Reputation
9,604
Top tag
Next privilege 10,000 Rep.
Access moderator tools
Badges
22 50
Newest
 Yearling
Impact
~223k people reached

Nov
11
comment Prove that $I$ is a maximal ideal
Based on this, and the other comments, you seem to have misunderstood my point - which was that it is enough to prove that $f$ is irreducible or to prove that $K[x]/I$ is a field (and each will imply the other). You proved that $f$ is irreducible, so you're done. You could instead have proved that $K[x]/I$ is a field, but you don't have to do both. There are almost certainly other reasonable approaches as well.
Nov
10
answered Prove that $I$ is a maximal ideal
Nov
10
comment What does it mean for two polynomials to be the same in this fundamental field extension theorem?
It would be interesting to know what it is, and to check that the image of $\alpha$ does indeed have the same minimal polynomial as $\alpha$... This is an important observation - the theorem you are quoting gives necessary and sufficient conditions for a particular map $F(\alpha)\to F(\beta)$ to be an isomorphism, but the two fields may still be isomorphic even if this map is not an isomorphism.
Nov
10
comment What does it mean for two polynomials to be the same in this fundamental field extension theorem?
Does he say what the isomorphism is between $\mathbb{Q}(\alpha)$ and $\mathbb{Q}(\beta)$? Because if it doesn't send $\alpha$ to $\beta$ then the theorem doesn't apply.
Nov
10
comment Given $x$ and $y$ in $\mathbb{Z}[i]$, find $q$ and $r$ such that $x=qy+r$.
@Kieran That was an aside though - mixedmath's comment is correct, you are being asked to do division with remainder.
Nov
10
revised Given $x$ and $y$ in $\mathbb{Z}[i]$, find $q$ and $r$ such that $x=qy+r$.
added 70 characters in body; edited title
Nov
10
comment Given $x$ and $y$ in $\mathbb{Z}[i]$, find $q$ and $r$ such that $x=qy+r$.
You are correct that there are very few possible $r\in\mathbb{Z}[i]$ with $|r|<2$; there are two more than you have written though.
Nov
10
comment What does it mean for two polynomials to be the same in this fundamental field extension theorem?
I think you meant to say that $\eta$ is a root of $x^2+x-1$...or I messed up my calculation.
Nov
10
comment Basis of image and kernel of a linear transformation
I am guessing (and it really is just a guess) that the "dimension theorem" that isn't supposed to be used is another name for the rank-nullity theorem. But on the other hand, it isn't so hard to show that this set is independent directly.
Nov
10
comment Basis of image and kernel of a linear transformation
Look at what kinds of polynomials appear in the image, in particular at their coefficients. Can you see how, for a polynomial in the image, the coefficient of $x^2$ is determined by the other two coefficients?
Nov
10
comment Prove that $(x^3-2)$ is a maximal ideal of $\Bbb Q[x]$
To get better answers, you should add some context to your question - what have you already tried? Do you know some theorems that might help? For example, would it help you to know that $x^3-2$ is irreducible? Can you prove that it is?
Nov
7
comment Proving that $f$ is a bijection.
@egreg I'm not sure I'm following - the OP has a definition ($g(y)=x\iff f(x)=y$) in the question.
Nov
6
comment Proving that $f$ is a bijection.
It doesn't appear to be the definition the OP is using, given the phrasing of both the problem they are trying to solve and the clause beginning "I basically need to show...".
Nov
6
answered Proving that $f$ is a bijection.
Nov
6
answered Semisimple submodule
Nov
6
reviewed Approve I love maths, but my school is limited in its teachings.
Nov
5
comment Give a counterexample to show that $(AB)^{-1} \neq A^{-1}B^{-1}$
To make the same point in another way - it is not true that $AB\ne BA$ for all $A,B$, but there do exist some $A,B$ for which $AB\ne BA$ - so find a pair (that are also invertible), and you have an explicit counterexample.
Nov
5
comment Shape of the visible part of the Moon
Pas de problème. ;) I have just discovered that both of these things are "ellipse" in French, which makes your mistake even more forgivable than it already was!
Nov
5
comment Shape of the visible part of the Moon
Since you asked for language corrections - I'm pretty certain that you meant "ellipse" in "other cases". This is the shape (singular of ellipses), whereas ellipsis refers to the three dots immediately preceding the word!
Nov
5
revised Shape of the visible part of the Moon
deleted 1 character in body