Error: TypeError: user.send is not a function

Question

I am trying a discord bot with a modal, I want to send a private message to the user, but an error appears like this

Error: TypeError: user.send is not a function

this is my code

const invoke = async (interaction) => {
    const user = interaction.user.id

    const modal = new ModalBuilder({
        customId: `createChar-${interaction.user.id}`,
        title: 'Create Character',
    });

    const CharacterName = new TextInputBuilder({
        customId: 'characterName',
        label: 'Input your (fisrtname_lastname)',
        style: TextInputStyle.Short,
    });

    const PasswordInput = new TextInputBuilder({
        customId: 'characterPassword',
        label: 'Input your Password',
        style: TextInputStyle.Short,
    });
    
    const firstActionRow = new ActionRowBuilder().addComponents(CharacterName);
    const secondActionRow = new ActionRowBuilder().addComponents(PasswordInput);

    modal.addComponents(firstActionRow, secondActionRow);

    await interaction.showModal(modal)

    const filter = (interaction) => interaction.customId === `createChar-${interaction.user.id}`;

    interaction
        .awaitModalSubmit({ filter, time:30_000 })
        .then((modalInteraction) => {
            const CharacterName = modalInteraction.fields.getTextInputValue('characterName')
            const PasswordInput = modalInteraction.fields.getTextInputValue('characterPassword')

            const user = interaction.user.id
            user.send(`ASDASD`) // <<<<< THIS IS MY CODE

            modalInteraction.reply(`Your character name: ${CharacterName}\nYour Password: ${PasswordInput}\nMy discord ${interaction.user.id}`)
        })
        .catch((err) => {
            console.log(`Error: ${err}`)
        });
    
    
}

What am I doing wrong?

I want to send a private message when submitting the button modal

Answer 1

send is a method of the user, not the user's id:

const user = interaction.user // was interaction.user.id in the question
user.send(`ASDASD`)