What are the fields produced around a current carrying conductor? If you consider a current carrying conductor, every instant an electron enters the conductor, another electron will be leaving the conductor. Thus, the current carrying conductor will not be charged (i.e. it would not have any net positive or negative charge). Remember a dipole has zero net charge, but it does have electric field around it (although the net electric field around a dipole drops to insignificance as the distance increases to multiples of the dipole separation). So, if net charge is zero, it doesn't mean there is no electric field, only that it is small (i.e. largely cancelled because every electron in the conductor is paired as a dipole), and gets rapidly smaller with distance.
It is important to notice that, if we assume only electrons to be moving, and kernels (positive nuclei) to be static, the magnetic field will be produced only due to electrons.
The speed at which energy or signals travel down a conductor is actually the speed of the electromagnetic wave, not the movement of electrons (this is an modified statement extracted from wiki encyclopedia-speed of electricity).
Does it mean that an electric field and magnetic field exist around the current carrying conductor?
Or
Does it mean that only a magnetic field exists around the current carrying conductor?
NOTE
By the discussion until now (2/11/2013), I have found a difference in answers with respect to AC and DC. Can everyone update their answers with respect to both the cases (AC and DC)?
LINKS

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*Relativistic electromagnetism-Wiki provides you an idea about fields around a current carrying conductor, with respect to different frame of reference.

*I went through quite similar question Does a current carrying wire produce electric field outside?
. I found the first answer was assuming electric field to be existing around the current carrying conductor, and the second answer was assuming electric field to be not existing around current carrying conductor.

*I went through the link: propagationtime.pdf, it expresses the presence of both electric and magnetic fields around a current carrying conductor.

*I took a look on this question, [Confusion between Electric field and Magnetic field of a charged particle].(Do moving charged particles have both magnetic and electric fields?), it makes a clear attempt to say that, there is no electric field around a current carrying conductor.

 A: For the magnetic field, the currents are one source of the magnetic, but this problem is more linked to the source of the current in the wire.  For a conductor with finite conductivity, an electric field is needed in order to drive a current in the wire.  
If we assume your wire is straight, this required field is uniform. One way to realize this field is by taking two oppositely charged particles and send them to infinity while increasing the magnitude of their charge to maintain the correct magnitude of electric field.  In this limit, you will obtain a uniform electric field through all space.  
Now, put your conductor in place along the axis between the voltage sources--a current will flow.  In the DC case, this gives rise to the magnetic field outside of the wire.  As for the electric field, a conductor is a material with electrons that can move easily in response to electric fields and their tendency is to shield out the electric field to obtain force balance.  Because the electrons can't just escape the conductor, they can only shield the field inside the conductor and not outside the conductor.  With this model, we see that the electric field is entirely set up by the source and placing the conductor in the field really just establishes a current.  Note here that if you bend the wire or put it at an angle relative to the field, surface charges will form because you now have a field component normal to the surface.
For the limit of an ideal conductor, no electric field is needed to begin with to drive the current and so there isn't one outside the wire.
For the AC case, solving for the fields becomes wildly complicated very fast as now the electric field driving particle currents has both a voltage source and a time-varying magnetic source through the magnetic vector potential.  The essential physics is the same, though, as the source will establish the fields (in zeroth order), and the addition of the conductor really just defines the path for particle currents to travel. In the next order, the current feeds back and produces electromagnetic fields in addition to the source(s) and will affect the current at other locations in the circuit.
I guess a short answer to your question is that there are always fields outside of the current-carrying wire and the electric field outside disappears only in the ideal conductor limit.  Conductors generally do not require very strong fields to drive currents anyway so that the electric field outside is usually negligible, but don't neglect it for very large potentials in small circuits.
A: It seems to be little known, but a static current inside a wire of finite conductivity does produce a static electric field outside. See "Application 9.2" in the new text by Zangwill (which is fantastic and will replace Jackson), or
http://www.ifi.unicamp.br/~assis/Found-Phys-V29-p729-753(1999).pdf
A: Yes there is an electric field outside of a current carrying wire, in a direction along the wire axis (i.e. parallel to the wire). This is true in both the AC and DC case. There is also of course a magnetic field in the azimuthal direction.
For a resistive wire oriented along the z-axis, the electric field inside the wire is given by Ohm's Law $E_y=\eta j_y$ where $\eta$ is the resistivity and $j_y$ is the current density. This current tends to flow near the surface of the wire (this is the skin effect).
In order that there is no discontinuity in the electric field parallel to the wire, the electric field $E_y$ in the vacuum at the edge of the wire has to match the field in the wire given by Ohm's Law. Further away from the wire, there must be a radial electric field needed to keep the total electric field curl-free.
The fact that this electric field is parallel to the wire is the reason why electromagnetic energy flows radially into the wire. This is somewhat counterintuitive, but can be seen by noting the direction of the Poynting vector $\bf E \times \bf B$.
This is mentioned briefly in Feynman's Lectures on Physics II chapter 27-5.
A: This isn't a direct answer to your question, but I believe goes to the heart of the issue.
When talking about energy and electrodynamics in a classical sense, it's often possible to interpret energy in two different ways. The first way is to think about the potential energy of a charge configuration. For example, if you have a fixed positive charge, and bring a second positive charge close, then you're pushing it uphill against the potential energy. The electric field from the first charge would normally repel the second. The work necessary to bring the second charge closer goes into the potential energy of the configuration.
The second way of thinking about the energy is a little more subtle. When you bring the two charges together, you get a very large electric field between them. You can actually consider the energy as stored in the electric field.
The big advantage of understanding the energy as stored in the electric field comes when you talk about light. Light has energy, but is also electromagnetic field. There aren't any charge configurations to store the energy, so the energy must be in the field itself.
Electrical energy and signals really do transmit at the speed of light in a conductor, like the Wikipedia article says. The electrons themselves travel much slower. This is clearer if you consider that the energy is stored in the electric field (and magnetic field), so the signal should travel at the speed of the propagating electric field (aka the speed of light).
A more direct answer to your question: Yes, there are both electric fields and magnetic fields around the conductor while there's current flowing through it. This is guaranteed by Maxwell's equations, and the fact that current is proportional to the electric field in a conductor.
A: There is no E field outside the conductor due to electrons because the conductor is not charged--this means there exist an equal number of protons to every electron in the conductor. If the conductor was charged, then there would be E field outside the conductor due to that charge. In fact, for a conductor to produce an E field outside of itself, it must be statically charged like a capacitor plate.
