Galois Correspondence as it relates to $\mathbb{Q}(\sqrt{1 + \sqrt{2}})$

Question

Tl;dr

Before getting into the details, let me just give a summary of the main source of my confusion. Basically, I currently believe the following list of statements are true:

1. The extension $\mathbb{Q}(\sqrt{1 + \sqrt{2}})$ is Galois
2. Every element of $\mathbb{Q}(\sqrt{1 + \sqrt{2}})$ can be written as $k_0 + k_1 \sqrt{2} + k_2 \sqrt{2}\sqrt{1 + \sqrt{2}} + k_3\sqrt{1 + \sqrt{2}}$ with $k_i \in \mathbb{Q}$
3. The automorphisms $\alpha$ of the extension are the four automorphisms defined by mapping $\sqrt{2}$ to $\pm \sqrt{2}$ and $\sqrt{1 + \sqrt{2}}$ to $\pm \sqrt{1 + \sqrt{2}}$
4. The set of members of $\mathbb{Q}(\sqrt{1 + \sqrt{2}})$ of the form $k_0 + k_3\sqrt{1 + \sqrt{2}}$ is not a field despite the fact that it is fixed by a subgroup of $Aut(\mathbb{Q}(\sqrt{1 + \sqrt{2}}))$

However, my reasoning below makes me to believe that at least one of these statements is not true, but I can't figure out which one it is. So my question boils down to, which one of these assumptions is wrong? If none of them are wrong, then what is wrong with the reasoning below?

Context

So, my understanding of the Galois Correspondence is that if $L/K$ is some Galois extension (i.e. seperable, and splits some polynomial in $K[x]$), and $G$ is some subgroup of $Aut(L/K)$, then the set of members of $L$ fixed by $G$ is a subextension of $L/K$.

Furthermore, any finite extension $L/K$ is a finite dimensional vector space over $K$. The Galois Automorphisms of $L/K$ being the automorphisms which are both field automorphisms and vector space automorphisms of $L/K$.

So, applying this to the extension $\mathbb{Q}(\sqrt{1 + \sqrt{2}})$, this can be thought of as a vector space over $\mathbb{Q}$, and we can write any member $x$ of this field as

$$x = k_0 + k_1 \sqrt{2} + k_2 \sqrt{2}\sqrt{1 + \sqrt{2}} + k_3\sqrt{1 + \sqrt{2}}$$

With $k_0, k_1, k_2, k_3 \in \mathbb{Q}$. Let $\alpha$ be a galois automorphism in $Aut(L/K)$, then

$$\alpha(x) = k_0 + k_1 \alpha(\sqrt{2}) + k_2 \alpha(\sqrt{2})\alpha(\sqrt{1 + \sqrt{2}}) + k_3\alpha(\sqrt{1 + \sqrt{2}})$$

Now, $\mathbb{Q}(\sqrt{1 + \sqrt{2}})$ can be created algebraically via adjoining a root of $x^2 - 2$ to $\mathbb{Q}$ resulting in $\mathbb{Q}(\sqrt{2})$, and then adjoining a root of $x^2 - (1 + \sqrt{2})$ to $\mathbb{Q}(\sqrt{2})$ resulting in this tower

$$\mathbb{Q} \subset \mathbb{Q}(\sqrt{2}) \subset \mathbb{Q}(\sqrt{1 + \sqrt{2}})$$

The point being that $\alpha$ above is defined via a decision to map $\sqrt{2}$ to $-\sqrt{2}$ and a decision to map $\sqrt{1 + \sqrt{2}}$ to $-\sqrt{1 + \sqrt{2}}$. Meaning there are 4 such automorphisms, and this makes sense since the degree of the extension is 4 and this extension is Galois. But let us choose the automorphism which sends $\sqrt{2}$ to $-\sqrt{2}$ but fixes $\sqrt{1 + \sqrt{2}}$.

I.e.

$$\alpha(x) = k_0 - k_1 \sqrt{2} - k_2 \sqrt{2}\alpha(\sqrt{1 + \sqrt{2}}) + k_3\sqrt{1 + \sqrt{2}}$$

This automorphism together with the identity automorphism forms a subgroup of $Aut(L/K)$, but the only elements fixed by this subgroup are of the form

$$k_0 + k_3\sqrt{1 + \sqrt{2}}$$

But this is not a field, as there is no inverse of $\sqrt{1 + \sqrt{2}}$ of the form $k_0 + k_3\sqrt{1 + \sqrt{2}}$ with $k_0, k_3 \in \mathbb{Q}$

So, something is wrong in my reasoning above, although I can't figure out what it is.