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17 votes
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Does there exist a pair of infinite fields, the additive group of one isomorphic to the multiplicative group of the other?

A pair of fields like this exists. As noted in the question, one may assume that $E$ is a field of characteristic zero. Therefore $E^+$ is an infinite, torsion-free, divisible, abelian group; i.e. an ...
Keith Kearnes's user avatar
12 votes
Accepted

Let $R$ be a commutative unital ring. Is it true that the group of units of $R$ is not isomorphic with the additive group of $R$?

Counterexample: $R=\mathbb{R}\times\mathbb{Z}_2$ satisfies $(R,+)\cong(R^\times,\cdot)$.
anon's user avatar
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11 votes
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Cyclic Artin-Schreier-Witt extension of order $p^2$

Let's begin with a usual Artin-Schreier extension $k(x)/k$ with $F(x)-x=f$ where $F$ is the endomorphism of $k$ raising everything to power $p$, and $f\in k$ is not of the form $g^p-g$ for any $g\in k$...
Jyrki Lahtonen's user avatar
9 votes
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Can a separable isogeny of elliptic curves have an inseparable dual?

Just so this has an answer, let me expand upon what I said in the comments slightly. So, let us begin with the following trivial observation: Observation: If $f:E\to E'$ is an isogeny of elliptic ...
Alex Youcis's user avatar
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9 votes
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Classifiying algebraic group extensions of $\mathbb{Z}/p\mathbb{Z}$ by $\mu_p$

I am not sure what Milne has taught you up until this point, but I will tell you how I would approach this problem. Cohomology and extensions The main point here is that calculating extensions is hard,...
Alex Youcis's user avatar
  • 54.3k
8 votes

Prove that a polynomial is irreducible or the field contains a $p$th root

One side: If $a=b^{p}$ for $b\in k$ then: $$x^{p^{n}}-a=x^{p^{n}}-b^{p}=\left(x^{p^{n-1}}-b\right)^{p}$$ showing that the polynomial is not irreducible. Let $k^{\mathbf{a}}$ denote an algebraically ...
drhab's user avatar
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8 votes
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Peirce decomposition of a ring: must the ideal generators be idempotent in characteristic 2?

Just observe that $ef=0$ means $e(1-e)=0$ so $e-e^2=0$ and $e^2=e$.
Eric Wofsey's user avatar
7 votes
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The group algebra is not semisimple if characteristic divides group order.

As for the question at hand, I can't think of anything better than what Matthew Towers has already suggested in the comment. What I suggest below is probably closely related to that observation. ...
rschwieb's user avatar
  • 154k
6 votes

Let $R$ be a commutative ring with unity of prime characteristic $p$. If $a\in R$ is nilpotent then is $1+a$ unipotent?

Yes, because we have $$(1+a)^{p^e} = 1+a^{p^e} = 1 \text{ for } e \gg 0.$$
MooS's user avatar
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6 votes

Peirce decomposition of a ring: must the ideal generators be idempotent in characteristic 2?

No, it has nothing to do with characteristic $2$. The representations of elements as $i+j$ in a direct sum are necessarily unique (that is an equivalent way of saying it.) So to continue from what ...
rschwieb's user avatar
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6 votes
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If the characteristic of a field is two, doesn't the quadratic form associated with a bilinear form exist?

If $b:V\times V\to K$ is a symmetric bilinear form, you can always associate a quadratic form $q_b:V\to K$ by setting $q_b(x)=b(x,x)$ for all $x\in V$. If $q:V\to K$ is a quadratic form, by very ...
GreginGre's user avatar
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6 votes

Example of ring can’t be defined over finite field.

You can take any central division $K$-algebra A of dimension $>1$ over $K$. Then it cannot be defined over a finite field. Otherwise, we would have $A\simeq A_0\otimes_FK$, and $A_0$ would be a ...
GreginGre's user avatar
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6 votes
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Example of ring can’t be defined over finite field.

Let $K = \mathbb{F}_p (X)$ and let $A = K [Y] / (Y^n - X)$, where $n$ is a prime other than $p$. So $A$ is a finite field extension of $K$. I say $A$ is not defined over any finite subfield of $K$. ...
Zhen Lin's user avatar
  • 90.4k
5 votes

Examples of uncountable fields of characteristic $p$?

A "natural" example is the field $k((x))$ of Laurent series over a field $k$ of characteristic $p$. (For example take $k = \mathbb{F}_p = \mathbb{Z}/p\mathbb{Z}$.) This is the fraction field ...
Ben Blum-Smith's user avatar
5 votes

Can a finite ring have a characteristic $0$?

I'm not sure whether your ring has unity, so here is a general argument. If a ring $R$ has characteristic zero, then, for each $n>0$, there is some $a\in R$ with additive order $>n$. This ...
Andrew Dudzik's user avatar
5 votes
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Do invariants and co-invariants for a cyclic group have the same dimension?

Well, $V^{C_p}$ and $V_{C_p}$ are the kernel and cokernel of $V\xrightarrow{g-1}V$, $g$ a generator of $C_p$, so they have the same dimension.
user10354138's user avatar
  • 33.3k
5 votes
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Why is the Levi-decomposition not available in positive characteristic?

The short answer is, that there are counterexamples in characteristic $p$, see for example this MO-post. Moreover, most results of classical Lie algebra theory fail in characteristic $p>0$, such as ...
Dietrich Burde's user avatar
4 votes

Canonical connection on the Frobenius pull-back

I haven't thought concretely about how this connection should look in your example, though I think it is uniquely characterized on local sections by the claim that it annihilates sections of $F^*L$ (...
Simone Weil's user avatar
  • 2,150
4 votes

Does $2x=0$ imply $x=0$ in a field with characteristic not equal 2.

Yes, just multiply both sides of the equation $2x = 0$ by the inverse of $2$ to get $x = 0$.
Alex Provost's user avatar
  • 21.1k
4 votes

Prove that a polynomial is irreducible or the field contains a $p$th root

Rename the field $k$ as $K$. Rename the indeterminate $x$ as $X$. Thus we must prove that either the polynomial $X^{p^n}-a$ is irreducible over $K$ or $a$ is a $b$-th power. In an extension field of $...
Angina Seng's user avatar
4 votes
Accepted

Characteristic of a finite field - equivalent formula

It follows immediately from the distributivity of multiplication over addition: $$a+a+\ldots+a=a\cdot(1+1+\ldots+1)=a\cdot0=0.$$
Servaes's user avatar
  • 63.4k
4 votes
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Why does the cardinality of the vector space over a finite field of characteristic $p$ have to be a power of $p$?

Suppose $F$ has basis $\{v_1, v_2, \ldots, v_m\}$ over $\mathbb{F}_p$. Then each element $x$ of $F$ is uniquely expressible as $$ x = c_1v_1 + c_2v_2 + \cdots +c_mv_m. $$ But there are $p=|\mathbb{F}...
Randall's user avatar
  • 19.1k
4 votes
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A homomorphism on a nontrivial commutative ring with trivial unit group

Let $r\in R$ be such that $f(r)=0$. Then $r^2=0$ and hence $$(r+1)^2=r^2+2r+1=1.$$ This shows that $r+1$ is a unit, so $r+1=1$ and hence $r=0$. This shows that $f$ is injective. To see that it is not ...
Servaes's user avatar
  • 63.4k
4 votes

Finite Characteristic of a Ring

Consider the ring of polynomials with coefficients in $\mathbb{Z}/p$ it has characteristic $p$ and it is infinite.
Tsemo Aristide's user avatar
4 votes
Accepted

Character group of diagonalizable group has no $p$-torsion

Since $K$ has characteristic $p$, it has no nontrivial $p$th roots of unity. The polynomial $x^p-1$ factors as $(x-1)^p$ so the only $p$th root of unity is $1$.
Eric Wofsey's user avatar
3 votes

Is the Polynomial $X^{p^n}-X$ the zero polynomial in characteristic $p$?

$X^{p^n}-X$ is clearly not the zero polynomial because it has nonzero coefficients. However, $X^{p^n}-X$ does induce the zero function. In fact, the polynomials that induce the zero function are ...
lhf's user avatar
  • 217k
3 votes

The local ring $(R,\mathfrak{m})$ contains a field if and only if $\mathrm{char}(R) = \mathrm{char} (R/\mathfrak{m})$.

The result you want is that $R$ contains a field $K$ iff $\operatorname{char}(R) = \operatorname{char}(R/\mathfrak{m}).$ $\implies$ If $K\subseteq R,$ then clearly $K$ and $R$ must have the same ...
Stahl's user avatar
  • 23.3k
3 votes
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About a refined Burnside's theorem in prime characteristic.

The version of Burnside's theorem given in B.A.F. Wehrfritz' book "Infinite Linear Groups" as Corollary 1.23 states that a completely reducible subgroup of $GL(n, F)$ of finite exponent $e$ has finite ...
j.p.'s user avatar
  • 1,176
3 votes
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Determinant of Killing form of sl_n

Here's a proof of your formula. Let's compute things in the standard (not adjoint) representation. For matrices $A,B$, define $s(A,B)=\mathrm{Tr}(AB)$. First, in the basis of all matrices, the ...
YCor's user avatar
  • 18k

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