I am studying Galois theory and have come across many questions regarding finding the splitting field of $x^n-a$ over $\mathbb{Q}$.   The specific question I was looking at was      Let $L$ be the splitting field of $x^6-2$ over $\mathbb{Q}$. Determine $[L:\mathbb{Q}]$ and describe the Galois group.    In this case I know that $L=\mathbb{Q}(\sqrt[6]{2}, i\sqrt{3})$. I think the degree would be $12$ since the minimal polynomial of $\sqrt[6]{2}$ is $x^6-2$ and the minimal polynomial of $i\sqrt{3}$ is $x^2+3$ and the former is nonreal, so we can multiply the field degrees. I don't know how I would find the Galois group. And are there any methods that would work more generally?
7115
I have some problems with this exercise. I don't know if it can be done. Consider the polynomial $ x^n - a \in \mathbb{Q} $ Can I compute the Galois group of this over $\mathbb{Q}$? Maybe having a nice "basis.  The splitting field is given by $\mathbb{Q}(\zeta_n,\alpha)$ , where $\zeta_n$ is a primitive root of unity , and $\alpha$ is some number such that $\alpha^n = a $. Well first of all, I want a $\underline{"good basis"}$ for the splitting field. In the sense that the minimal polynomials, of the adjoined elements, are different (in this case the computation of the galois group is very simple).  For example one easy case, it's when $a&gt;0$, then $(a)^{\frac{1}{n}} \in \mathbb{R}$ , so clearly the minimal polynomial of $(a)^{\frac{1}{n}} , \zeta_n$ are distincs, and I'm done. If $n$ is odd then , it's also easy, since one root it's also real, for example $x^3-3 $, the real root is $ \root 3 \of { - 3}  =  - \root 3 \of 3  $ , so I can consider the splitting field as $\mathbb{Q}(-\root 3 \of 3 , \zeta_3 )=\mathbb{Q}(\root 3 \of 3 , \zeta_3 )$. The difficult case is when $n$ is even and $a&lt;0$ , for example  $x^8+20$ or $x^4+20$ in some cases as in the second, there are particular cases since there exist algorithms for the Galois group of quartics, but in general. It can be done?  $\underline{Remark}$ I'm searching a $\underline{"good basis"}$ for the splitting field. In the sense that the minimal polynomials, of the adjoined elements, are different since in this case the computation of the galois group is very simple.
99290
The question:     Find all the middle fields of the splitting field of $x^4-2$ over $\mathbb{Q}$.   My attampt:  Let $E$ be the splitting field of $f(x):=x^4-2$ over $F:=\mathbb{Q}$.  The roots of $f$ are $\pm\alpha,\pm\alpha i$ where $\alpha =\sqrt[4]{2}$. I know that $$ G:=\text{Aut}(E/F)=\{\sigma^i\tau^j:0\leq i\leq 1,0\leq j\leq3\} $$ where $$  \sigma: \alpha \mapsto i\alpha\\ \tau: i\mapsto -i$$ And because $\sigma\tau\sigma=\tau$, we get $G\cong D_8$.  The subgroups of $G$ are $$\begin{align} &amp;H_1=\{1,\sigma^2\},&amp;&amp;H_2=\{1,\tau\}\\&amp;H_3=\{1,\sigma^2\tau\},&amp;&amp;H_4=\{1,\sigma\tau\}\\&amp;H_5=\{1,\sigma^3\tau\},&amp;&amp; H_6=\{1,\tau,\sigma^2,\sigma^2\tau\}\\&amp;H_7=\{1,\sigma\tau,\sigma^2,\sigma^3\tau\},&amp;&amp;H_8=\{1,\sigma,\sigma^2,\sigma^3\} \end{align}$$ The required middle fields are $K_i=E^{H_i}$.  Observe that $K_1=\mathbb{Q}(i)$ and $K_2=\mathbb{Q}(\alpha)$, but I don't think I calculate $K_i$ correctly. For example: I want to calculate $K_3$: Lex $K_3\ni x=a+\alpha b+ic$. We know $\sigma^2\tau(x)=x$ then $$a=b\alpha i\\ \Rightarrow x=b(\alpha(1+i))+ci$$ I don't know how to calculate $K_3$ from here.  Another example: Let $K_4\ni x=a+\alpha b+ci$. Then $$x=\sigma\tau(x)\\ \Rightarrow x=a+b\alpha-b({i-\alpha\over 1+i})\\ =a+{1\over 2}b(5\alpha-\alpha i-i-1)$$.  I'm not sure how to proceed and I think that's not the way to solve the question. Thanks in advence.
43942
Galois Group of $\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}$ is cumbersome to computing. Is easy to find the all four possible candidates but is cumberstone to show that they are automorphisms:  For multiplication is too long showing that  $f((a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})(a'+b'\sqrt{2}+c'\sqrt{3}+d'\sqrt{6}))=f(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})f(a'+b'\sqrt{2}+c'\sqrt{3}+d'\sqrt{6})$  For some non trivial candidate since the left hand side expands in $16$ terms. What about $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}):\mathbb{Q}$? there are $7$ non trivial candidates and multiplication expands in $64$ terms!   How to do it in a tricker form?  Edit(Let me be more precise in my question):   I know that if $f\in\Gamma(\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q})$, then $f(\sqrt{2})=\pm\sqrt{2}$, $f(\sqrt{3})=\pm\sqrt{3}$ so there are $4$ candidates for $\mathbb{Q}$-Automorphisms. Candidates are given by:  $f_1(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})=a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}$  $f_2(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})=a-b\sqrt{2}+c\sqrt{3}-d\sqrt{6}$  $f_3(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})=a+b\sqrt{2}-c\sqrt{3}-d\sqrt{6}$  $f_4(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})=a-b\sqrt{2}-c\sqrt{3}+d\sqrt{6}$  But I need to check that they are $\mathbb{Q}$-Automorphisms in $\mathbb Q(\sqrt{2},\sqrt{3})$. All candidates $f_i$ fixes $\mathbb{Q}$ and is obvious that all preserves sums, $f_1$ obviously preserves product also since is the identity, but is cumberstone to show that $f_i(xy)=f_i(x)_if(y)$ to conclude $f_i\in\Gamma(\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q})$.
75324
I'm reading a brief introduction on Galois theory, it talks about splitting field. The idea is genius, however I have some doubts on whether in general a splitting field always exists?  By definition, a splitting field of a polynomial $p(X)$ over a field $K$ is a field extension $L$ of $K$ over which $p$ factors into linear factors $p(X) = \prod_{i=1}^{deg(p)} (X-a_i)$ where for each $i$ we have $(X - a_i) \in L[X]$ and such that the roots $a_i$ generate $L$ over $K$.  If $K$ is $\mathbb R$, the result is straightforward due to  as it states that every non-constant single-variable polynomial with complex coefficients has at least one complex root.  But if $K$ is a general field, does such an extension $L$ always exist? How to prove that?
32199
Question is to solve for Galois Group of $x^4-2x^2-2$ over $\mathbb{Q}$.  I know the roots of this polynomial are $\sqrt{1+\sqrt{3}},-\sqrt{1+\sqrt{3}},\sqrt{1-\sqrt{3}},-\sqrt{1-\sqrt{3}}$.  But, $\sqrt{2}i=\sqrt{1+\sqrt{3}}.\sqrt{1-\sqrt{3}}$. So, I concluded that splitting field would be $\mathbb{Q}(\sqrt{1+\sqrt{3}},\sqrt{2}i)$ (for some reason i am not very sure about, i have not written splitting field to be $\mathbb{Q}(\sqrt{1-\sqrt{3}},\sqrt{2}i)$).  So, i have splitting field as $\mathbb{Q}(\sqrt{1+\sqrt{3}},\sqrt{2}i)$.   What i do usually after i know that extension is of the form $F(a,b)$ with $F(a)\cap F(b)=F$, I calculate Galois group of $F(a,b)/F(b)$ and Galois group of $F(a,b)/F(a)$ and write their product as in $G=Gal(F(a,b)/F)$  I do the same here for $Gal(\mathbb{Q}(\sqrt{1+\sqrt{3}},\sqrt{2}i)/\mathbb{Q}(\sqrt{1+\sqrt{3}}))$, this is of order 2, sending $\sqrt{2}i\rightarrow -\sqrt{2}i$ (fixing $\sqrt{1+\sqrt{3}}$) So, I have $Gal(\sqrt{1+\sqrt{3}},\sqrt{2}i)/\mathbb{Q}(\sqrt{1+\sqrt{3}}))\cong \mathbb{Z}_2$.  The problem is with $Gal(\mathbb{Q}(\sqrt{1+\sqrt{3}},\sqrt{2}i)/\mathbb{Q}(\sqrt{2}i))$, I Have to prove that Galois group of $x^4-2x^2-2$ is dihedral group of order 8.I already have an element of order 2, I have to get an element of order 4 from $Gal(\mathbb{Q}(\sqrt{1+\sqrt{3}},\sqrt{2}i)/\mathbb{Q}(\sqrt{2}i))$ which is becoming more cumbersome.  Any help would be appreciated. Thank You.
58050
Let $E=\mathbb{Q}(\sqrt{2},\sqrt{3})$ and $$ L = E \left( \sqrt{ ( \sqrt{2}+2 ) ( \sqrt{3} + 3)} \right) \ . $$ I want to show that $L/\mathbb{Q}$ is a Galois extension with the Quaternion group as its Galois group.   I know $E/\mathbb{Q}$ is Galois and $L/E$ is also Galois, but it is not true in general that if $K_1/K_0$ is Galois and $K_2/K_1$ is Galois then $K_2/K_0$ is Galois (take $K_0 = \mathbb{Q}$, $K_1 = \mathbb{Q}(\sqrt{2})$ and $K_2 = \mathbb{Q}(\sqrt[4]{2})$ as a counter-example).
51163
I want to compute the Galois group of some polynomials, but I want to see some examples first. For example this proposition could be helpful. I don't know how to prove it &lt;.&lt;  Let's consider a polynomial $f(x)= x^4+ax^2+b \in \mathbb{Z} $. Let $ \pm\alpha , \pm\beta$ denote the roots of $f(x)$. Note that $f$ is irreducible iff $\alpha^2, \alpha\pm\beta \notin \mathbb{Q}$. So let's suppose that $f(x)$ is irreducible. Let's denote $G$ the galois group of $f$ (the Galois group of the splitting field $\mathbb{Q}(\alpha,\beta)/\mathbb{Q}$)  $i)$ $G\cong V$ the Klein 4-group iff $\alpha\beta \in \mathbb{Q}$  $ii)$ $G\cong \mathbb{Z}_4$ iff $\mathbb{Q}(\alpha\beta)=\mathbb{Q}(\alpha^2)$  $iii)$ $G\cong D_8$ the dihedral group of order 8, iff $\alpha\beta \notin \mathbb{Q}(\alpha^2)$  I need help with the proofs of this propositions... Well I'm very lost with the computation of Galois group, some care must be exercised, I don't have it. For example I know that Galois group acts transitively, in the sense that $u\in K/F$ a galois extension, the other roots of the minimal polynomial of $m_u(x)\in F[x]$ are precisely the elements $ \sigma(u)$ , where $\sigma \in Gal(K/F)$. And clearly any automorphism, is determined by only the value of a basis. In this case we are working on $\mathbb{Q}(\alpha,\beta)$, so I have to know the values of both. But I have to respect the algebraic relations involving $\alpha , \beta$ I proved that in $i),ii)$ $\mathbb{Q}(\alpha,\beta) = \mathbb{Q}(\alpha) $ So I have to know the values on $\alpha$.   My first question is: If in both cases $i),ii)$, I have an extension of degree 4 (so $|G|=4$) I have to map $\alpha$ to determine the automorphism, but I have also to map $\alpha$ to all of it's four roots. Why the galois group in this case are different?  My second question is about $iii)$ I have no idea How to attack it.
120574
So I have seen the similar question and answers on here for $x^4 +1$, but I am having trouble extending anything there to this polynomial... I understand it is fairly trivial with Galois theory, but my class has just barely covered Field Extensions, so suffice it to say we have no Galois theory to play with.  I managed to prove it for the primes such that $p \equiv 1, 7 \pmod 8$, by noting that $2$ is a square modulo those primes and thus $x^4 - 2x^2 +1 = (x^2 -1 + 2qx)(x^2 - 1 - 2qx)$ for those $\mathbb{Z}_p$... however, trying to get a similar result for $3 \pmod 8$ and $5 \pmod 8$ has been stumping me for a long time, I am having a hard time making $q^2 = -1$ and $q^2 = -2$ give me something factorable...  I guess the worst part of all of this is that I don't think this solution is even particularly enlightening, in terms of abstract algebra. It's really just some number theory trickery. I don't think my course has prepared me theoretically for this problem, does anyone have an elementary approach to it?
26346
Let $K$ be a Galois extension of $F$, and let $a\in K$. Let $n=[K:F]$, $r=[F(a):F]$, and $H=\mathrm{Gal}(K/F(a))$. Let $z_1, z_2,\ldots,z_r$ be left coset representatives of $H$ in $G$. Show that    $$\min(F,a)=\prod_{i=1}^r\left( x - z_i(a)\right).$$   In this product $\min(F,a)$ is of degree $r$. That is true from fundamental theorem.  And if one of $z_i$ is the identity, then $a$ satisfies the polynomial.  My question is:  What is the guarantee that $z_i(a)$ belongs to $F$ for all $i$?  What if I choose a representative which is not the identity of $\mathrm{Gal}(K/F)$?
21628
I've been learning about Galois theory recently on my own, and I've been trying to solve tests from my university. Even though I understand all the theorems, I  seem to be having some trouble with the technical stuff. A specific example would be how to find the Galois group of a given polynomial. I know some tricks, and I manage to solve some of those questions, but some not.  For example, one of the tests asks to find the Galois group of $x^{4}-4x+2$.  I can see that it is irreducible over $\mathbb Q$ (Eisenstein), but I have no clue as to how to find its Galois group over $\mathbb Q$. Can someone tell me how to do this? General techniques concerning this sort of problems are also welcome :).  Thanks!
49801
Can someone please give me a tip how to show that for distinct prime $p,q$ that $\mathbb{Q}(\sqrt{p},\sqrt[3]{q})=\mathbb{Q}(\sqrt{p}\sqrt[3]{q})$ and how to find the minimal polynomial of $\sqrt{p}\sqrt[3]{q}$ over $\mathbb{Q}$ ?
48253
I had an example in the book given as follows:        Find the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$ .       Solution:  $~~~~~$$(\sqrt{2}+\sqrt{3})^2=5+2 \sqrt6$      $(\sqrt{2}+\sqrt{3})^4=49+20 \sqrt6$       Then $(\sqrt{2}+\sqrt{3})^4-10(\sqrt{2}+\sqrt{3})^2+1=0.$       Thus $a=\sqrt{2}+\sqrt{3}$ satisfies $f(x)=x^4-10x^2+1$ over $\mathbb Q$.       Let $p(x)$ be the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$.Then $(\sqrt{2}-\sqrt{3}),(-\sqrt{2}+\sqrt{3}),(-\sqrt{2}-\sqrt{3})$ are also roots of $p(x).$   So degree of $p(x)$ is atleast $4$.      But $f(a)=0 $ and $f(x) \in \mathbb Q[x] \implies p(x)$ divides $f(x)$ .      So $f(x)$ is minimal polynomial of $\sqrt{2}+\sqrt{3}.$   But I can't get the step  how are $(\sqrt{2}-\sqrt{3}),(-\sqrt{2}+\sqrt{3}),(-\sqrt{2}-\sqrt{3})$ also roots of $p(x).$   Kindly help with this.
27428
Let $K$ be a field and $f \in K[X]$ of degree $n$ with Galois group $S_n$. Let $a$ be a root of $f$, $L = K(a)$, and let $E$ be a intermediate field of the extension $K \subset L$. Prove that $E = K$ or $E = L$.   Any suggestions on where to start? Thanks.
66935
Okay, so I mostly worked this out, and I even created lattice diagrams as shown below. But I have a specific question about finding intermediate fields, which I will ask shortly. Let $\alpha = \sqrt[4]{2}$ and $\omega = e^{\frac{\pi}{4}i} = i$.  Then $L = \mathbb{Q}(\alpha, i)$ is the splitting field of $x^4 -2$ over $\mathbb{Q}$. Also, the Galois group $\Gamma_\mathbb{Q}(x^4 - 2) = D_8$ acts on the roots $\alpha, \alpha i, -\alpha,$ and $-\alpha i$, and is generated by rotation $\sigma$ and reflection $\tau$, where $\sigma(i) = i, \sigma(\alpha) = \alpha i$ and $\tau(\alpha) = \alpha, \tau(i) = -i$. To find the intermediate fields between $L$ and $\mathbb{Q}$, find the subgroups of $D_8$ instead with the idea that finding subgroups is easier and better understood than finding intermediate fields. Then from the subgroups, use the Galois correspondence to get all the intermediate fields. There are 10 subgroups of $D_8$ which must correspond to 10 intermediate fields.  Well, I pieced together 8 obvious candidates for intermediate fields, and in the end, I had to look up the other 2 which were $\mathbb{Q}(\alpha(1 + i))$ and $\mathbb{Q}(\alpha(1 - i))$. Those two seemed strange until I realized that $\sqrt{8\alpha^2 i} = \alpha(1 + i)$. Finally, I was able to check fixed fields to verify the exact correspondence, and come up with the diagrams. Question: Is there a systematic approach to finding and connecting up the corresponding intermediate fields once all the subgroups are known? I'm guessing, in general and maybe in this example with $D_8$, there isn't a good, canonical way to anticipate and construct the field extensions?  The structure of groups and subgroups, as stated earlier, is easier and better understood than the structure of field extensions. Maybe this makes sense because the groups are finite and have only one operation, and fields are often infinite and have two operations.  and  UPDATE: In , Richard Borcherds clearly describes how to obtain the two nonobvious intermediate fields from the subgroups. Specifically, add the roots $\alpha$ and $\alpha i$ to fix by reflection one way, and then add the roots $\alpha$ and $-\alpha i$ to fix by reflection the other way.
114611
I need to show that the extension $\mathbb{Q}(\sqrt{2+\sqrt{2}})$ is Galois over $\mathbb{Q}$, and compute its Galois group.   I am learning Galois theory by myself and got stuck in this exercise. I know the fundamental theorem of Galois theory. Any help would be useful  Thanks
15972
I have to find the minimal polynomial of $\sqrt 2 + \sqrt[3] 2$ over $\mathbb Q$. The suggested way of doing it is to prove that $\mathbb Q[\sqrt 2 + \sqrt[3] 2]=\mathbb Q[\sqrt 2,\sqrt[3] 2]$ first.  I can prove that. It's enough to prove that $\sqrt 2,\sqrt[3] 2\in \mathbb Q[\sqrt 2 + \sqrt[3] 2]$ and that $\sqrt 2 + \sqrt[3] 2\in \mathbb Q[\sqrt 2,\sqrt[3] 2]$. The latter is obvious and $\sqrt[3] 2\in \mathbb Q[\sqrt 2 + \sqrt[3] 2]$ will follow from $\sqrt 2\in \mathbb Q[\sqrt 2 + \sqrt[3] 2].$ It remains to prove $\sqrt 2\in \mathbb Q[\sqrt 2 + \sqrt[3] 2].$  For $\alpha=\sqrt 2 + \sqrt[3] 2,$ we have   $$ \begin{eqnarray} &amp;\alpha^0&amp;=1,\\ &amp;\alpha^1&amp;=\sqrt 2 + \sqrt[3] 2,\\ &amp;\alpha^2&amp;=2+\sqrt[3]4+\sqrt 2\sqrt[3]2\\ &amp;\alpha^3&amp;=2+2\sqrt 2+6\sqrt[3]2+3\sqrt 2\sqrt[3]4\\ &amp;\alpha^4&amp;=4+4\sqrt 2+12\sqrt[3]4+8\sqrt 2\sqrt[3]2\\ &amp;\alpha^5&amp;=32+4\sqrt 2+20\sqrt[3]2+4\sqrt 2\sqrt[3]2 \end{eqnarray} $$  If I can express $\sqrt 2$ as a linear combination of $\{\alpha^0,\alpha^1,\cdots,\alpha^5\}$, I'm done. I don't know how to find out whether $$\{1,\sqrt 2,\sqrt[3]2,\sqrt[3]4,\sqrt 2\sqrt[3]2,\sqrt 2\sqrt[3]4\}$$ are linearly independent over $\mathbb Q$ but I think I don't have to care. I can use Gaussian elimination anyway. I write the matrix  $$\left[ \begin{array}{rrrrrr|r} 1 &amp; 0 &amp; 2 &amp; 2 &amp; 1 &amp; 8 &amp; 0\\ 0 &amp; 1 &amp; 0 &amp; 2 &amp; 1 &amp; 1 &amp; 1\\ 0 &amp; 1 &amp; 0 &amp; 6 &amp; 0 &amp; 5 &amp; 0\\ 0 &amp; 0 &amp; 1 &amp; 0 &amp; 3 &amp; 0 &amp; 0\\ 0 &amp; 0 &amp; 1 &amp; 0 &amp; 2 &amp; 1 &amp; 0\\ 0 &amp; 0 &amp; 0 &amp; 3 &amp; 0 &amp; 5 &amp; 0 \end{array}\right]$$  and by performing row operations I obtain a row echelon form, which proves that I'll obtain the reduced row echelon form if I want, which will give me the desired coefficients $a_0,\cdots,a_5$ such that   $$\sqrt 2=\sum_{i=0}^5a_i\alpha^i.$$  I don't have to find the coefficients -- I'm just satisfied with their existence.  If this is correct, I have proven that   $$\mathbb Q[\sqrt 2 + \sqrt[3] 2]=\mathbb Q[\sqrt 2,\sqrt[3] 2].$$  But I don't know how to proceed further. Of course, I can find the minimal polynomials of $\sqrt 2$ and $\sqrt[3] 2$ over $\mathbb Q$ but I don't know how to find the minimal polynomial of $\sqrt 2$ over $\mathbb Q[\sqrt[3]2]$ or the minimal polynomial of $\sqrt [3]2$ over $\mathbb Q[\sqrt 2].$  What I did is this. I wrote the equation  $$x=\sqrt 2+\sqrt [3]2$$  and by squaring and cubing obtained  $$W(x):=x^6-6x^4-4x^3+12x^2-24x-4=0.$$  I belive this is the required polynomial but I would need to prove that it's irreducible over $\mathbb Q.$ Eisenstein's criterion doesn't work, which means I'm lost.  Could you please help me with this? I would like to know three things.  1) Is my proof of $\mathbb Q[\sqrt 2 + \sqrt[3] 2]=\mathbb Q[\sqrt 2,\sqrt[3] 2]$ correct?  2) How can I use this fact? I believe it's supposed to make the solution easier to find.  3) How can I tell whether $W(x)$ is reducible over $\mathbb Q$ or not?  EDIT: 1) has been answered by André Nicolas in a comment. All of this answers 3) but I would like to ask whether we could know that the polynomial is irreducible without knowing that $\sqrt 2+\sqrt[3]2$ is its zero. Also, it would be to know whether  4) there is a theorem that would give me $\mathbb Q[\sqrt 2 + \sqrt[3] 2]=\mathbb Q[\sqrt 2,\sqrt[3] 2]$ without the need of performing the Gaussian elimination?  For these numbers it was possible to do, but a bit higher dimension and it would turn out impossible without a computer.
13252
Let's start with a definition:     Definition: Let $F$ be a field, $I$ a nonzero ideal in $F[x]$, and $g(x)$ an element of $F[x]$. Then, $I= \langle g(x)\rangle$ if and only if $g(x)$ is a nonzero polynomial of minimum degree in $I$.   Now it gets completely foggy what is meant with "minimum degree".I constantly come across the "minimal polynomial". This seems to be linked, but it seems strange that there is a division here.   Could someone give a pointer to what the second line actually means. Maybe some example or diagram (if possible) would help a lot here.