Artem

 1d comment Decomposing an affine algebraic set into irreducible onesThank you! Unfortunately this is at such a point in the course where we've just defined what an algebraic subset of an affine space is and looked briefly at the correspondence between sets of points and ideals in the ring of polynomials. So I'm guessing something very elementary is expected. 2d comment Does there exist a field $K$ such that $\mathbb R \subsetneq K \subsetneq \mathbb C$?A completely elementary proof would be to assume that K has an element with a nonzero imaginary part and show that in this case K contains $i$ and therefore contains all of $\mathbb C$. 2d answered Does there exist a field $K$ such that $\mathbb R \subsetneq K \subsetneq \mathbb C$? 2d asked Decomposing an affine algebraic set into irreducible ones May 3 comment Is $Y$ open in $X\cup_f Y$?Thank you! I couldn't pick a better title that would fit the rules. Can you help with the apparent contradiction which I outlined in the question? May 3 asked Is $Y$ open in $X\cup_f Y$? Apr 5 awarded Yearling Apr 5 awarded Yearling Mar 3 awarded Nice Question Jan 24 comment Existence of an irreducible polynomial over $\mathbb F_p$.@GerryMyerson Yes, but the question you cited was answered based on the assumption that the roots of $x^{p^n} - x$ form a field. I am looking to establish that via the answer to my question. Jan 24 comment Existence of an irreducible polynomial over $\mathbb F_p$.@MartinBrandenburg I know how $F_{p^n}$ is obtained as a splitting field. The point of the exercise is to show the existence of such fields using only the concept of simple extension. Jan 24 asked Existence of an irreducible polynomial over $\mathbb F_p$. Jan 24 answered Unable To Understand The Difference Between $(-3)^4$ And $-3^4$ Jan 14 comment Is the class of algebraic extensions distinguished?Thank you, @DonAntonio! Jan 14 accepted Is the class of algebraic extensions distinguished? Jan 14 comment Is the class of algebraic extensions distinguished?In the example you gave above Langs property (2) states that $\mathbb Q\cdot\mathbb Q(\pi)/\mathbb Q(\pi)$ is algebraic. It does not require that $\mathbb Q\cdot\mathbb Q(\pi)/\mathbb Q$ be algebraic. Jan 14 comment Is the class of algebraic extensions distinguished?I sent it unfinished by mistake. There is now a quote. The property (2) does not require that both $E,F$ are members of an element of $\mathcal C$, only that they are both subfields of some other field. Jan 14 comment Is the class of algebraic extensions distinguished?My book (revised 3rd edition, don't know which printing, but I got it new a couple of months back from amazon, if that's any help) "(2) If $k\subset E$ is in $\mathcal C$, if $F$ is any extension of $k$, and E,F are both contained in some field, then $F\subset EF$ is in $\mathcal C$". Jan 14 comment Is the class of algebraic extensions distinguished?That $F(\alpha)$ is finitely generated is clear. Why can it be expressed in terms of finitely many elements from $E$? And how? Jan 14 asked Is the class of algebraic extensions distinguished?