Why isn't there a minus sign in Ohm's law, $V = IR$? Suppose current runs through a resistor from left to right, and we define the left-to-right direction as positive.
Then from left to right, the potential decreases. So $V,$ the voltage across the resistor (defined as the electric potential at the right of the resistor minus the electric potential at the left of the resistor), is negative, while $I,$ the current through the resistor, is positive. So it would seem $V = -IR$ is the correct relationship, but I don't recall ever seeing Ohm's law written this way. Why not?
(For example, in the introductory textbook I have on hand, Ohm's law is derived using $E = V/\Delta L,$ even though, in the previous chapter, the book writes $E_x = -\Delta V_x/\Delta x.$ [1] The intermediate book I have on hand argues that $V$ and $I$ must be proportional, saying "$V$ is the line integral of $\mathbf{E}$ on a path through the conductor...", but it would seem more appropriate to say "$V$ is the negative of the line integral of $\mathbf{E}$..." The book defines $V$ as "the difference in potential between those terminals" without specifying the direction. [2])
[1] Physics, Resnick, Halliday, and Krane, 5ed. equations 29-12 and 28-37
[2] Electricity and Magnetism, Purcell and Morin, 3ed. text before equation 4.12
 A: In the comments, Hal Hollis pointed to the "passive sign convention".
The passive sign convention is defined like this:

So it is simply taken as a definition that the positive direction for current is the opposite direction as the positive direction for voltage. With this convention, $V=IR$ is the correct equation.
A: I think that the answer lies in the fact that resistors are not the only components in an electronic circuit.  
Consider the following circuit.  
 
Following the path of the electric current (positive charges), $V=\int \vec E \cdot d \vec l$ is negative for component $A$ and positive for component $B$.  

What does that mean?

It means that when positive charges go through component $A$, work is done on the positive charges and when the positive charges go through component $B$, work is done by them.
In simple terms, this can be interpreted as $A$ being a source of electrical energy, and $B$ being a sink of electrical energy. If $A$ is a photocell, the light is being converted into electrical energy, and if $B$ is an LED, then electrical energy is being converted into light. 

For both components, the power is $|VI|$, so how does one
  differentiate between these two components?

A sign convention is introduced which is usually the passive sign convention, which then results in the electrical power produced in component $A$ being negative (a source of electrical energy) and that in component $B$ being positive (a sink of electrical energy). 
There is another convention which is used and that is the active sign convention which makes the power of sources of electrical energy positive and that of sinks of electrical energy negative. For example, this is used when generators are sold; you buy a $(+)1\,\rm kW$ generator not a $-1\,\rm kW$ one.  
The difficulty with running both conventions together, the passive for sinks and the active for sources, is that some circuit elements can be a source (battery connected to a bulb) or a sink (battery being recharged).  
The passive sign convention is the one most commonly used, and then for a resistor, $V=IR$ with the current flowing from a high potential to a lower potential where $V$ is the drop in potential (in the direction of the current) across the resistor, i.e. a positive quantity. 
A: I think it's all a matter of definition. Electric current's direction is just defined as the direction in which (effective) positive charge carriers move. Positive charges move from the higher voltage points to lower-voltage points, so a flux variable (current) defined in that direction would be positive. You can, for example, define "current"'s direction as the direction in which electrons travel, and write V = -IR.
A: There is a minus sign in $\vec J = \sigma \vec E$ and $\vec E = - \vec \nabla V$: the current goes down the voltage gradient. However, V in Ohm's law is not the potential but the potential drop caused by the resistor. I am all in favour of $\Delta V = -IR$. Yes, I changed my story ...
A: You're right. $R$ is a constant, and we have variables $V \in \mathbb{R}$ and $I \in \mathbb{R}$ where $V$ stands for the change in potential between two points and $I$ is the current from the current vector $\vec{I} = I \;\hat{u}$. So $V$ and $I$ are just some numbers, positive or negative. Ohm's law should be stated $V = -IR$ as you say.
Draw a number line on a sheet of paper as you would in algebra class. The positives are pointing in some direction. You could have drawn the positives in another tilted direction. You could even draw a curvilinear number line. A curvilinear number line has no fixed direction (each point on the line has an instantaneous positive direction), but it still has the sense of a single orientation. A curved number line is still one-dimensional from this standpoint.
Because of this (circuits are curvy), my guess is that Ohm's law is written as $|V| = |I|R$ (from my definitions of $V,I$ above) because there is no fixed direction (current is moving in this direction, then this direction, then down, etc).
