Subring of Real numbers Let $ S= \{ i +j 2^{1/4} +k 4^{1/4} + l 8^{1/4} |i,j,k,l \in \mathbb {Z}\}$ 
Decide if S is a subring of the real numbers with its usual operations of addition and multiplication. if it is give a proof and if not explain why.
Well i mean this is a Ring, not every element is a unit but it is a ring that is defined for multiplication it has identity and no zero divsiors so its rather well behaved.
we dont really have theorems for sub-rings ( only groups in my intro class)  how do i prove that this ring is a subring of the the field of reals?
 A: 
Theorem (Subring Criterion): Let $(R,+,\times)$ be a ring and consider $S\subseteq  R$. Then for $S$ to be a during off $R$, it is necessary and sufficient that:
  
  
*
  
*For all $a,b\in S, a-b\in S$, where $-$ is the inverse operation of $+$
  
*For all $a,b\in S, a\times b\in S$
  
*(Sometimes) $S$ contains the identity element of $\times$ in $R$.

Wether the fourth is included depends on if your definition of a ring requires that it contains a multiplicative identity. Some people do require this, others don't.
The first,and third are trivial in your example, and so people in the comments are suggesting that you test for closure under multiplication because that's all you need to do for this theorem to verify that you are looking at a subring. This is the standard way to check that something is a subring, and should be your go-to approach.
The first alone forms what is known as the subgroup criterion and is all that is necessary to check to see if a set is a subgroup of a group.
A: Let $\alpha = 2^{1/4} \in \mathbb R$. Then, by definition, $\mathbb Z[\alpha]$ is the smallest subring of $\mathbb R$ that contains $\alpha$.
$\mathbb Z[\alpha]$ also coincides with the set of all polynomial expressions in $\alpha$ with coefficient in $\mathbb Z$.
Since $\alpha^4=2 \in \mathbb Z$, all terms of degree $4$ or higher can be replaced by lower degree terms.
Therefore, 
$\mathbb Z[\alpha]= \{ i +j \alpha +k \alpha^2 + l \alpha^3 : i,j,k,l \in \mathbb {Z}\}=S$, and $S$ is a ring.
