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May 26 |
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How to establish an isomorphism between these two tensor products? Zhen is absolutely correct. You do not want to prove isomorphisms between tensor product spaces and other vector spaces (including other tensor product spaces) using bases, unless you're a physicist I guess. :) You shouldn't avoid trying to use the univ. mapping property but figure out how to use it, because that's the only way you're going to get so comfortable with it that you become unafraid of that (mathematically) essential feature of tensor products. |
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May 26 |
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How to establish an isomorphism between these two tensor products? Can you do the case $k = 3$ and $l = 1$? You don't write anything about the universal mapping property of tensor products. Are you comfortable proving any other isomorphism theorems about tensor products (e.g., $V \otimes (W \oplus W') \cong (V \otimes W) \oplus (V \otimes W')$)? |
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May 26 |
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What does the discriminant of an algebraic number field mean intuitively? The question in the subject line and the question you actually ask are different. The intuition behind a concept may not be closely related to the information it conveys, so clarify what it is you really want to know. |
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May 25 |
answered | What does the discriminant of an algebraic number field mean intuitively? |
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May 25 |
revised |
What's the difference between 'any', 'all' and 'some'? added 599 characters in body |
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May 25 |
answered | What's the difference between 'any', 'all' and 'some'? |
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May 21 |
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Why $F(\alpha_1)∩F(\alpha_2)=F$ is false Your example is false. One that works is $F = {\mathbf Q}$, $\alpha_1 = \sqrt[4]{2}$, and $\alpha_2 = i\sqrt[4]{2}$. Then $F(\alpha_1) \cap F(\alpha_2) = {\mathbf Q}(\sqrt{2})$, which is larger than $F$. |
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May 16 |
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Residue fields of gaussian primes Another way: think about the tower of ideals $(p) \subset (a+bi) \subset {\mathbf Z}[i]$. I hope you can see that $(p)$ has index $p^2$ in ${\mathbf Z}[i]$ (think about what a congruence mod $p{\mathbf Z}[i]$ really means on the real and imaginary parts). Therefore $(a+bi)$ has index 1, $p$, or $p^2$ in ${\mathbf Z}[i]$. Since $a$ and $b$ are not 0, $(a+bi) \not= (p)$. Since $a+bi$ is not a unit (it has norm $p$), $(a+bi) \not= {\mathbf Z}[i]$. So the only choice is for $(a+bi)$ to have index $p$ in ${\mathbf Z}[i]$. |
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May 16 |
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Residue fields of gaussian primes of $f$ includes two square roots of $-1$ inside ${\mathbf Z}[i]/(a+bi)$. Those are the only ones possible because ${\mathbf Z}[i]/(a+bi)$ is a field (the number $a+bi$ is prime in the PID ${\mathbf Z}[i]$, so working mod $a+bi$ is a field). Therefore $i \bmod a+bi$ has to be in the image of $f$, so $f$ is surjective and thus an isomorphism. (Wait, an easier soln: it's easy to solve for $i$ in $a+bi \equiv 0 \bmod a+bi$ because $b$ has an inverse mod $p$.) Thus ${\mathbf Z}[i]/(a+bi)$ has order $p$. The other cases of primes in ${\mathbf Z}[i]$ are simpler. |
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May 16 |
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Residue fields of gaussian primes If $p = 1 \bmod 4$ and $a^2 + b^2 = p$, then $a+bi$ is a factor of $p$, so the ring homomorphism ${\mathbf Z} \rightarrow {\mathbf Z}[i]/(a+bi)$ kills $p$ and thus induces a ring homomorphism $f \colon {\mathbf Z}/(p) \rightarrow {\mathbf Z}[i]/(a+bi)$. It's injective since any ring homomorphism out of a field is injective. The meaty part is showing this is surjective, and for that it's enough to see that $i \equiv c \bmod a+bi$ for an integer $c$. Well, $i \bmod a+bi$ is a square root of $-1$, and $-1$ is a square in ${\mathbf Z}/(p)$ because $p \equiv 1 \bmod 4$. Therefore the image [...] |
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May 16 |
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Irreducibility of $x^n-x-1$ over $\mathbb Q$ @user14284: the congruence $k^{\varphi(m)} \equiv 1 \bmod m$ is not true for all integers $k$, only those $k$ that are rel. prime to the modulus $m$, Your polynomial congruence $x^{\varphi(m)} \equiv 1 \bmod m{\mathbf Z}[x]$ is just false. Try it when $m = 3$: is $x^2 \equiv 1 \bmod 3{\mathbf Z}[x]$? Nope.... |
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May 16 |
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Irreducibility of $x^n-x-1$ over $\mathbb Q$ I would be completely shocked that an elementary argument would succeed. I don't believe there's any known proof of the irreducibility (for general $n$) other than the proof Selmer gave back in the 1950s. Selmer himself points out in his paper that no known irreducibility tests at that time seemed to work on the family of polynomials $x^n-x-1$ for general $n$, and that's why he was led to figure out his proof. |
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May 16 |
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Irreducibility of $x^n-x-1$ over $\mathbb Q$ Not only is the proof of irreducibility of $x^n-x-1$ over ${\mathbf Q}$ nontrivial, but Selmer points out in Section 6 of his paper that the method that he developed to prove irreducibility of these polynomials doesn't seem viable as an irreducibility test for other families of polynomials, aside from the one other family he considers in his paper. |
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May 14 |
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Is there a general algorithm to find a primitive element of a given finite extension (with a finite number of intermediate fields)?) If you read the proof of the primitive element theorem you'll understand why Qiaochu's comment is accurate. |
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May 14 |
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Prime norm ideals that are also principal @Siitd: Read about the Chebotarev density theorem and the Hilbert class field. It is deeper than a "lemma". |
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May 14 |
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Prime norm ideals that are also principal That would be $1/h$, where $h$ is the class number. This comes from class field theory. |
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May 14 |
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Do we really need polynomials (In contrast to polynomial functions)? A noncommutative polynomial in two variables is not a map $K^2 \rightarrow K$, since $xy^2, yxy$, and $y^2x$ are different. You know examples of noncommutative polynomials (well, a quotient ring of the noncommutative polynomials): differential operators with polynomial coefficients. If $D = d/dx$ on smooth functions, and polynomials act on functions as multiplication (so "$x$" means the function $f \mapsto xf$ for smooth $f$), then the product rule $D(xf) = xD(f) + f$ is the same as $Dx = xD + 1$. Polynomial differential operators form a ring ${\mathbf R}[x,D]$ where $Dx = xD + 1$. |
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May 13 |
awarded | Nice Answer |
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May 13 |
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Prime norm ideals that are also principal The prime ideals whose norm is not a prime number have natural (or Dirichlet) density 0. |
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May 13 |
revised |
Do we really need polynomials (In contrast to polynomial functions)? added 25 characters in body |
