Zero of Subfield is Zero of Field
Theorem
Let $\struct {F, +, \times}$ be a field whose zero is $0$.
Let $\struct {K, +, \times}$ be a subfield of $\struct {F, +, \times}$.
The zero of $\struct {K, +, \times}$ is also $0$.
Proof 1
By definition, $\struct {F, +, \times}$ and $\struct {K, +, \times}$ are both rings.
Thus $\struct {K, +, \times}$ is a subring of $\struct {F, +, \times}$
The result follows from Zero of Subring is Zero of Ring.
$\blacksquare$
Proof 2
By definition, $\struct {K, +, \times}$ is a subset of $F$ which is a field.
By definition of field, $\struct {K, +}$ and $\struct {F, +}$ are groups such that $K \subseteq F$.
So, by definition, $\struct {K, +}$ is a subgroup of $\struct {F, +}$.
By Identity of Subgroup, the identity of $\struct {F, +}$, which is $0$, is also the identity of $\struct {K, +}$.
$\blacksquare$