From my answer on stack overflow: https://stackoverflow.com/a/29772759/2534472

Ok, let's have a look at your 2NF attempt:

Student

studentID | studentName | courseCode | modCode | resultCode

Course

courseCode | courseTitle

Module

modCode | modTitle | credits

Let's talk about Student first.

Your key cannot be studentID, because the resultCode depends on studentID and courseCode (you have one result for every course). But your studentName depends only on studentID, so a part of the key (studentID,courseCode). So 2NF is violated. You need to do something like this:

Student

studentID | studentName | courseCode | modCode

Course

courseCode | courseTitle

Module

modCode | modTitle | credits

Result

studentID | courseCode | resultCode

But please keep in mind, that this respects the 2NF, but still seems not correct, because modules and courses are now completely unrelated. So try this:

Student

studentID | studentName | modCode

Module

modCode | modTitle | credits

Course

courseCode | courseTitle | modCode

Result

studentID | courseCode | resultCode

A course belongs to a module (a module has many courses) - from my point of view. This is automatically in 2NF (because the key is always just one attribute (except for the result, but there is just one non-key-attribute always depending on both key attributes), so nothing can depend on a "part" of the key). And it is in 3NF, too, because every "physical" entity has it's logical representation in the data model (more a rule of thumb than a formalism).

Now let's have a look at your 3NF attempt. I think you've got the module<->course dependency the wrong way round, but let's just concentrate on the normalization.

Student

studentID | studentName | courseCode

Course

courseCode | courseTitle

Module

modCode | modTitle | credits | courseCode

Results

studentID | modCode | resultCode

This is correct 3NF, because there is simply no other key candidate than the key itself. So there can't be any transitive dependency.

To clarify this: a key candidate is one of the many minimal sets of possible keys. In your relation you have found at least one key candidate with one element (except for the results relation). So any other key candidate cannot have more than one element. That means, that you can simply look at every single attribute and decide "can this be a key or not?" And in your example you find, that no other attribute can be a key - so it's automatically in 2NF and 3NF.

Your relation is in 3NF, (and not only in 2NF), since as you say the only non prime attribute is Grade, which only appears on the right hand side of your FDs.

The relation is not in BCNF, because the left hand side of the two small FDs is not a superkey.

You can, however, losslessly decompose the relation to (SubjectCode, SubjectName) and either (StudentName, SubjectCode, #Exam, Grade) or (StudentName, SubjectName, #Exam, Grade)

This decomposition gives you two BCNF relations and preserves all functional dependencies. This isn't always possible (you can always decompose a relation to 3NF, but not necessarily to BCNF).

2NF

If you want an example of 2NF (and not 3NF), your relation needs to contain transitive dependencies.

For instance, say you have a Score column. Intuitively Score->Grade since all exams with the same score should get the same grade (it would be rather unfair otherwise), but note that we cannot say Grade->Score since several scores can have the same grade (11% and 12% would likely be "Fail", for instance).

Now your relation is:

Gradings(StudentName, SubjectCode, SubjectName, #Exam, Score, Grade)

and you have a new form of redundancy since every time you enter a result with the same score as another Gradings record you also have to repeat the corresponding Grade. To get to 3NF you could therefore decompose to

ScoreGrades(Score,Grade)

with Score as the key, and

Scores(StudentName, SubjectCode, SubjectName, #Exam, Score)