Proving that $p(x)$ is the minimum polynomial

Question

I want to show that the minimum polynomial of $\sqrt{2} + i$ over $\mathbb{R}$ is $p(x) = x^2-2\sqrt{2}x + 3$. I've shown that $\sqrt{2} + i$ is a root of $p(x)$. I'm done if I show that $p(x)$ is irreducible, but am having trouble proving this—any strategies here?

I was also thinking of getting around this by saying that, reducing over $\mathbb{C}$, we have $p(x) = (x - (\sqrt{2} + i))(x - (\sqrt{2} - i))$. Because $\mathbb{C}$ contains $\mathbb{R}$, and because polynomials are uniquely factorized, $p(x)$ could not possibly be reduced $\mathbb{R}$. Is this reasoning ok?

Answer 1

Yes, your reasoning is okay. However, you should be aware that you are making an implicit degree argument of the sort that Dustan explicitly calls out in the comments. The reason $p(x)$ could not possibly be reduced in $\mathbb R$ is that no proper non-empty subset of the complex-valued factors can be combined to form a polynomial with real coefficients, and that's happening because the factors have degree $1$ and the original polynomial has degree $2$.

In the degree $3$ case, the analogous claim also holds because if some two factors multiply to a real polynomial, then the remaining factor (being the quotient) must also be real. However, for a degree $4$ polynomial you would want to check that no pair of linear factors combine to form a degree $2$ polynomial with real coefficients.

Now you may have already had this in mind when you wrote your argument. But identifying this sort of unspoken assumption is a really vital skill in writing mathematics, and probably the fact that you didn't identify it explicitly contributed to your uncertainty about whether your argument is actually complete.

Answer 2

A general result about polynomials with coefficients in $\mathbb{R}$ is the following:

A polynomial in $\mathbb{R}[x]$ is irreducible if and only if it has degree $1$ or it has degree $2$ and negative discriminant.

This is a consequence of the so-called “fundamental theorem of algebra”: every polynomial of positive degree in $\mathbb{C}[x]$ has a root.

So, suppose $p(x)\in\mathbb{R}[x]$ has degree $\ge3$. If it has a real root, then it is reducible. Suppose $\alpha$ is a non real root (which exists because of the FTA); then also $\bar{\alpha}$ (the conjugate of $\alpha$) is a root of $p(x)$, since $p$ has real coefficients. By hypothesis, $\alpha\ne\bar{\alpha}$, so $p(x)$ is divisible by $$ (x-\alpha)(x-\bar{\alpha})=x^2-(\alpha+\bar{\alpha})x+\alpha\bar{\alpha} $$ which has real coefficients. Thus $p(x)$ is not reducible. So, if $p(x)$ is irreducible in $\mathbb{R}[x]$, then it must have degree $1$ or $2$. If it has degree $2$, then its discriminant must be negative, or $p(x)$ would have a real root and so it would be reducible.

The converse is obvious.

However, in your case you need not appeal to the FTA. You found a polynomial of degree $2$ which has $\sqrt{2}+i$ as a root. Since this root is non real, the polynomial is irreducible, because otherwise it would be a product of degree $1$ polynomials in $\mathbb{R}[x]$ that so would have real roots: contradiction.