# minimal polynomial for $\sqrt{-3}+\sqrt{2}$ over $\mathbb{Q}$

Almost have the answer. Let $$a = \sqrt{-3}+\sqrt{2} \implies (\sqrt{-3}+\sqrt{2})(\sqrt{2}-\sqrt{-3}) = 5 \therefore$$ a is a root of $$x^4+2x^2+25$$.

$$\sqrt{2} = \frac{1}{2}(a+5/a) \in \mathbb{Q}(a), \sqrt{-3} = a-\sqrt{2}\in \mathbb{Q}(a)\implies \mathbb{Q}(\sqrt{2}),\mathbb{Q}(\sqrt{-3})\subset \mathbb{Q}(a)$$ and thus $$\mathbb{Q}(\sqrt{2},\sqrt{-3})\subset \mathbb{Q}(a)$$

$$[\mathbb{Q}(a):\mathbb{Q}] \geq [\mathbb{Q}(\sqrt{2},\sqrt{-3}):\mathbb{Q}] = [\mathbb{Q}(\sqrt{2},\sqrt{-3}):\mathbb{Q}(\sqrt{2})][\mathbb{Q}(\sqrt{2}):\mathbb{Q}] \geq 4$$.

Thus the degree of the minimal polynomial for $$a$$ over $$\mathbb{Q} \geq 4$$. So I'm not sure how to go from there to showing degree of the minimal polynomial for $$a$$ over $$\mathbb{Q} = 4$$? Because I think $$x^4+2x^2+25$$ should be the minimal polynomial. Thanks.

• If $a$ is a root of a 4th degree polynomial, then $[\mathbb Q[a]:\mathbb Q] \le 4$ – J. W. Tanner Apr 17 at 0:00
• Because $\mathbb{Q}(a):\mathbb{Q}$ is the degree of the minimal polynomial? Thanks. – manifolded Apr 17 at 0:02
• Even if it's not, if $a^4=c_3a^3+c_2a^2+c_1a+c_0$ then all elements of $\mathbb Q[a]$ can be expressed in the form $q_0+q_1a+q_2a^2+q_3a^3$ with $q_0,q_1,q_2,q_3 \in \mathbb Q,$ so $[\mathbb Q[a]:\mathbb Q]\le 4$ – J. W. Tanner Apr 17 at 0:09

Note that $$\sqrt{-3}\notin\mathbb{Q}(\sqrt{2})$$, so $$[\mathbb{Q}(\sqrt{2},\sqrt{-3}):\mathbb{Q}(\sqrt{2})]=2$$. Hence $$[\mathbb{Q}(\sqrt{2},\sqrt{-3}):\mathbb{Q}]= [\mathbb{Q}(\sqrt{2},\sqrt{-3}):\mathbb{Q}(\sqrt{2})] [\mathbb{Q}(\sqrt{2}):\mathbb{Q}]=4$$ Now note that $$\mathbb{Q}(a)\subseteq\mathbb{Q}(\sqrt{2},\sqrt{-3})$$, which together with what you proved yields equality.

Anyway, when you have proved that the degree is at least $$4$$ and $$a$$ is the root of a degree $$4$$ polynomial, you are finished.

If $$a$$ is a root of a 4th degree polynomial, then the minimal polynomial for $$a$$ divides a 4th degree

polynomial, so the degree of the minimal polynomial (or the extension field) is at most $$4$$.

On the other hand, you have shown that the degree of the extension field

(or the minimal polynomial) for $$a=\sqrt{-3}+\sqrt2$$ over $$\mathbb Q$$ is at least $$4$$.

Therefore, the degree of the minimal polynomial for $$a$$ is exactly $$4,$$

and $$x^4+2x^2+25$$ is itself indeed the minimal polynomial for $$a$$.