1d
comment The union of finite field extensions is a finite field extension
Also, the set of rational functions (with scalars from a field) is a field, so there will be no proper nonzero ideals to quotient it by. You need to think about quotienting rings of polynomials.
1d
comment The union of finite field extensions is a finite field extension
Being a finite extension is not equivalent to adjoining a finite set of elements - the elements need to be algebraic, in which case they must already exist within some extension of the base field. Note that "the union" of extensions is ill-defined: they need to exist within a common larger field in order to take the union. Even in that case, taking the union of two fields will not be a field generally (for not closed under the operations). You're speaking of the compositum of extensions, which is the join in some lattice of extensions ordered by inclusion.
2d
awarded  Enlightened
2d
awarded  Nice Answer
Nov
25
comment Why $F[x]/p(x)$ would contain $F$?
It doesn't make sense to me to consider them as the same - so, how do you plan on interpreting $F[x]/(p)$ as an extension of $F$? There is a common theme in abstract algebra of identification. Better get used to it.
Nov
24
comment Finding $ \dfrac{\partial z}{\partial x} \text{and}\dfrac{\partial z}{\partial y} $ if $ F(cx - az, cy-bz) = 0 $
$a\frac{\partial z}{\partial x}=b\frac{\partial z}{\partial y}=c$ does not follow from $F(cx-az,cy-bz)=0$. Take $F(u,v)=u+v$ for instance. Why do you think we should be able to get rid of $F_u$ and $F_v$? Do you have any evidence whatsoever? There's a good possibility there are things you aren't telling us, or else you've otherwise incorrectly communicated the question to us.
Nov
24
comment Finding $ \dfrac{\partial z}{\partial x} \text{and}\dfrac{\partial z}{\partial y} $ if $ F(cx - az, cy-bz) = 0 $
Assume $x$ and $y$ are independent and $z$ is a function of them. You said you "don't know how to differentiate this," but really you do know how to differentiate it, you've just been too worried to actually do it!
Nov
24
comment Finding $ \dfrac{\partial z}{\partial x} \text{and}\dfrac{\partial z}{\partial y} $ if $ F(cx - az, cy-bz) = 0 $
Did you try differentiating?
Nov
23
answered Is the map from representation ring to class functions a isomorphism?
Nov
23
comment The converse of Factor Group criterion
In order to speak of whether or not $G/N$ is a group, there needs to be a binary operation on it in the first place. The putative operation $(aN)(bN)=abN$ is ill-defined if $N$ is not normal, since we can have $aN=a'N$ and $bN=b'N$ but $abN\ne a'b'N$.
Nov
21
awarded  Enlightened
Nov
21
awarded  Nice Answer
Nov
13
comment Should it stand that $\gcd(f(x), g(x))=1$?
The gcd of $2$ and $x$ cannot be $2$, since $2$ is not a divisor of both.
Nov
13
comment Quotientring-Show the isomorphism
Note it's injective and surjective and you're done. For future reference, isomorphism theorems are friends.
Nov
12
comment Polynomials close to idempotents in quotient ring of $\Bbb R[x_1,x_2,\dots,x_n]$
If $x_i\mid(p^2-tp)$ and $(x_i-1)\mid(p^2-tp)$ then $x_i(x_i-1)\mid(p^2-tp)$, so you don't need to write the "where $x_i\in\{0,1\}$" condition.
Nov
12
comment Polynomials close to idempotents in quotient ring of $\Bbb R[x_1,x_2,\dots,x_n]$
Yes, evaluation at a point is an algebra homomorphism.
Nov
12
comment Polynomials close to idempotents in quotient ring of $\Bbb R[x_1,x_2,\dots,x_n]$
Is that ideal generated by $x_i^2-x_i$ for a specific $i$ or for $i=1,\cdots,n$? Also by "when $x_i\in\{0,1\}$" do you mean the equality is true when pushed forward to the image of $S$ mod $x_i$ and mod $x_i-1$?
Nov
12
comment Find the square root of a matrix
@Pubbie Haha, wow. Well, I don't really know a good reference to give you, sorry.
Nov
11
comment Find the square root of a matrix
@Pubbie In any paper I'd write I wouldn't bother with a citation for a fact like this.
Nov
11
comment Find the square root of a matrix
@Pubbie What do you want a source for? This answer seems self-contained to me.