rings?
Well, every integral domain embeds into a field (via the field of fractions). So all you need to do is classify all subrings of all fields.
Apr
21
comment What does it mean by a set is bounded.
You should probably ask a friend to get the definition that was given in class, since there are multiple definitions.
Apr
21
revised Under what conditions there exists a topology making a bijection into a homeomorphism?
added 126 characters in body
Apr
21
answered Under what conditions there exists a topology making a bijection into a homeomorphism?
Apr
21
comment Modulo arithmetic proof
No problem. ${}{}$
Apr
21
comment Modulo arithmetic proof
You have $1 \le j-i < p$, now look at your assumption.