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A wizard that possesses the feat Arcane Mastery (Complete Arcane 73) employs the special ability share spells so that both the wizard and the wizard's familiar gain the benefits of the 6th-level Sor/Wiz spell eyebite [necro] (PH 228-9).1 The familiar (that does not possess the feat Arcane Mastery) targets with the eyebite spell's effect a foe that possesses spell resistance.

Is it possible to take 10 on the caster level check needed to bypass this foe's spell resistance?


1 Maybe the spell eyebite is invalid for the special ability share spells. In my campaign the spell that was actually shared for several levels was the 4th-level Sor/Wiz spell energy spheres [evoc] (Spell Compendium 80) until I reread the spell and found it was likely ineligible. However, the crux of the question remains Does the familiar benefit from its master's feats (or, at least, the feat Arcane Mastery in particular) when sharing spells? not Can the spell [whatever] be shared using the special ability share spells? Answers are free to use as examples more appropriate spells.

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Yes. A caster level check is rolled by the caster, not by the subject of the spell.

From the definition of Spell Resistance:

To affect a creature that has spell resistance, a spellcaster must make a caster level check (1d20 + caster level) at least equal to the creature’s spell resistance. (The defender’s spell resistance is like an Armor Class against magical attacks.) If the caster fails the check, the spell doesn’t affect the creature. The possessor does not have to do anything special to use spell resistance. The creature need not even be aware of the threat for its spell resistance to operate.

So it doesn't matter whether the familiar does or doesn't have access to Arcane Mastery; the wizard having it is necessary and sufficient. You don't need Share Spells for this either (assuming you pick a non-personal-range spell). e.g. Alice the wizard casts Glorious Master of the Elements (Sor/Wiz7, Dragon Magic) on Bob, granting Bob the ability to shoot elemental ray attacks; Bob's rays benefit from Alice's Arcane Mastery, and wouldn't benefit if Bob himself had the feat instead of Alice.

The same applies to most other spell-affecting feats, since they're phrased as affecting "your spells" or "your caster level" or something like that. There are a few exception, however: if Alice has Weapon Focus [ray], then Bob doesn't benefit from that, since Weapon Focus cares who's making the attack roll rather than who casts the spell.

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  • \$\begingroup\$ I'm okay with this, but it seems a really narrow reading. I mean, what if the spell resistance needs to be overcome by a creature that's not a spellcaster proper, like a creature that's instead using a spell-like ability or an effect that mimics a spell that's created by a magic item? \$\endgroup\$ Commented Dec 20, 2016 at 2:13
  • \$\begingroup\$ @HeyICanChan If you are using magic item, you aren't a caster. Most of your feats don't apply. \$\endgroup\$ Commented Dec 22, 2016 at 15:40
  • \$\begingroup\$ @annoyingimp O, I know. I'm concerned that the answer's case for reading a spellcaster must make a caster level check as specific (rather than as a creature generating the effect must make a caster level check) means some snarky lawyer (but I repeat myself) then reads this as The magic item's casting the spell, and the magic item is not a spellcaster, so I guess the magic item's effect overcomes spell resistance automatically! or something. (By the way, I know that's a terrible and adversarial reading--and I'd totally sock that dude--, but it is potentially valid.) \$\endgroup\$ Commented Dec 22, 2016 at 20:13
  • \$\begingroup\$ @annoyingimp If you're using a magic item, you're not necessarily a caster, and you aren't necessarily the caster of the spell, even if you make decisions for it. There's quite a lot to this. \$\endgroup\$
    – Firebreak
    Commented Dec 23, 2016 at 23:50
  • \$\begingroup\$ @HeyICanChan I don't think it is potentialy valid. Probably, it is a spellcaster who created the item, who is making the check. \$\endgroup\$ Commented Dec 24, 2016 at 10:51

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