I have a simple question:
If I cast a nondetection spell on me can I also cast spells like detect magic on myself while I'm under the nondetection effect?
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2\$\begingroup\$ Are you trying to detect magic that is on you or just in general can you use that spell if you have non-detection running? \$\endgroup\$ – NautArch Jan 30 '19 at 17:30
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\$\begingroup\$ Hello and welcome! You can take the tour to learn about the site. I made some changes to the question to hopefully make it more clear but feel free to revert the changes or edit yourself if you don't like the change or if I changed what you wanted to ask. Happy gaming! \$\endgroup\$ – Sdjz Jan 30 '19 at 17:30
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No, you can't
It says clearly in nondetection's spell description:
The target can't be targeted by any divination magic
"Any" includes your own, so no divination spells would affect the target, unless that spell says otherwise.
Detect magic is an AOE spell that targets the caster as the point of origin as stated in the rules for targeting:
A spell's description tells you whether the spell targets creatures, objects, or a point of origin for an area of effect
And thus the spell cannot target the caster.
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3\$\begingroup\$ Might be worth mentioning explicitly that detect magic is a "self" spell and thus would unequivocally target the caster. \$\endgroup\$ – Rubiksmoose♦ Jan 30 '19 at 17:40
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1\$\begingroup\$ Excellent assessment. Pretty sure my comment in the question isn't necessary anymore :) \$\endgroup\$ – NautArch Jan 30 '19 at 17:42
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\$\begingroup\$ @Rubiksmoose That sounds odd to me, so I Googled it, and found this argument to the contrary: reddit.com/r/dndnext/comments/61kgpb/… \$\endgroup\$ – Brilliand Jan 30 '19 at 21:49
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\$\begingroup\$ @Brilliand the first 2 sentences of the post are correct, but everything else is not. The origin of an AOE is a target of the spell. In this case the origin is the caster and thus the caster is a target. From the Targeting rules: "A spell's description tells you whether the spell targets creatures, objects, or a point of origin for an area of effect " \$\endgroup\$ – Rubiksmoose♦ Jan 30 '19 at 21:52
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\$\begingroup\$ I added a small part to your answer. Feel free to revert if you disagree but I think it makes you case clearer (especially since there seems to be confusion about the matter). It seemed too small to write my own answer with and it fits perfectly with what you have already said. \$\endgroup\$ – Rubiksmoose♦ Jan 30 '19 at 22:02
