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Feb
19
comment Splitting field of a separable polynomial is separable
@caffeinemachine - Your answer looks very nice to me! I upvoted it.
Feb
19
comment Splitting field of a separable polynomial is separable
@caffeinemachine - Thanks! I'll read your answer carefully, but please be patient: I'm very slow! About your bilinear form $f$, let's assume to simplify that $f$ is symmetric and that $\dim V<\infty$ (this suffices for our purpose). We can also assume that $f$ is non-degenerate, so that the map $g:V\to V^*$ defined by $g(v)(v')=f(v,v')$, which is clearly injective, is also surjective. Let $h$ be a nonzero linear form on $V$ which vanishes on $W$. Then there is a $v\in V$ such that $g(v)=h$, that is $v\in W^\perp$ (and obviously $v\neq0$).
Jan
29
comment Case analysis in theorem 4.1.3 of the HoTT book
By Lemma 2.3.9, Lemma 2.10.1 and Axiom 2.10.3 (univalence axiom) in the HoTT Book, it suffices to show $f\circ g=g\circ f$ for all invertible maps $f,g:\mathbf2\to\mathbf2$.
Jan
17
comment Holomorphic function of a matrix
@loupblanc - Thank you very much for your interesting comment!
Jan
13
answered HOTT proof of transitivity of ordering of natural numbers
Jan
8
comment A question from Bourbaki's Theory of Sets
C27 says: If $R$ is a theorem of a logical theory $\mathcal T$ in which the letter $x$ is not a constant, then $(\forall x)R$ is a theorem of $\mathcal T$. It does not say that $R\implies(\forall x)R$ is a theorem of $\mathcal T$ (supposing that $R$ is a relation of $\mathcal T$ and $x$ a letter which is not a constant). The assumption that $x$ is not a constant is crucial. (+1 for your nice question!)
Dec
17
comment Finding a type such that $X + 1 \not\simeq X$ and $X+2\simeq X$
@KevinQuirin - You wrote "$X+1\not\simeq1$ and $X+2\simeq X$", but I suspect you mean "$X+1\not\simeq X$ and $X+2\simeq X$".
Nov
27
comment Atiyah-Macdonald Exercises 5.16-5.19
@NikolasKuhn - I think you're right. Thanks for your comment!
Nov
19
comment Exercise 4.6 (iii) in the HoTT Book
Reading notes about the HoTT Book.
Nov
12
awarded  Nice Question
Nov
6
comment Given a commutative ring $R$ and an epimorphism $R^m \to R^n$ is then $m \geq n$?
@RobertCardona - The assumption of Atiyah-MacDonald' exercise is that $A$ is a nonzero commutative ring and $\phi:A^m\to A^n$ is an $A$-linear map. (There is no $R$.) Sorry, I don't understand your answer.
Nov
6
comment Given a commutative ring $R$ and an epimorphism $R^m \to R^n$ is then $m \geq n$?
@RobertCardona - Which isomorphism are you referring to?
Oct
2
awarded  Custodian
Oct
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reviewed Close Visualization of subspaces
Sep
30
reviewed Approve Lipschitz condition absolute value
Sep
20
reviewed Approve With indices and without indices.
Sep
19
reviewed Approve make a salad with 4 toppings
Sep
19
reviewed Approve Is $1+T$ a topological generator for $Z_{p}[[T]]$?
Sep
13
reviewed Approve How to solve this pde equation: $(p^2 + q^2)y = qz$