In the previous post in this series I promised to do something with fields with characteristic two, and instead I did weird things with surreal numbers and ordinals. Neither of them has characteristic two, because we used the wrong arithmetic. In this post, I will give three new definitions of addition and multiplication in On, and prove that they are actually the same. This will turn On into a field of characteristic two, which we shall call On₂. From know on, we distinguish the ordinary operations from those in On₂ by the use of square brackets. All expressions between $[$ and $]$ are meant in the sense of ordinary arithmetic.

Arithmetic in On₂

Simplicity rules

The most obvious and at the same time unusual way of defining an addition is by starting from zero and working up. We will find the simplest addition and multiplication which make On into a field.

There is no reason why $0+0$ cannot be $0$, because there are fields (any field) with an element satisfying $x+x=x$. This is the first entry in our addition-table. This implies that $0$ must be the zero element, so we must have $0+\alpha=\alpha+0=\alpha$ for all $\alpha$. The first row and column are already filled. What about $1+1$? The least possible answer is $0$, which gives us characteristic two. Next is $1+2$. This cannot be $0$, $1$ or $2$, so we must take $3$. We can go on like this, and make sure that $\alpha + \beta$ is compatible with $\alpha’ + \beta$, $\alpha + \beta’$ and $\alpha’ + \beta’$ (with $\alpha’ < \alpha$ and $\beta’ < \beta$).

We do the same for multiplication. $0.\alpha$ can be $0$, so $0$ must be the zero of the field. Because $1.1 = 1$ is possible, $1$ is the one. The first two rows and columns are filled with $0.\alpha$, $\alpha.0$, $1.\alpha$ and $\alpha.1$. It is obvious that $2.2$ cannot be $0$, $1$ or $2$. Since there are fields (e.g. $\mathbb{F}_4$) with elements that satisfy $x^2 = x + 1$, $3$ is possible. Note that the product has to be compatible with previous entries and with the whole addition-table.

Remark

This definitions are rather difficult to work with, because we must prove a theorem every time we want to fill in an entry. Besides, it is not obvious that these definitions really define a field.

Inductive definitions

Define the minimal excluded number $\DeclareMathOperator{\mex}{mex}\mex(S)$ as the least ordinal not in the set $S$. This can be used for the following inductive definitions:

• $\alpha + \beta = \mex(\alpha’ + \beta, \alpha + \beta’)$
• $\alpha\beta = \mex(\alpha’\beta + \alpha\beta’ + \alpha’\beta’)$

It is easy to verify that $\alpha + \alpha = 0$, because $\alpha + \alpha’$ cannot be zero. We can now prove that this definitions are equivalent to the former.

If $\alpha + \beta < \mex(\alpha’ + \beta, \alpha + \beta’)$, there exists an $\alpha’ < \alpha$ such that $\alpha + \beta = \alpha’ + \beta$. This implies $\alpha = \alpha’$, which is impossible. Therefore, $$\alpha + \beta \geq \mex(\alpha’ + \beta, \alpha + \beta’).$$

If $\alpha\beta < \mex(\alpha’\beta + \alpha\beta’ + \alpha’\beta’)$, there exist $\alpha’$ and $\beta’$ so that $\alpha\beta = \alpha’\beta + \alpha\beta’ + \alpha’\beta’$. This is equivalent to $$\begin{gather}\alpha\beta + \alpha’\beta + \alpha\beta’ + \alpha’\beta’ = 0 \\ (\alpha + \alpha’)(\beta + \beta’) = 0 ,\end{gather}$$ which implies $\alpha = \alpha’$ or $\beta = \beta’$. Both are impossible. It follows that $$\alpha\beta \geq \mex(\alpha’\beta + \alpha\beta’ + \alpha’\beta’).$$

If we can prove that these inductive definitions form a field, it must be the smallest possible field, as defined before. This is a standard verification.

Nim-arithmetic

The inductive definition of the sum is known as nim-addition (frequently used in the theory of the game of Nim). An easy rule to perform nim-addition is:

1. The nim-sum of a number of distinct $2$-powers is their ordinary sum.
2. The nim-sum of two equal numbers is 0.

This rule allows us to compute the nim-sum of finite and infinite ordinals. A similar rule for nim-multiplication is:

1. The nim-product of a number of distinct Fermat $2$-powers (numbers of the form $2^{2^n}$) is their ordinary product.
2. The square of a Fermat $2$-power is its sesquimultiple (multiplying by $\frac{3}{2}$ in the ordinary sense).

Unfortunately, this rule applies only to finite ordinals. A more general rule is explained at neverendingbooks.

Groups in On₂

The ordinals that are groups are precisely the $2$-powers. This can be proved with the simplest extension theorems.

Theorem 1. If $\Delta$ is not a group (under addition), then $\Delta = \alpha + \beta$, where $(\alpha, \beta)$ is any lexicographically earliest pair of numbers in $\Delta$ whose sum is not in $\Delta$.

Theorem 2. If $\Delta$ is a group, we have $[\Delta\alpha] + \beta = [\Delta\alpha + \beta]$, for all $\alpha$, and all $\beta \in \Delta$.

If $\Delta$ is a group, and $\Gamma$ is a group with $\Delta < \Gamma < [\Delta.2]$, we can write $\Gamma = [\Delta + \delta]$ with $\delta < \Delta$. This is a contradiction, because $\Gamma > \Delta + \delta$ follows from the inductive definitions, and $[\Delta + \delta]  = \Delta + \delta$ according to Theorem 2.

If $\alpha, \beta \in [\Delta.2]$, there are three possible cases.

1. $\alpha < \Delta$ and $\beta < \Delta$. Then $\alpha + \beta < \Delta < [\Delta.2]$
2. $\alpha \geq \Delta$ and $\beta < \Delta$. By Theorem 2:
   $\alpha + \beta = [\Delta + \delta] + \beta = \Delta + \delta + \beta = \Delta + \delta’ = [\Delta + \delta'] < [\Delta.2]$
3. $\alpha \geq \Delta$ and $\beta \geq \Delta$. By Theorem 2 and $\alpha + \alpha = 0$:
   $\alpha + \beta = [\Delta + \delta] + [\Delta + \delta'] = \Delta + \Delta + \delta + \delta’ = \delta + \delta’ < \Delta < [\Delta.2]$

This proves that if $\Delta$ is any group, then the next group is $[\Delta.2]$. Because $2$ is a group, it follows that the groups are the $2$-powers. This justifies the rule for the calculation of nim-sums.

Fields in On₂

Similar theorems exist for fields in On₂. Complete proofs can be found in Conway’s On Numbers and Games.

Theorem 3. If $\Delta$ is a group but not a ring, then $\Delta = \alpha\beta$, where $(\alpha, \beta)$ is any lexicographically earliest pair of numbers in $\Delta$ whose product is not in $\Delta$.

Theorem 4. If $\Delta$ is a ring but not a field, then $\Delta = \alpha^{-1}$, where $\alpha$ is the earliest non-zero number in $\Delta$ which has no inverse in $\Delta$.

Theorem 5. If $\Delta$ is a field but not algebraically closed, then $\Delta$ is a root of the lexicographically earliest polynomial having no root in $\Delta$.

Finite ordinals

We will prove by induction that the finite ordinals that are fields are precisely the Fermat $2$-powers. We suppose that the following statements are true for $n$, and prove them for $n + 1$:

1. $[2^{2^n}]$ is a field
2. $[2^{2^{n-1}}]^2 = [\frac{3}{2}2^{2^{n-1}}]$
3. $x^2 + x$ takes precisely the values $0, 1, \dotsc, [2^{2^n-1}-1]$ as $x$ varies in $[2^{2^n}]$

The lexicographically earliest irreducible polynomial over $[2^{2^n}]$ is $x^2 + x = [2^{2^n-1}]$, because $x^2 = \alpha$ always has a root in finite field of characteristic $2$, and $x^2 + x = \alpha$ has a root for earlier $\alpha$ according to statement 3. We know by Theorem 5 that $[2^{2^n}]$ is a root of $x^2 + x = [2^{2^n-1}]$, hence $$\textstyle [2^{2^n}]^2 = [2^{2^n}] + [2^{2^n-1}] = [2^{2^n} + 2^{2^n-1}] = [\frac{3}{2}2^{2^n}].$$ We obtain the field $[2^{2^{n+1}}]$ as a vector space over $[2^{2^n}]$ with typical element $X = [2^{2^n}]x + y$. We examine the polynomial $$\begin{align}X^2 + X &= ([2^{2^n}]x + y)^2 + [2^{2^n}]x + y \\ &= [2^{2^n}]^2 x^2 + y^2 + [2^{2^n}]x + y \\ &= [2^{2^n}](x^2 + x) + ([2^{2^n-1}]x^2 + y^2 + y).\end{align}$$ By induction,  $x^2 + x$ can take any value in $[2^{2^n-1}]$. Note that $x^2 + x$ remains unchanged when we replace $x$ by $x + 1$. The same is true for $y^2 + y$. It follows that $[2^{2^n-1}]x^2 + y^2 + y$ can be made to take any value in $[2^{2^n}]$ without affecting the value of $x^2 + x$. This implies that the values of $X^2 + X$ can be written as $[2^{2^n}]\alpha + \beta$, where $\alpha < [2^{2^n-1}]$ and $\beta < [2^{2^n}]$, which are precisely the values less than $[2^{2^{n+1}-1}]$.

This and Theorem 6 justify the rule for the calculation of nim-products.

Infinite ordinals

Consider the sequence $$[\omega^{\omega^k}], [\omega^{\omega^k p_k}], [\omega^{\omega^k p_k^2}], \dotsc, [\omega^{\omega^k p_k^n}], \dotsc$$ where $p_k$ is the $(k+1)$’st prime. Then the following statements are true for each $k > 0$:

1. Each term in the sequence is a field
2. The field $[\omega^{\omega^k p_k^n}]$ is the union of all finite fields $\mathbb{F}_{2^{p_0^{n_0} p_1^{n_1} \dotsm p_k^{n_k}}}$ with $n_i < \omega$ for $0 \leq i \leq k – 1$ and $n_k \leq n$
3. Each term is the $p_k$’th power of its successor, and $[\omega^{\omega^k}]$ is the $p_k$’th root of $\alpha_{p_k}$, which is the least number in $[\omega^{\omega^k}]$ with no $p_k$’th root in $[\omega^{\omega^k}]$.

We will prove this by induction on $k$.

$\boldsymbol{n = 0}$

$[\omega^{\omega^{k+1}}]$ is the union of all fields $[\omega^{\omega^k p_k^n}]$. It is obvious that this defines a field, and there are no fields in between. This proves statement 1, and statement 2 follows immediately. Because of Theorem 5, $[\omega^{\omega^{k+1}}]$ is the root of the lexicographically earliest polynomial having no root in $[\omega^{\omega^{k+1}}]$. If $f(x)$ is a polynomial of degree $d < p_{k+1}$, all coefficients are contained in a finite field $\mathbb{F}_{2^{p_0^{n_0} \dotsm p_k^{n_k}}}$. Therefore, the root of $f(x)$ is an element of the field $\mathbb{F}_{2^{p_0^{n_0} \dotsm p_k^{n_k} d}} = \mathbb{F}_{2^{p_0^{m_0} \dotsm p_k^{m_k}}}$, which is a subfield of $[\omega^{\omega^{k+1}}]$. It follows that the earliest irreducible polynomial is $x^{p_{k+1}} = \alpha_{p_{k+1}}$ with $\alpha_{p_{k+1}}$ as defined in statement 3.

$\boldsymbol{n > 0}$

Assume $\Gamma = [\omega^{\omega^{k+1} p_{k+1}^{n-1}}]$ is a field, and $\Delta$ is the lexicographically earliest algebraic extension. The field $\Gamma$ is not closed for polynomials of degree $p_{k+1}$, because $\mathbb{F}_{2^{p_{k+1}^n}}$ is a field extension of $\mathbb{F}_{2^{p_{k+1}^{n-1}}} \subset \Gamma$ of degree $p_{k+1}$, and $\mathbb{F}_{2^{p_{k+1}^n}}$ is not contained in $\Gamma$. This means that $[\Delta:\Gamma]$ is at most $p_{k+1}$. Therefore, every element $\alpha \in \Delta$ is the root of a polynomial $f(x)$ of degree $d \leq p_{k+1}$. By induction, all coefficients of $f(x)$ are contained in a field $\mathbb{F}_{2^{p_0^{n_0} \dotsm p_{k+1}^{n_{k+1}}}}$ with $n_{k+1} \leq n – 1$. It follows that the root of $f(x)$ is contained in $\mathbb{F}_{2^{p_0^{n_0} \dotsm p_{k+1}^{n_{k+1}}d}} = \mathbb{F}_{2^{p_0^{m_0} \dotsm p_{k+1}^{m_{k+1}}}}$ with $m_{k+1} \leq n$. If $d < p_{k+1}$, this is a subfield of $\Gamma$ and $f(x)$ is not irreducible. We can conclude that $[\Delta:\Gamma] = p_{k+1}$, and $\Delta = [\omega^{\omega^{k+1} p_{k+1}^{n}}]$. This proves statements 1 and 2.

If $f(x)$ is a polynomial $x^{p_{k+1}} = \alpha$ with $\alpha \in [\omega^{\omega^{k+1} p_{k+1}^{n-1}}]$, we know by induction that $\alpha$ is contained in $\mathbb{F}_{2^{p_0^{n_0} \dotsm p_{k+1}^{n_{k+1}}}}$ with $n_{k+1} \leq n – 1$. The root of $f(x)$ is thus contained in $\mathbb{F}_{2^{p_0^{m_0} \dotsm p_{k+1}^{m_{k+1}}}}$ with $m_{k+1} \leq n$, which is a subfield of $[\omega^{\omega^{k+1} p_{k+1}^{n}}]$. It follows that the earliest irreducible polynomial is $x^{p_{k+1}} = [\omega^{\omega^{k+1} p_{k+1}^{n-1}}]$. By Theorem 5, $[\omega^{\omega^{k+1} p_{k+1}^n}]$ is a root of this polynomial. This proves statement 3.

From this, we can conclude that $[\omega^{\omega^\omega}]$ is the algebraic closure of $2$.

Remark

The computation of $\alpha_p$ is not a trivial task. Conway did stop at $\alpha_7$. Hendrik Lenstra described an effective method in his paper On the algebraic closure of two, and computed $\alpha_p$ for $p \leq 43$. Lieven Le Bruyn extended the list.

About Pieter Goetschalckx

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Last time, I proved that the Hasse principle is valid for Kronecker-Weber Theorem. So in order to prove Kronecker-Weber, the local version still has to be proved. As noted before, you only need to prove it for Galois extensions $K/\mathbb{Q}_p$ with Galois group $\mathbb{Z}/q^m\mathbb{Z}$, where $q$ is a prime number that can be equal to $p$. There will be 3 cases to prove:

• The extension is unramified.
• The extension is ramified, but the ramification degree $e$ is not divisible by $p$ (also known as tamely ramified).
• The extension is ramified with ramification degree $e$ divisible by $p$ (wildly ramified).

In all these cases, we will have need for Hensel’s Lemma:

Hensel’s Lemma Let $L$ be a local field, with ring of integers $\mathcal{O}_L$ and maximal ideal $\mathfrak{b}$. Let $l$ be it’s residue field. Suppose $f \in \mathcal{O}_L [x]$, monic with restriction $\bar{f}$ in $l[x]$. Suppose$\bar{f}$ has a root $\alpha$ in $l$, such that $\bar{f}’(\alpha) \neq 0$. Then there exists a root $\beta \in \mathcal{O}_L$ of $f$, with $\beta \equiv \alpha \bmod \mathfrak{b}$.

Let us start the unramified case, which will be the easiest. We will use the notation from the previous post.

$K/\mathbb{Q}_p$ is unramified

There is a much stronger result for this, which I will prove here. Taking $K = \mathbb{Q}_p$ will give what we want.

Theorem Suppose $L/K$ is an unramified, finite Galois extension, where $L$ and $K$ are finite extensions of $\mathbb{Q}_p$ (not necessarily Galois over $\mathbb{Q}_p$). Then there exists an $n \in \mathbb{N},p \nmid n$ so that $L = K(\zeta_n)$. Furthermore, $Gal(L/K)$ will be cyclic.

Proof Take $L/K$ like in the theorem, with prime $\mathfrak{p}$ in both extensions, since $e=1$. Let $l$ and $k$ be the residue fields of respectively $L$ and $K$. Since $e = 1$, $Gal(L/K) = Gal(l/k)$. The right hand side is a Galois group of an extension in finite fields, hence it is cyclic. Since $l/k$ is a Galois extension, there exists a primitive element $\alpha$ so that $l = k[\alpha]$. Consequently, there exists an $n$, with $\gcd(n,p)=1$, such that $X^n-1$ has a root in $l$, namely $\alpha$. Using Hensel’s Lemma, we find a $\beta \in \mathcal{O}_L$ as a root of $X^n-1$ and $\beta \equiv \alpha \bmod \mathfrak{p}$. Since $\beta$ is a root of $1$ and, due to being equivalent to $\alpha$, has order at least $n$, we have $$\left[K(\beta):K\right] \geq \left[k[\alpha]:k\right] = \left[l:k\right] = \left[L:K\right].$$ But $\beta \in L$, so we get $L = K(\beta)$ and $\beta = \zeta_n$.

This was the easy one, the next two cases are harder.

$K/\mathbb{Q}_p$ is tamely ramified.

This case follows from 3 lemmas. I will give the proofs depending on how difficult or technical it is.

Lemma Let $p\neq 2$ be a prime. Then $\mathbb{Q}_p$ contains the $p-1$-th roots of unity, but none other.

Proof Let $\mu_p$ be the set of roots of unity in $\mathbb{Q}_p$. Thanks to Hensel’s Lemma, we have a surjective morphism $\phi:\mu_p \rightarrow \mathbb{F}_p^*$, so we need to prove that it is injective. So suppose we have a root of unity of $(1+tp)^n$. Using Newton’s binomial formula, we get $$\sum_{i = 1}^n \binom{n}{i} (tp)^{i}=0.$$ Now, there are 2 cases: $t = 0$ and we are done, or $$ \sum_{i = 1}^n \binom{n}{i} (tp)^{i-1}=0,$$ but then, since $p|0$, $p|\binom{n}{1}=n.$ Since $\mathbb{Q}_p$ contains the $n$-th root of $1$ and $p|n$, $\mu_p$ must contain the $p$-th roots of unity. So we are reduced to $(1+tp)^p = 1$. But then we have $$\sum_{i = 1}^p \binom{p}{i} (tp)^{i}=0.$$ If $t \neq 0$, we have $$\sum_{i = 1}^p \binom{p}{i} (tp)^{i-1}=0.$$But every term is divisible by $p^2$ except for the first one, which is only divisible by $p$, so we have a contradiction. This concludes the proof.

In case of $p=2$, $\mathbb{Q}_2$ only has $-1,1$ as roots of unity. This is proved by noticing that $(\mathbb{Z}/8\mathbb{Z})^* = \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$.

Lemma Let $L/K/\mathbb{Q}_p$ be a tower of finite extensions, with $L/K$ Galois. Let $\mathfrak{p}_K$ be the unique maximal ideal of $K$ and suppose that that $L/K$ is totally ramified of degree $e$. Then there exists a $\pi \in K$ a generator of $\mathfrak{p}_K$ and a root $\alpha$ of the polynomial $f(X) = X^e – \pi$, such that $L = K(\alpha)$.

Proof Use of the non-archimedean absolute value, choosing a uniformizing parameter and using the fact that elements of the Galois group preserve absolute values, this is a long but not very difficult proof and will be omitted.

Lemma $\mathbb{Q}_p((-p)^{\frac{1}{p-1}}) = \mathbb{Q}_p(\zeta_p)$

Proof It can be proved that the maximal ideal of $\mathbb{Q}_p(\zeta_p)$ is given by $(1-\zeta_p)$ and that $(p) = (1-\zeta_p)^{p-1}$. Consequently, $(1-\zeta_p)^p = (p(1-\zeta_p))$. Take the polynomial $$g(X) = \frac{(X+1)^p-1}{X}=\sum_{i=1}^{p} \binom{p}{i}X^{i-1},$$ which is the minimal polynomial of $\zeta_p-1$, so we get $$0 = g(\zeta_p) \equiv (\zeta_p-1)^{p-1}+p \bmod (\zeta_p-1)^p.$$ This can be modified to $$u = \frac{(\zeta_p-1)^{p-1}}{-p} \equiv 1 \bmod (\zeta_p-1).$$ Now look at the polynomial $f(X) = X^{p-1}- u$. Then $f(1) \equiv 0 \bmod (\zeta_p – 1)$ and $(\zeta_p – 1)\nmid f’(1)$. Hensel’s Lemma (again) implies that there exists an $\alpha \in \mathbb{Q}_p(\zeta_p)$ as a root of $f$. But $(-p)^{\frac{1}{p-1}} = \frac{\zeta_p-1}{\alpha}$, so $(-p)^{\frac{1}{p-1}} \in \mathbb{Q}_p(\zeta_p)$. Now, the minimal polynomial of $(-p)^{\frac{1}{1-p}}$ is $X^{p-1}+p$, since it is irreducible due to Eisenstein’s criterion, so $(-p)^{\frac{1}{1-p}}$ and $\zeta_p$ have the same degree over $\mathbb{Q}_p$. This gives $\mathbb{Q}_p((-p)^{\frac{1}{1-p}}) = \mathbb{Q}_p(\zeta_{p})$.

Now, for the proof of the tamely ramified case. Let $L/\mathbb{Q}_p$ be a tamely ramified abelian extension of ramification degree $e, p\nmid e$, with $K/\mathbb{Q}_p$ the maximal unramified subextension. We already know that $K \subset \mathbb{Q}_p(\zeta_n)$ for some $n$. We also have $L/K$ is totally ramified with degree $e$. From one of the above lemmas, we know that $L = K(\pi^{\frac{1}{e}})$ with $\pi$ a generator of the unique maximal ideal $\mathfrak{p}_K$ of $\mathcal{O}_K$. $K/\mathbb{Q}_p$ is unramified, so $p$ is also a generator of $\mathfrak{p}_K$. This gives $\pi = -up$ for some unit $u$ of $\mathcal{O}_K$. $u$ is a unit, $p \nmid e$, so the discriminant of $f(X) = X^e-u$ is not divisible by $p$. This implies that $K(u^{\frac{1}{e}})/K$ is unramified. $K \subset \mathbb{Q}_p(\zeta_n)$, so $$K(u^{\frac{1}{e}}) \subset K(\zeta_M) \subset \mathbb{Q}_p(\zeta_{Mn}),$$ with $M$ an integer. Let $T$ be the compositum of $\mathbb{Q}_p(\zeta_{Mn})$ and $L$. $T$ is an abelian extension of $\mathbb{Q}_p$ and $u^{\frac{1}{e}},\pi^{\frac{1}{e}} \in T$, which gives $(-p)^{\frac{1}{e}} \in T$. Since $\mathbb{Q}_p((-p)^{\frac{1}{e}}) \subset T$ and $T/\mathbb{Q}_p$ is abelian, it is an Galois extension over $\mathbb{Q}_p$. Since all the roots of $X^e +p$ are $\zeta_e^k (-p)^{\frac{1}{e}}, k=0 \ldots e-1$, we know that $\zeta_e \in \mathbb{Q}_p((-p)^{\frac{1}{e}})$. $\mathbb{Q}_p((-p)^{\frac{1}{e}})$ is totally ramified and since every subfield of a totally ramified subfield is totally ramified in case of local fields, $\mathbb{Q}_p(\zeta_e)$ is also totally ramified. But $p \nmid e$, so it can’t ramify, except when it’s the trivial extension, so $\zeta_e \in \mathbb{Q}_p$ and $e | p-1$. We have a tower of extensions $$\mathbb{Q}_p((-p)^{\frac{1}{e}}) \subset \mathbb{Q}_p((-p)^{\frac{1}{p-1}}) \subset \mathbb{Q}_p(\zeta_p).$$ Consequently, $$L = K(\pi^{\frac{1}{e}}) \subset K(u^{\frac{1}{e}},(-p)^{\frac{1}{e}}) \subset \mathbb{Q}_p(\zeta_{Mnp}),$$ which concludes this case.

$K/\mathbb{Q}_p$ is widely ramified

This is the hardest case, so I won’t give the entire proof, but I will describe the way the proof works. We know that we just have to work with extensions with Galois group $\mathbb{Z}/q^{m}\mathbb{Z}$, with $q$ a prime. Since we work in the widely ramified case, this breaks down to $q = p$. There are 2 cases to consider: $p = q = 2$ or $p = q \neq 2$ (2 is always a bad prime). Let us first proof the last case.

Take an abelian widely ramified extension $L/\mathbb{Q}_p$ of degree $p^m$, with cyclic Galois group. We first construct 2 Galois extensions of $\mathbb{Q}_p$ of degree $p^m$, one totally ramified, the other unramified:

• Consider the extension $\mathbb{Q}_p(\zeta_{p^{m+1}})$ with Galois group $\mathbb{Z}/p^{m+1}\mathbb{Z}$ and let $K_r$ be the fixed field by the subgroup of order $p-1$. This will be a totally ramified extension of degree $p^m$ with cyclic Galois group.
• Take an irreducible polynomial in $\mathbb{F}_p[X]$ of degree $p^m$, lift it to an irreducible polynomial in $\mathbb{Z}_p[X]$ and construct the corresponding Galois extension. This will give an unramified extension $K_u$ with Galois group $\mathbb{Z}/p^m\mathbb{Z}$.

As we know from the unramified case, $K_u \subset \mathbb{Q}_p(\zeta_n)$ for some $n$. Since one extension is unramified and the other is totally ramified, we necessarily have $K_u \cap K_v = \mathbb{Q}_p$ and consequently, $Gal(K_uK_v/\mathbb{Q}_p) = (\mathbb{Z}/p^m\mathbb{Z}^2)$. Suppose that $L \not\subset \mathbb{Q}_p(\zeta_n,\zeta_{p^{m+1}})$, then we should have $$Gal(L(\zeta_n,\zeta_{p^{m+1}})) = (\mathbb{Z}/p^m\mathbb{Z})^2 \times\mathbb{Z}/p^k\mathbb{Z},$$ with $k > 0$. This implies that there is a Galois extension of $\mathbb{Q}_p$ with Galois group $(\mathbb{Z}/p\mathbb{Z})^3$. The next part is to prove that there doesn’t exist such an extension, but this is very hard and requires knowledge of Kummer Theory..

The case $p=q=2$ is similar. One concludes that there should exist a Galois extension of $\mathbb{Q}_2$ with Galois group $(\mathbb{Z}/2\mathbb{Z})^4$ or $(\mathbb{Z}/4\mathbb{Z})^3$ and once again it is proved that this is impossible.

About Kevin De Laet

Interested in mathematics, specifically Group & Representation theory, currently studying in Antwerp (and sometimes Brussels)

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Finite fields might seem to be easy to understand, but they are definitely not. Some examples of our limited knowledge are:

• We represent the field $\mathbb{F}_{p^n}$ by $\mathbb{F}_p[x]/(f(x))$ whith $f(x)$ a monic irreducible polynomial of degree $n$. The polynomial $f$ is not unique, and there is no best option.
• We can’t extend $\mathbb{F}_{p^n}$ to $\mathbb{F}_{p^{n+1}}$ with compatible addition and multiplication.
• Simple problems like placing knights at a table are too difficult.
• We know what the algebraic closure of a finite field is, but how can we do calculations in such an exotic thing?

Conway polynomials are a good example of the danger of solving these problems.

Therefore a better description of finite fields is necessary. For fields with characteristic two there is a good candidate using ordinal numbers with nim-addition and nim-multiplication. In this series I will work out this example. Let’s start by defining numbers.

Numbers

In ONAG, Conway defines (surreal) numbers as follows:

Definition. If $L$, $R$ are any two sets of numbers, and no member of $L$ is $\geq$ any member of $R$, then there is a number $\{L \vert R\}$. All numbers are constructed in this way.

The last sentence is somewhat informal. It means there is no sequence of numbers $x_i = \{L_i \vert R_i\}$ with $x_{i+1} \in L_i \cup R_i$ for all $i \in \mathbb{N}$.

Notation. If $x = \{L \vert R\}$ we write $x^L$ and $x^R$ for the typical member of $L$ resp. $R$. For $x$ itself we then write $\{x^L \vert x^R\}$.

Note that all the definitions are inductive, and don’t need a basis because they start with the empty set. Each element of the empty set has all desired properties because the empty set has no members. This allows us to make a lot of proofs very short. If we can proof that for a property $P$, $P(x)$ is implied by $P(x^L)$ and $P(x^R)$, then $P$ must be true for all numbers. Otherwise, there would be a number $x$ for which $P(x)$ is false. Hence, there is a $x^L$ or $x^R$ for which $P(x^\bullet)$ is false. We can go on like this, but since that would create an infinite descending sequence of numbers, $P(x)$ has to be true.

The relations on numbers are defined as:

• $x \leq y$ iff $x < y^R$ and $x^L < y$
• $x \geq y$ iff $x > y^L$ and $x^R > y$
• $x = y$ iff $x \leq y$ and $x \geq y$

This is a total order. A remarkable property is that $x^L < x < x^R$ for all numbers $x$. We now define an addition and multiplication. We know that $x + y$ must lie between both $x^L + y$ and $x + y^L$ (on the left) and $x^R + y$ and $x + y^R$ (on the right). From $x – x^L > 0$ and $y – y^L > 0$ we can deduce $(x – x^L)(y – y^L) > 0$, so that we must have $xy > x^Ly + xy^L – x^Ly^L$. This motivates following definition:

• $x + y = \{x^L + y, x + y^L \vert x^R + y, x + y^R \}$
• $xy = \{ x^Ly + xy^L – x^Ly^L, x^Ry + xy^R – x^Ry^R \vert x^Ly + xy^R – x^Ly^R, x^Ry + xy^L – x^Ry^L \}$

With this addition and multiplication, the Class No of all surreal numbers forms a Field, but not a field (because No is not a set).

Examples of numbers

According to the definition, every number is constructed as two sets of earlier constructed numbers. The only way to get it off the ground is by using empty sets. We call $\{ \vert \}$ the number $0$, born on the zeroth day.

Starting from $0$, the next generation of numbers can be constructed. We obtain $1 = \{ 0 \vert \}$ and $-1 = \{ \vert 0 \}$. It is easy to verify that $\{ 0 \vert 0 \}$ is not a number. We have $-1 < 0 < 1$. This was the first day.

The number $\{ 0 \vert 1 \}$ lies somewhere between $0$ and $1$. We call this number $\frac{1}{2}$. What about $\{ -1, 0 \vert 1 \}$? We can verify the inequalities $\{ -1, 0 \vert 1 \} \leq \{ 0 \vert 1 \}$ and $\{ -1, 0 \vert 1 \} \geq \{ 0 \vert 1 \}$, so we have two expressions for the same number. They are equal, but not identical.

We can go on like this with $\frac{1}{4} = \{ 0 \vert \frac{1}{2} \}$ and $\frac{3}{4} = \{ \frac{1}{2} \vert 1 \}$. The whole number $n$ is born on day $n$ as $\{ n-1 \vert 0 \}$. All dyadic rationals and whole numbers are born on finite days. On day $\omega$, $L$ and $R$ can be infinite sets. Examples are:

• $\frac{1}{3} = \{ \frac{1}{4}, \frac{1}{4} + \frac{1}{16}, \frac{1}{4} + \frac{1}{16} + \frac{1}{64}, \ldots \vert \frac{1}{2}, \frac{1}{2} – \frac{1}{8}, \ldots \}$
• $\omega = \{ 0, 1, 2, 3, \ldots \vert \}$
• $\frac{1}{\omega} = \{ 0 \vert 1, \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \ldots \}$

On day $\omega + 1$ it’s starting to get a little strange. The number $x = \{ 0, 1, 2, 3, \ldots \vert \omega \}$ satisfies $n < x < \omega$ for all finite integers $n$, and $x = \omega – 1$.

Subsets

The surreal numbers are the largest possible totally ordered field. The rationals, the reals, the ordinals and all other ordered fields are subsets of No. An interesting subclass of No is On, the class of all ordinal numbers.

Definition. $\alpha$ is an ordinal number if $\alpha$ has an expression of the form $\alpha = \{L\vert \}$.

No is a proper Class (not a set), but for every ordinal $x$ the subclass $\{ a : a < x \}$ is a set. Because $\alpha = \{ \beta : \beta < \alpha \vert \}$, we can treat $\alpha$ as the set of all lesser ordinals.

About Pieter Goetschalckx

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