How to make custom list_display in django admin?

Question

Django

Good morning!
For example, there are models:
models.py

class Library(models.Model):
    title = models.CharField()
    cover = models.ImageField()
    description = models.TextField()

class Book(models.Model):
    material = models.ForeignKey(Library)
    user = models.ForeignKey(User)

How to display a separate Book column in the admin panel in the listing of the Library model?
I understand that something like this is needed:
admin.py

class LibraryAdmin(admin.ModelAdmin):
    list_display = ('id', 'title', 'get_book', )
    model = Library

    def __init__(self, *args, **kwargs):
        super(LibraryAdmin, self).__init__(*args, **kwargs)
        self.list_display_links = ('id', 'get_book')

And in the Library model, how, more precisely, how to implement feedback ??.
models.py.Library

class Library(models.Model):
    title = models.CharField()
    cover = models.ImageField()
    description = models.TextField()

    def get_book(self):
        if self.book: # Как тут осуществить обратную связь?
            return self.book
        else:
           return None
    get_book.short_description = "Книга"

class Book(models.Model):
    material = models.ForeignKey(Library)
    user = models.ForeignKey(User)

Answer 1

In my opinion, you can have more than one Book instance correspond to one Library instance ? Why then is it written "book" and not "books"? Every Library instance has a book_set property that contains a list of all the books that match the given object. You can rename it by specifying related_name :
Now you can list books like this:

class LibraryAdmin(admin.ModelAdmin):
    list_display = ('id', 'title', 'book_list', )

    def book_list(self, obj):
        return ', '.join(book.name for book in self.books)

Or do you have one library corresponds to only one book, and vice versa?