POST-запрос: как записать данные в таблицу из select? Spring boot + freemarker

Есть две таблицы: Doctor с личными данными и Speciality с названием специальностей, связаны они через speciality_id в таблице Doctor При создание нового аккаунта для доктора специальность выбирается из списка select и выдаёт ошибку при сохранении. Не понимаю, как мне реализовать запоминание выбора из select и записать speciality_id в таблицу Doctor?

Controller

@GetMapping("/profile")
    public String getProfile(Model model, @AuthenticationPrincipal User user, Personal personal) {
        model.addAttribute(user);
        model.addAttribute("personal", userService.findBio(user));
        model.addAttribute("specialities", userService.specialityList());
        return "user-profile";
    }

@PostMapping("/activate/{id}")
public String doctorUser(@PathVariable("id") Long id, Doctor doctor, Speciality speciality) {
    userService.createDoctor(id, doctor, speciality);
    return "redirect:/users";
}

UserService

public List<Speciality> specialityList(){
    return specialityRepository.findAll();
}

public void createDoctor(Long id, Doctor doctor, Speciality speciality) {
    User user = userRepository.findById(id).orElse(null);
    doctor.setUser(user);
    doctor.setSpeciality(specialityRepository.getById(speciality.getId()));
    doctorRepository.save(doctor);
    user.getRole().clear();
    user.getRole().add(Role.DOCTOR);
    userRepository.save(user);
}

HTML

<form action="/activate/${user.id}" method="post">
    <div class="select_wrp">
    <select class="js-select2" name="speciality" placeholder="Выберите специальность">
        <option value=""></option>
        <#list specialities as speciality>
        <option value="speciality">${speciality.name}</option>
        </#list>
    </select><br><br>
    </div>

    <input type="hidden" name="_csrf" value="${_csrf.token}">
    <input type="submit" value="Активировать"/>
</form>

Speciality

Entity
@Table(name ="speciality")
public class Speciality {

    @Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "id")
private Long id;

@Column(name = "name", length = 200) 
private String name;

@OneToMany(cascade = CascadeType.ALL, fetch = FetchType.EAGER, mappedBy = "speciality")
private List<Doctor> doctors = new ArrayList<>();
}

Doctor

@Entity
@Table(name ="doctors")
public class Doctor{
    
    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    @Column(name = "id")
    private Long id;
    
    @OneToOne
    @JoinColumn(name = "user_id")
    private User user;
    
    @ManyToOne(cascade = CascadeType.REFRESH, fetch = FetchType.LAZY)
    @JoinColumn(name = "speciality_id")
    private Speciality speciality;}

Ошибка

2022-03-22 15:24:38.555  WARN 12452 --- [io-8080-exec-10] .w.s.m.s.DefaultHandlerExceptionResolver : Resolved [org.springframework.validation.BindException: org.springframework.validation.BeanPropertyBindingResult: 1 errors<EOL>Field error in object 'doctor' on field 'speciality': rejected value [speciality]; codes [typeMismatch.doctor.speciality,typeMismatch.speciality,typeMismatch.com.example.hospital.models.Speciality,typeMismatch]; arguments [org.springframework.context.support.DefaultMessageSourceResolvable: codes [doctor.speciality,speciality]; arguments []; default message [speciality]]; default message [Failed to convert property value of type 'java.lang.String' to required type 'com.example.hospital.models.Speciality' for property 'speciality'; nested exception is org.springframework.core.convert.ConversionFailedException: Failed to convert from type [java.lang.String] to type [java.lang.Long] for value 'speciality'; nested exception is java.lang.NumberFormatException: For input string: "speciality"]]