Does the degree of a projective curve depend on the embedding into projective space?

I'm studying the classification of curves in $P^3$ in chapter IV of Hartshorne.

I've already studied chapter I, chapter II (sections 1 to 6), and chapter III (sections 1 to 7) of Hartshorne in my algebraic geometry courses.

I have a problem understanding a basic definition of curves. According to Hartshorne, a curve is an integral scheme of dimension 1, proper over $K$ all of whose local rings are regular. Such a curve is necessarily projective.

My problem is with the definition of the degree of a curve. I didn't find a precise definition for it in chapter IV. This is what I thought: Since a curve is projective, there is always a closed immersion of the curve in $P^n$ for some $n$.

So the degree of a curve could be the degree of its image as a closed of $P^n$ as defined in (I,7) (with the Hilbert polynomial).

is it correct? It seems that with this definition, the degree of a curve depends on the embedding one can consider.

Question: "So the degree of a curve could be the degree of its image as a closed subscheme of $\mathbb{P}^n$ as defined in (I,7) (with the Hilbert polynomial) is it correct? It seems that with this definition, the degree of a curve depends on the embedding one can consider."

Answer: Yes, the degree of a projective variety/scheme depends on the embedding. By HH.Prop I.7.6 it folllows $deg(\mathbb{P}^1)=1$ and if you consider the 2-uple embedding

$$v_2: \mathbb{P}^1 \rightarrow \mathbb{P}^2$$

defined by

$$v_2(a_0:a_1):=(a_0^2:a_0a_1:a_1^2),$$

its image $Im(v_2)$ is defined by the polynomial $f:=xz-y^2$ which has degree $2$, hence $deg(Im(v_2))=deg(f)=2$ by the same HH.Prop.I.7.6d.

Note: The degree is defined in terms of the Hilbert polynomial $P_S(t)$ and this polynomial is defined in terms of the graded coordinate ring $S$ of $C$, and $S$ depends on the embedding. The ring $S$ is not an invariant of $Proj(S)$: There are $S \neq S'$ with $Proj(S) \cong Proj(S')$.