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May
7
answered Is an $R$-module $A$ a module over the image of a homomorphism $f:R\rightarrow{f(R)}?
May
7
awarded  Informed
May
7
comment Prove a specific basis exists satisfying certain conditions with an endomorphism
For intuition's sake, have you tried writing $T$ as a matrix with respect to $\{e_1, \dots, e_n\}$? Can you compute the eigenvalues of $T$?
May
7
comment What is the Dedekind cut for 2?
You're right, $\beta$ does have the property that there is no largest element.
May
2
comment Euclid's proof for the existence of infinitely many prime
$p_i$ divides both $q$ (by definition of $p_i$) and $P$ (because we have assumed that $p_i$ is one of the factors making up $P$), so it divides $q-P$, which is equal to $1$. But there isn't a prime number that divides $1$. This contradicts our assumption (that $p_i$ is a factor of $P$), so the assumption must be false. So $p_i$ must be a new prime number we didn't already have in our list.
May
2
comment Split extension for semi-direct and direct products. Can a split extension be exact?
I don't know what you mean by either "equivalent" or "isomorphic" here, because elements can't be those things. You are right that they have to be equal. We have a composite map $H\to N\times H \to H$ which sends $h\mapsto (1,h) \mapsto h$, and $h$ is definitely very much equal to $h$.
May
2
comment Split extension for semi-direct and direct products. Can a split extension be exact?
No, that is correct. The map $N\times H\to H$ is the obvious one, $(n,h)\mapsto h$, and as a 'splitting' map, we may take $H\to N\times H$ to be $h\mapsto (1_N,h)$.
May
2
comment Split extension for semi-direct and direct products. Can a split extension be exact?
The extension that we are testing for split-ness is $1\to K\to G\to H\to 1$, not $H\to G\to H$ (which is not an extension anyway).
May
2
answered Split extension for semi-direct and direct products. Can a split extension be exact?
Apr
14
answered Prove that $\mathbb{Q}(r+s\sqrt{t})=\mathbb{Q}(\sqrt{t})$.
Apr
14
comment Prove that $\mathbb{Q}(r+s\sqrt{t})=\mathbb{Q}(\sqrt{t})$.
I see. That object that user222031 is defining isn't a priori a field (note the use of square brackets rather than round brackets - in general, $F(u)$ and $F[u]$ are different things), but it happens to be in this case. I'll write up an answer.
Apr
14
comment Defining an operation on a quotient set
Oh, strange. I don't know the book. In any case, if f isn't a homomorphism, there's no guarantee that Q (or Gf) is a group, so that must be what he means.
Apr
14
comment Prove that $\mathbb{Q}(r+s\sqrt{t})=\mathbb{Q}(\sqrt{t})$.
@Skull-Face You're going to have to tell us what definitions your class is working from. As far as I'm concerned, what user222031 gave is the definition of $\mathbb{Q}[x]$ - but your class might be doing things in a different order. What do you understand $\mathbb{Q}(x)$ to mean?
Apr
14
comment Defining an operation on a quotient set
No, it means that f is a group homomorphism. (That is, if * is the operation on G and # is the operation on G', we have f(g * h) = f(g)#f(h) for all g and h in G.) But group homomorphisms do have the property that images are still groups.
Apr
14
comment Defining an operation on a quotient set
1. G' also has a binary operation. 2. Isn't Gf the image of G under the group map f? In which case, it's a subgroup of G'.
Apr
14
comment Proving $(a_1,b_1)\times (a_2,b_2)\times\cdots\times(a_n,b_n)$ is open in $\mathbb{R}^n$.
Using the distance formula makes it harder than it needs to be. In particular, $y\in B(x;r)$ implies that $|x_i-y_i| < r$ for each $i$ separately. This reduces it to the one-dimensional problem: does an open interval of radius $r$ about $x_i$ lie inside the interval $[a_i, b_i]$? (Answer: yes, by your choice of $r$.)
Apr
13
awarded  Nice Answer
Apr
11
answered At what points of r in real number is f continuous?
Apr
11
reviewed Approve Right and left hand limits.
Apr
8
comment What does this notation mean? $x \mapsto f(x)$
@YoTengoUnLCD: if you consider it tautologous, fair enough. (But there are people who would write their functions as $x\mapsto xf$ or $x\mapsto x^f$ or similar. Not stating this can lead to genuine confusion - does $fg$ mean "do $f$ then $g$" or "do $g$ then $f$"? Writing explicitly that you will denote the image of $x$ under $f$ by $f(x)$ avoids this potential confusion. Also, for instance, with more arguments, you might write $g(x,y)$, but would prefer to write $x*y$ rather than $*(x,y)$; writing this out explicitly avoids confusing your reader.)