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In the book of Linear Algebra by Werner Greub, whenever we choose a field for our vector spaces, we always choose an arbitrary field $F$ of characteristic zero, but to understand the importance of the this property, I am wondering what would we lose if the field weren't characteristic zero ?

I mean, right now I'm in the middle of the Chapter 4, and up to now we have used the fact that the field is characteristic zero once in a single proof, so as main theorems and properties, if the field weren't characteristic zero, what we would we lose ?

Note, I'm asking this particular question to understand the importance and the place of this fact in the subject, so if you have any other idea to convey this, I'm also OK with that.

Note: Since this a broad question, it is unlikely that one person will cover all the cases, so I will not accept any answer so that you can always post answers.

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    $\begingroup$ It's perhaps surprising when you first encounter vector spaces over finite fields how much /doesn't/ change. One perhaps obvious difference is that fields of positive characteristic can have only finitely many elements, and hence vector spaces that have only finitely many elements. Thus, the familiar statement from real linear algebra that a linear equation $A {\bf x} = {\bf b}$ has $0$, $1$, or infinitely many solutions is no longer true. $\endgroup$ – Travis Jun 30 '17 at 13:02
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    $\begingroup$ Arguably the most interesting differences show up when you use vector spaces over finite fields to do representation theory. In the generic case, when the dimension of the representation is not divisible by the characteristic, a lot of the theory formally looks like what happens in the real/complex case. But the behavior of representations whose dimension is divisible by the characteristic can be comparatively strange. $\endgroup$ – Travis Jun 30 '17 at 13:05
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    $\begingroup$ "The theory of quadratic forms over a field of characteristic 2 has important differences and many definitions and theorems must be modified." Wikipedia. $\endgroup$ – lhf Jun 30 '17 at 13:13
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    $\begingroup$ @Travis That's (as you also mention) more about finite vs. infinite fields, not about the characteristic $\endgroup$ – Hagen von Eitzen Jun 30 '17 at 13:26
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    $\begingroup$ @Travis "fields of positive characteristic can have only finitely many elements": I think you know what you meant, but this can easily be interpreted as a false statement. Might I suggest that "might" would be a better verb to use here than "can"? $\endgroup$ – Marc van Leeuwen Jun 30 '17 at 14:27
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The equivalence between symmetric bilinear forms and quadratic forms given by the polarization identity breaks down in characteristic $2$.

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    $\begingroup$ Your answer is very short; the link says (quote): The polarization identities are not restricted to inner products. If $B$ is any symmetric bilinear form on a vector space, and $Q$ is the quadratic form defined by $$Q(v) = B(v,v),\,\!$$ then $$\begin{align} 2 B(u,v) &= Q(u+v) - Q(u) - Q(v), \\ 2 B(u,v) &= Q(u) + Q(v) - Q(u-v), \\ 4 B(u,v) &= Q(u+v) - Q(u-v). \end{align}$$ (end quote). I think in characteristic other than two, one would divide both sides of this equation by $2$, and conclude that the bilinear form $B$ can be reconstructed from the "norm" $Q$. Read more at the link. $\endgroup$ – Jeppe Stig Nielsen Jul 1 '17 at 22:10
  • $\begingroup$ (continued) But when you cannot divide by two, it will not work. $\endgroup$ – Jeppe Stig Nielsen Jul 1 '17 at 22:12
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Many arguments using the trace of a matrix will no longer be true in general. For example, a matrix $A\in M_n(K)$ over a field of characteristic zero is nilpotent, i.e., satisfies $A^n=0$, if and only if $\operatorname{tr}(A^k)=0$ for all $1\le k\le n$. For fields of prime characteristic $p$ with $p\mid n$ however, this fails. For example, the identity matrix $A=I_n$ then satisfies $\operatorname{tr}(A^k)=0$ for all $1\le k\le n$, but is not nilpotent.
The pathology of linear algebra over fields of characteristic $2$ has been discussed already here.

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One important difference (which I don't see in any other answer) is that in fields of non-zero characteristic, we can't have a "norm" or "inner product" the way we might over $\Bbb R, \Bbb C,$ or even $\Bbb Q$. In particular: in order to make sense of conditions like $$ \|\alpha x\| = |\alpha| \cdot \|x\|\\ \langle x,x \rangle \geq 0 $$ It is important to have a notion of "positive numbers" (i.e. we must have an ordered subfield) which we lack for non-zero characteristics.

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    $\begingroup$ It should be noted that $\mathbb{F}_q((t))$ can be equipped with a nontrivial norm. Even if it is ultrametric. $\endgroup$ – Robert Wolfe Jun 30 '17 at 18:22
  • $\begingroup$ @Robert how does that work? $\endgroup$ – Omnomnomnom Jun 30 '17 at 18:34
  • $\begingroup$ If a field has an absolute value defined on it then one can talk about normed vector spaces over this field. This is presumably what Robert is referring to for $\mathbb{F}_q((t))$. $\endgroup$ – carmichael561 Jun 30 '17 at 19:40
  • $\begingroup$ @carmichael561 so how does one define absolute value over a field of finite characteristic? $\endgroup$ – Omnomnomnom Jun 30 '17 at 19:43
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    $\begingroup$ There are tons of them, defined much like the $p$-adic absolute value. For instance, pick a monic irreducible polynomial $f(x)\in\mathbb{F}_q[x]$, and define the absolute value $|r(x)|=q^{-v_f(r)}$ on $\mathbb{F}_q(x)$, where $v_f(r)$ is the power of $f$ dividing the rational function $r$. $\endgroup$ – carmichael561 Jun 30 '17 at 19:47
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When doing basic linear algebra, there is no real advantage for the theory in assuming a field of characteristic zero. (Nor, I should add, is there any real advantage in assuming commutativity: until doing eigenvalue problems, working over a division ring is perfectly fine. Indeed not assuming commutativity is a very good exercise in mental discipline, keeping scalars to one side and matrices to the other.)

There is a practical advantage that in examples one can write down explicit scalars that are obviously unequal; without any assumption on the characteristic, any integer except $-1,1$ might fail to be nonzero, and beginning students might be surprised e.g. that $\frac{13}{16}=\frac9{14}$ when the characteristic is $19$.

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  • $\begingroup$ Thanks for you answer, but about your last example, if the scalar set is characteristic 19, how does $\frac{13}{16}=\frac9{14}$ ? I mean as far as I know, in this case $\frac{13}{16} * 19$ should be zero ?, but it is not and neither the other one, and $\frac{13}{16}* 19\not =\frac9{14} *19$ ? $\endgroup$ – onurcanbektas Jun 30 '17 at 14:54
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    $\begingroup$ In fact, there is a real advantage to not assuming commutativity, even for people who care only about matrices over fields -- when working with block matrices it can be quite convenient to view them as matrices over a matrix ring, and it would be a pain to have to relearn the subject to take advantage of this! $\endgroup$ – Hurkyl Jun 30 '17 at 15:07
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    $\begingroup$ @Leth: Use cross multiplication. $13\times14 = 182 \equiv 11 \equiv 144 = 9\times16 \pmod{19}.$ $\endgroup$ – Ilmari Karonen Jun 30 '17 at 15:30
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    $\begingroup$ @Leth: In the usual way, by treating $\frac ab$ as a shorthand for $a \cdot b^{-1}$, where $b$ is any invertible element. They're not really "rational numbers", just expressions denoting the division one ring element by another, but (by definition) they obey the usual algebraic rules like $\frac ab \cdot b = a$. $\endgroup$ – Ilmari Karonen Jun 30 '17 at 17:04
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    $\begingroup$ @Leth A fraction $a/b$ writable with denominator coprime to $p$ has a natural interpretation as $ab^{-1}\in \Bbb F_p.\,$ The set of all such fractions form a subring of $\Bbb Q$ that has a natural image in $\,\Bbb F_p.\,$ In particular, any algebraic computation perfomed with such fractions maps naturally into $F_p$, e.g. $$\begin{align}\frac{1}2 -\frac{1}3 &= \frac{1}6\\[.5em] \mapsto\ \ 3\ -\ 2\, &=\, 1\ \ {\rm in}\ \ \Bbb F_{\large 5}\\[.5em] \mapsto\ \ 4\ -\ 5\, &=\, 6\ \ {\rm in}\ \ \Bbb F_{\large 7}\\[.5em] \mapsto\ \ 6\ -\ 4\, &=\, 2\ \ {\rm in}\ \ \Bbb F_{\large 11} \end{align}$$ $\endgroup$ – Bill Dubuque Jun 30 '17 at 17:05
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When dealing with inner products, you need to use inequalities, so you have to work with an ordered field (in general $\Bbb R$). Thus anything that is proved using to inner products need not be true in a field of positive characteristic; for example, a symmetric matrix over a finite field is not necessarily diagonalisable. For example, in a field of characteristic $2$ the matrix $$\begin{pmatrix}1 & 1 \\ 1 & 1\end{pmatrix}$$ is nilpotent, but not zero, and thus it is not diagonalizable.

In fact, your previous question is another example (perhaps that's the case you mention in your question): it doesn't hold in characteristic $2$, because in the proof you need to divide by $2$. For example, the bilinear form $$\phi :\Bbb F_2^2\times \Bbb F^2_2\to \Bbb F_2:((x_1,x_2),(y_1,y_2))\mapsto x_1y_1+x_2y_2$$ is skew-symmetric in the sense that $\phi(\vec{x},\vec{y})=\phi(\vec{y},\vec{x})=-\phi(\vec{y},\vec{x})$, but $\phi((1,0),(1,0))\neq 0$.

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    $\begingroup$ ordered is not the same as of characteristic zero, though. $\endgroup$ – Hagen von Eitzen Jun 30 '17 at 13:27
  • $\begingroup$ Yes, I agree that it is not a perfect example, but I thought it could be useful to point it out. $\endgroup$ – Arnaud D. Jun 30 '17 at 13:35
  • $\begingroup$ I didn't get the part "you need to divide by 2" $\endgroup$ – onurcanbektas Jun 30 '17 at 14:22
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"Skew-symmetric" is different from "alternating" in characteristic $2$, but this is more of a multilinear (or exterior) algebra issue than a linear algebra issue, and it's more of a "why characteristic $2$ matters" example than a "why characteristic $0$ matters" one. But it gives a flavor of what can go wrong in positive characteristic: you might not be able to divide by a constant you'd like to divide by. This is the same problem that arises in lhf's example of the polarization identity.

(More precisely, we have alternating $\implies$ skew-symmetry in all characteristics, but the converse holds only for characteristic not equal to $2$.)

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Two problems that I have myself come across are:

  1. The invertibility of a matrix can change. For example, the integer matrix $$\left[\begin{matrix}4 & 1\\2 & 2\end{matrix}\right]$$ is invertible over the real field because its determinant is nonzero (it is six). Considering the same matrix over the field $\mathbb{F}_3$ (with characteristic three), the determinant is zero and the same matrix is now singular. In this case the matrix should be written as $$\left[\begin{matrix}1 & 1\\2 & 2\end{matrix}\right]$$ and it is easy to see that one row is the multiple of the other.
  2. This one is the shocking one that you can have a nonzero vector with norm zero. For example, consider $$x=[1,1,1]$$ which is obviously a nonzero vector over the field $\mathbb{F}_3$ but using the usual inner-product and norm definitions $$||x||^2=|<x,x>|^2=0.$$ So this nonzero vector is orthogonal to itself. Over the real and the complex field, only the zero vector is orthogonal to itself. Now just see how many of the standard linear algebra algorithms fail because of this fact. For one, QR decomposition won't work because Gram-Scmidt won't work here. Something as simple as normalizing a vector would fail.
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One important thing is there is no geometry of physical space when we work over finite fields. The interpretation for $\det A$ as the scaling factor of the "volume" is not available any more in finite fields. This is for the simple reason that elements of real field are measurement of quantities (numbers) whereas elements of finite field are not actually numbers, they don't measure any quantities, they happen to satisfy all the abstract axioms of a field.

So rotations, reflections etc which are geometric linear transformations do not have any intuitive explanation over finite fields. In a field with $q$ elements, for any eigenvector $v$ of any matrix $A$, we have $A^{q-1}v=v$.

In real field $A^{q-1} v$ would be a vector for away from $v$ (if the eigenvalue is of magnitude > 1).

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    $\begingroup$ I think the intuition about reflection remains the same, as long as the field characteristic is not 2. $\endgroup$ – user1551 Jul 1 '17 at 10:15
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    $\begingroup$ In fact the intuition for $\det A$ as the scaling factor of the "volume" is not available even for complex matrices. So the problem is not really one of characteristic. $\endgroup$ – Marc van Leeuwen Jul 1 '17 at 19:29
  • $\begingroup$ @MarcvanLeeuwen: Yes, that it is correct. I was obsessed with real numbers! So my answer contrasts real vs finite field, rather than zero vs non-zero characteristic. $\endgroup$ – P Vanchinathan Jul 3 '17 at 5:34
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Nowhere is the difference between characteristic $0$ and characteristic $p>0$ as stark as in the representation theory of finite groups.

Let me try to illustrate this with a baby example:

Let $V$ be a vector space together with a linear map $f\colon V\to V$ such that $f^p=1$ (where $1$ denotes the identity on $V$). This is precisely a representation of the cyclic group $\mathbb Z/p\mathbb Z$ but you can safely ignore that if you don't know about groups.

Assume we have a subspace $W\subset V$ which is stable under $f$, in the sense that $f(w)\in W$ for all $w\in W$. Then we could ask, if we can find a complement $X$ of $W$ in $V$ which is also stable under $f$. In other words, we want to know, whether we can "break up" $V=X\oplus W$ such that $f$ "acts separately on the two parts", i.e. we can write $f(x,w)=(f(x),f(w))$.

It turns out that you can always do this in characteristic zero! (you can look this up under "Maschke 's theorem"; don't let the name scare you, it is not a difficult proof)

But consider for instance the puny $2$-dimensional vector space $V= F^2$, where $F=\mathbb F_p$ is the field with $p$ elements. Consider $f$ given by the matrix $\left(\begin{matrix}1& 1\\ 0 &1\end{matrix}\right)$. Clearly $f^p=1$ (because $p\cdot 1=1+1+\dots+1=0$ in $F$). Let $W$ be subspace of $W$ spanned by the first basis vector $e_1=\left(\begin{matrix}1\\ 0\end{matrix}\right)$; clearly $W$ is $f$-stable because $f(e_1)=e_1$. But a complement of $W$ would have to be spanned by a vector of the form $x=\left(\begin{matrix}a\\ b\end{matrix}\right)$ with $b\neq 0$; hence $f(x)=\left(\begin{matrix}a+b\\ b\end{matrix}\right)$ is not a multiple of $x$. So $W$ has no $f$-stable complement.

It turns out that this seemingly small issue has enormous ripples. For instance, the representation theory of the symmetric group is extremely well understood in characteristic zero (you can find everything you need to know in any introductory textbook); on the other hand, understanding the representation theory of the symmetric group in positive characteristic is one of the big open problems of modern mathematics.

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    $\begingroup$ For aid in searching, I'll point out to the OP that representation theory where the characteristic divides the order of the group is called 'modular representation theory'. It's significantly harder and less well-understood than the ordinary case (for example, as pointed out here, Maschke's theorem fails). $\endgroup$ – anomaly Jul 12 '18 at 18:31

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