Alternative proofs (algebra) I wrote some short alternative proofs (sketches mostly) to my book, can someone tell me if they are okay.

The unity of a subfield is the unity of the whole field.

Let $H \subset F$ with $F$ being the field. Let $1_H=hh^{-1}$ for every $h$, and let $1_F = ff^{-1}$ for all $f \in F$. Let $f =h$, by uniqueness, we must have $1_H = 1_F$

The multiplicative inverse of a unit in a ring with unity is unique.

Basically $xy = 1$ and $xx^{-1} = 1$ with $y \neq x$. Substracting the equations yields $x(y-x^{-1}) =0$, for nonzero elements we must have $y = x^{-1}$. If we consider zero elements, there is nothing to prove.

Intersection of a subring/subfield is  ring/field respectively.

Let $H_i \subset F$ be the subring of a field $F$. We first show $\cap_i H_i$ is a subring. Let $a-b \in H_i$ for all $a,b \in H_i$, as $H_i$ is an additive group, we get all the group axioms and since this is true for every $i$, it is true for $\cap_i H_i$. We get a similar result for the multiplication of$H_i$. Now consider $H_i' \subset F$ where $H_i'$ is a subfield, as every element of $H_i$' has an inverse for each $i$, so does $\cap_i H_i'$. Finally $(H_i,+)$ i is abelian for each $i$, so we prove the result.
 A: *

*proof 1:
What is your definition of a subfield. You assume that the multiplicative inverse is the same in F and H. Is this part of your definition of a subfield? I don't think so. So this should be proven, too, if you need it for your proof.

*proof 2:
which nonzero/zero elements? If $ab=0$ then in a ring this does not mean that either $a$ or $b$ is $0$.

*proof3:
This is not well stated. "the intersection of a subring..." does not make sense. "The intersection of a set of subrings..." is what you want to say. 
Also you start with "Let $H_i \subset F$ be the subring of a field $F$". This should be "Let $H_i \subset F$ be a subring of a ring $F$".
In the statement "Let $a-b \in H_i$ for all $a,b \in H_i$" the word "let" does not make sense.
"$a-b \in H_i$ for all $a,b \in H_i$" because $H_i$ is a group. The formulation of the remaining proof is so sloppy that I cannot decide if it is right. "every element of $H_i'$ has an inverse for each $i$, so does $\cap_i H_i'$" is not precise enough. For $h \in \cap_i H_i'$ the inverse $h^{-1}$ of $h$ (in $F$) is in all $H_i'$, so it is in $\cap_i H_i'$, too.
A: The shortest proof of the first fact is the following:
$1_H$ is an element of $F \setminus \{0\}$ with $1_H^2=1_H$. Multiplication with $1_H^{-1}$ (the inverse being with respect to $1_F$) shows $1_H=1_F$.
A: *

*Depending on how much is proven prior it might be okey but it's not pretty

*Ring with unity does not exclude the possibility that $ab=0$ with neither $a$ nor $b$ being $0$. Better way is that we have $x^{-1}x=xx^{-1}=1$ and $xy=yx=1$, then we have $xy=xx^{-1}$ and my left multiplication of $x^{-1}$ we get $x^{-1}xy=y=x^{-1}xx^{-1}=x^{-1}$

*It is extremely convoluted and poorly phrased, you want to say "The intersection of a set of subrings..." along with you don't say let $a-b\in X$ as that is a result, you can say let elements be in it but not a subtraction in that way. As for proof a better method is
$$H=\bigcap_i H_i$$
and let $a,b\in H$, then by definition of intersection that means $a,b\in H_i$, as $H_i$ is a subring we have $a-b\in H_i$ and therefore $a-b\in H$, similarly can be done for multiplication and it's inverse etc.

