TL;DR - Can a 6th level wizard research (not cast) an existing 4th level spell?

As a wizard of the 13th circle, I have a need for some spells of the 4th tier. I have been unable to find scrolls for purchase, and I have limited time to spend researching. I have a small team of personal simulacra, which are each as powerful as a wizard of the 6th circle. As such, they can cast spells of the 3rd tier easily, but 4th tier is beyond them.

As the effects are known, and only the formulae need to be worked out, are a sufficiently high skill in Spellcraft and proper expenditure of resources all that is required to add the spell to a spellbook?


Research a Spell

The standard rules allow you to perform spell research, either to create a new spell or learn an existing spell from another source. In the downtime system, the steps for spell research each day are as follows.

  1. Pay 100 gp × the spell’s level for research costs and rare ingredients. You may spend Goods or Magic toward this cost.
  2. Determine the total days of progress required to complete the research, which is 7 × the spell level.
  3. Determine the spell research DC, which is 10 + twice the spell’s level.
  4. Attempt a Spellcraft check and a Knowledge check (arcana for an arcane spell, religion for a divine spell) against the spell research DC. You can’t take 10 on these checks. You may spend Magic to modify a check result, with 1 point of Magic adding 2 to your total (maximum +10). If both checks succeed, you make 1 day’s progress toward completing the spell. When your days of progress equal the total number of days needed, the spell is completed and added to your spellbook or list of spells known.

If either or both spell research checks fail by 4 or less, you make no progress. For each check that fails by 5 or more, your research has led to poor results and you lose a day of progress toward completing the spell.


Independent Research

A wizard can also research a spell independently, duplicating an existing spell or creating an entirely new one. The cost to research a new spell, and the time required, are left up to GM discretion, but it should probably take at least 1 week and cost at least 1,000 gp per level of the spell to be researched. This should also require a number of Spellcraft and Knowledge (arcana) checks.

  • \$\begingroup\$ Welcome to RPG.SE! Take the tour if you haven't already, and check out the help center for more guidance. \$\endgroup\$
    – V2Blast
    Mar 14, 2019 at 1:08

2 Answers 2


Yes, they can

The only limitation for a wizard is to prepare and cast a spell, they must have an open slot of that spell's level or higher. But otherwise, they could copy, learn or even research spells of any level, they are basically scientific texts for them, which they can only put in practice once they have enough experience and power to reproduce the written text into reality.

Note that researching is far more expensive than paying for another wizard to let you borrow their spellbook so you may copy that spell into yours. But the result is that you came up with your own version of the spell, like doing your own research based on a theory you only heard about.



This is an interesting question, but by RAW, the short answer is, No.

The basic proof is right there in the Wizard's own CRB entry (emphasis my own):

In addition to having a high ability score, a spellcaster must be of a high enough class level to be able to cast spells of a given spell level.

That's that.

However, Pathfinder is all about finding one's own path--if this is a truly vital part of the adventure, i.e. it's a make the plot or break the plot sort of thing, one could get creative, but one would have to work with the GM in order to do so.

Some options:

First: Copy the Spell into your Spellbook via Independent Research

You are right, you can independently research the spell OR create a new one. In Pathfinder, when it comes to writing a spell into your spellbook there are no restrictions on spell level. So, a Wizard can copy any spell into his spellbook by following all the appropriate checks, BUT he cannot prepare it until he meets the level requirement.

Second: Scribe a Scroll

This is the tricky/ not RAW part. A lower level wizard may directly cast a scroll of a higher caster level than his own, by making a caster level check. However, a lower level Wizard is unable to actually scribe the scroll because the spell must be prepared in order to do so...and (evidence above) the wizard is simply unable to prepare it. However, you could speak with your GM and decide on a fair, and most likely difficult, roll for the Wizard to make in order to be successful in this.

Originally I suggested raising the wizard's caster level, but it has been pointed out (see comments) that this is actually incorrect. It has been deleted.

Third: Create your own spell at your own caster level

You are limited only by your own creativity and, of course, the GMs discretion. If you need a particular spell, can't access it, but can afford to create your own, albeit a slightly less effective version, then why not? You may be able to work with your GM to come up with something that might work.

  • \$\begingroup\$ That's a great in-depth answer with alternatives. I'm not trying to have the lower level caster actually prepare or cast the spell, just research for the higher level caster. Sorry if I wasn't clear. \$\endgroup\$ Mar 14, 2019 at 2:31
  • \$\begingroup\$ Ah, in that case @Shadowkras is correct below :) \$\endgroup\$
    – TigerDM
    Mar 14, 2019 at 10:05
  • \$\begingroup\$ Note that improving your caster level does not allow you to cast higher level spells. \$\endgroup\$
    – ShadowKras
    Mar 14, 2019 at 11:30
  • \$\begingroup\$ Really? That's interesting...got a reference or is it just common knowledge? \$\endgroup\$
    – TigerDM
    Mar 14, 2019 at 12:30
  • \$\begingroup\$ @TigerDM Spells that a creature can cast are typically determined by class level; caster level determines a spell's efficacy. The two are often the same number—a typical Wiz20 has a caster level of 20—, but they don't have to be. \$\endgroup\$ Mar 14, 2019 at 13:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .