# Lüroth’s theorem (a “constructive” proof)

Lüroth’s theorem (Lüroth 1876 for $$k = \mathbb{C}$$, Steinitz 1910 in general). If $$k \subseteq K$$ are fields such that $$k \subseteq K \subseteq k(x)$$, where $$x$$ is an indeterminate over $$k$$, then $$K = k(g)$$ for some rational function $$g$$ of $$x$$ over $$k$$.

I am going to present a “constructive” proof of Lüroth’s theorem due to Netto (1895) that I learned from Schinzel’s Selected Topics on Polynomials (and give some applications to criteria for proper polynomial parametrizations). The proof uses the following result which I am not going to prove here:

Proposition (with the set up of Lüroth’s theorem). $$K$$ is finitely generated over $$k$$, i.e. there are finitely many rational functions $$g_1, \ldots, g_s \in k(x)$$ such that $$K = k(g_1, \ldots, g_s)$$.

The proof is constructive in the following sense: given $$g_1, \ldots, g_s$$ as in the proposition, it gives an algorithm to determine $$g$$ such that $$K = k(g)$$. We use the following notation in the proof: given a rational function $$h \in k(x)$$, if $$h = h_1/h_2$$ with polynomials $$h_1, h_2 \in k[x]$$ with $$\gcd(h_1, h_2) = 1$$, then we define $$\deg_\max(h) := \max\{\deg(h_1), \deg(h_2)\}$$.

## Proof of Lüroth’s theorem

It suffices to consider the case that $$K \neq k$$. Pick $$g_1, \ldots, g_s$$ as in the proposition. Write $$g_i = F_i/G_i$$, where

• $$\gcd(F_i, G_i) = 1$$ (Property 1).

Without loss of generality (i.e. discarding $$g_i \in k$$ or replacing $$g_i$$ by $$1/(g_i + a_i)$$ for appropriate $$a_i \in k$$ if necessary) we can also ensure that

• $$\deg(F_i) > 0$$ and $$\deg(F_i) > \deg(G_i)$$ (Property 2).

Consider the polynomials $H_i := F_i(t) – g_iG_i(t) \in K[t] \subset k(x)[t], i = 1, \ldots, s,$ where $$t$$ is a new indeterminate. Let $$H$$ be the greatest common divisor of $$H_1, \ldots, H_s$$ in $$k(x)[t]$$ which is also monic in $$t$$. Since the Euclidean algorithm for computing $$\gcd$$ respects the field of definition, it follows that:

• $$H$$ is also the greatest common divisor of $$H_1, \ldots, H_s$$ in $$K[t]$$, which means, if $$H = \sum_j h_j t^j$$, then each $$h_j \in K$$ (Property 3).

Let $$H^* \in k[x,t]$$ be the polynomial obtained by “clearing the denominator” of $$H$$; in other words, $$H = H^*/h(x)$$ for some polynomial $$h \in k[x]$$ and $$H^*$$ is primitive as a polynomial in $$t$$ (i.e. the greatest common divisor in $$k[x]$$ of the coefficients in $$H^*$$ of powers of $$t$$ is 1). By Gauss’s lemma, $$H^*$$ divides $$H^*_i := F_i(t)G_i(x) – F_i(x)G_i(t)$$ in $$k[x,t]$$, i.e. there is $$Q_i \in k[x,t]$$ such that $$H^*_i = H^* Q_i$$.

Claim 1. If $$\deg_t(H^*) < \deg_t(H^*_i)$$, then $$\deg_x(Q_i) > 0$$.

Proof of Claim 1. Assume $$\deg_t(H^*) < \deg_t(H^*_i)$$. Then $$\deg_t(Q_i) > 1$$. If in addition $$\deg_x(Q_i) = 0$$, then we can write $$Q_i(t)$$ for $$Q_i$$. Let $$F_i(t) \equiv \tilde F_i(t) \mod Q_i(t)$$ and $$G_i(t) \equiv \tilde G_i(t) \mod Q_i(t)$$ with $$\deg(\tilde F_i) < \deg(Q_i)$$ and $$\deg(\tilde G_i) < \deg(Q_i)$$. Then $$\tilde F_i(t)G_i(x) – F_i(x) \tilde G_i(t) \equiv 0 \mod Q_i(t)$$. Comparing degrees in $$t$$, we have $$\tilde F_i(t)G_i(x) = F_i(x) \tilde G_i(t)$$. It is straightforward to check that this contradicts Propeties1 and 2 above, and completes the proof of Claim 1.

Let $$m := \min\{\deg_\max(g_i): i = 1, \ldots, s\}$$, and pick $$i$$ such that $$\deg_\max(g_i) = m$$. Property 2 above implies that $$\deg_t(H^*_i) = \deg_x(H^*_i) = m$$. If $$\deg_t(H^*) < m$$, then Claim 1 implies that $$\deg_x(H^*) < \deg_x(H^*_i) = m$$. If the $$h_j$$ are as in Property 3 above, it follows that $$\deg_\max(h_j) < m$$ for each $$j$$. Since $$H^* \not\in k[t]$$ (e.g. since $$t-x$$ divides each $$H_i$$), there must be at least one $$h_j \not \in k$$. Since adding that $$h_j$$ to the list of the $$g_i$$ decreases the value of $$m$$, it follows that the following algorithm must stop:

### Algorithm

• Step 1: Pick $$g_i := F_i/G_i$$, $$i = 1, \ldots, s$$, satisfying properties 1 and 2 above.
• Step 2: Compute the monic (with respect to $$t$$) $$\gcd$$ of $$F_i(t) – g_i G_i(t)$$, $$i = 1, \ldots, s$$, in $$k(x)[t]$$; call it $$H$$.
• Step 3: Write $$H = \sum_j h_j(x) t^j$$. Then each $$h_j \in k(g_1, \ldots, g_s)$$. If $$\deg_t(H) < \min\{\deg_\max(g_i): i = 1, \ldots, s\}$$, then adjoin all (or, at least one) of the $$h_j$$ such that $$h_j \not\in k$$ to the list of the $$g_i$$ (possibly after an appropriate transformation to ensure Property 2), and repeat.

After the last step of the algorithm, $$H$$ must be one of the $$H_i$$, in other words, there is $$\nu$$ such that $\gcd(F_i(t) – g_i G_i(t): i = 1, \ldots, s) = F_{\nu}(t) – g_{\nu}G_{\nu}(t).$

Claim 2. $$K = k(g_{\nu})$$.

Proof of Claim 2 (and last step of the proof of Lüroth’s theorem). For a given $$i$$, polynomial division in $$k(g_\nu)[t]$$ gives $$P, Q \in k(g_\nu)[t]$$ such that $F_i(t) = (F_{\nu}(t) – g_{\nu}G_{\nu}(t))P + Q,$ where $$\deg_t(Q) < \deg_t(F_{\nu}(t) – g_{\nu}G_{\nu}(t))$$. If $$Q = 0$$, then $$F_i(t) = (F_{\nu}(t) – g_{\nu}G_{\nu}(t))P$$, and clearing out the denominator (with respect to $$k[g_\nu]$$) of $$P$$ gives an identity of the form $$F_i(t)p(g_\nu) = (F_{\nu}(t) – g_{\nu}G_{\nu}(t))P^* \in k[g_\nu, t]$$ which is impossible, since $$F_{\nu}(t) – g_{\nu}G_{\nu}(t)$$ does not factor in $$k[g_\nu, t]$$. Therefore $$Q \neq 0$$. Similarly, $G_i(t) = (F_{\nu}(t) – g_{\nu}G_{\nu}(t))R + S,$ where $$R, S \in k(g_\nu)[t]$$, $$S \neq 0$$, and $$\deg_t(S) < \deg_t(F_{\nu}(t) – g_{\nu}G_{\nu}(t))$$. It follows that $F_i(t) – g_iG_i(t) = (F_{\nu}(t) – g_{\nu}G_{\nu}(t))(P – g_iR) + Q – g_iS.$ Since $$F_{\nu}(t) – g_{\nu}G_{\nu}(t)$$ divides $$F_{i}(t) – g_{i}G_{i}(t)$$ in $$k(x)[t]$$ and since $$\deg_t(Q – g_iS) < \deg_t(F_{\nu}(t) – g_{\nu}G_{\nu}(t))$$, it follows that $$Q = g_iS$$. Taking the leading coefficients (with respect to $$t$$) $$q_0, s_0 \in k(g_\nu)$$ of $$Q$$ and $$S$$ gives that $$g_i = q_0/s_0 \in k(g_\nu)$$, as required to complete the proof.

## Applications

The following question seems to be interesting (geometrically, it asks when a given polynomial parametrization of a rational affine plane curve is proper).

Question 1. Let $$k$$ be a field and $$x$$ be an indeterminate over $$k$$ and $$g_1, g_2 \in k[x]$$. When is $$k(g_1, g_2) = k(x)$$?

We now give a sufficient condition for the equality in Question 1. Note that the proof is elementary: it does not use Lüroth’s theorem, only follows the steps of the above proof in a special case.

Corollary 1. In the set up of Question 1, let $$d_i := \deg(g_i)$$, $$i = 1, 2$$. If the $$\gcd$$ of $$x^{d_1} – 1, x^{d_2} – 1$$ in $$k[x]$$ is $$x – 1$$, then $k(g_1, g_2) = k(t)$. In particular, if $$d_1, d_2$$ are relatively prime and the characteristic of $$k$$ is either zero or greater than both $$d_1, d_2$$, then $k(g_1, g_2) = k(x)$.

Remark. Corollary 1 is true without the restriction on characteristics, i.e. the following holds: “if $$d_1, d_2$$ are relatively prime, then $k(g_1, g_2) = k(x)$.” François Brunault (in a comment to one of my questions on MathOverflow) provided the following simple one line proof: $$[k(x): k(g_1, g_2)]$$ divides both $$[k(x): k(g_i)] = d_i$$, and therefore must be $$1$$.

My original proof of Corollary 1. Following the algorithm from the above proof of Lüroth’s theorem, let $$H_i := g_i(t) – g_i(x)$$, $$i = 1, 2$$, and $$H \in k(x)[t]$$ be the monic (with respect to $$t$$) greatest common divisor of $$H_1, H_2$$.

Claim 1.1. $$H = t – x$$.

Proof. It is clear that $$t-x$$ divides $$H$$ in $$k(x)[t]$$, so that $$H(x,t) = (t-x)h_1(x,t)/h_2(x)$$ for some $$h_1(x,t) \in k[x,t]$$ and $$h_2(x) \in k[x]$$. It follows that there is $$Q_i(x,t) \in k[x,t]$$ and $$P_i(x) \in k[x]$$ such that $$H_i(x,t)P_i(x)h_2(x) = (t-x)h_1(x,t)Q_i(x,t)$$. Since $$h_2(x)$$ and $$(t-x)h_1(x,t)$$ have no common factor, it follows that $$h_2(x)$$ divides $$Q_i(x,t)$$, and after cancelling $$h_2(x)$$ from both sides, one can write $H_i(x,t)P_i(x) = (t-x)h_1(x,t)Q’_i(x,t),\ i = 1, 2.$ Taking the leading form of both sides with respect to the usual degree on $$k[x,t]$$, we have that $(t^{d_i} – x^{d_i})x^{p_i} = a_i(t-x)\mathrm{ld}(h_1)\mathrm{ld}(Q’_i)$ where $$a_i \in k \setminus \{0\}$$ and $$\mathrm{ld}(\cdot)$$ is the leading form with respect to the usual degree on $$k[x,t]$$. Since $$\gcd(x^{d_1} – 1, x^{d_2} – 1) = x – 1$$, it follows that $$\mathrm{ld}(h_1)$$ does not have any factor common with $$t^{d_i} – x^{d_i}$$, and consequently, $$t^{d_i} – x^{d_i}$$ divides $$(t-x)\mathrm{ld}(Q’_i)$$. In particular, $$\deg_t(Q’_i) = d_i – 1$$. But then $$\deg_t(h_1) = 0$$. Since $$H = (t-x)h_1(x)/h_2(x)$$ is monic in $$t$$, it follows that $$H = t – x$$, which proves Claim 1.1.

Since both $$H_i$$ are elements of $$k(g_1, g_2)[t]$$, and since the Euclidean algorithm to compute $$\gcd$$ of polynomials (in a single variable over a field) preserves the field of definition, it follows that $$H \in k(g_1, g_2)[t]$$ as well (this is precisely the observation of Property 3 from the above proof of Lüroth’s theorem). Consequently $$x \in k(g_1, g_2)$$, as required to prove Corollary 1.

## References

• Andrzej Schinzel, Selected Topics on Polynomials, The University of Michigan Press, 1982.